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opencompass/configs/datasets/MMLUArabic/README.md 0 → 100644
View file @ be3dfa50
# MMLUArabic
## Dataset Description
MMLUArabic is a benchmark for the assessment of knowledge in Arabic and covers a wide range of topics and aspects, consisting of multiple-choice questions in various branches of knowledge.
## How to Use
Download file from [link](https://github.com/FreedomIntelligence/AceGPT/tree/main/eval/benchmark_eval/benchmarks/MMLUArabic)
```python
val_ds = load_dataset("MMLUArabic", header=None)['validation']
test_ds = load_dataset("MMLUArabic", header=None)['test']
# input, option_a, option_b, option_c, option_d, target
print(next(iter(val_ds)))
```
## Citation
```
@misc{huang2023acegpt,
title={AceGPT, Localizing Large Language Models in Arabic},
author={Huang Huang and Fei Yu and Jianqing Zhu and Xuening Sun and Hao Cheng and Dingjie Song and Zhihong Chen and Abdulmohsen Alharthi and Bang An and Ziche Liu and Zhiyi Zhang and Junying Chen and Jianquan Li and Benyou Wang and Lian Zhang and Ruoyu Sun and Xiang Wan and Haizhou Li and Jinchao Xu},
year={2023},
eprint={2309.12053},
archivePrefix={arXiv},
primaryClass={cs.CL}
}
```
opencompass/configs/datasets/MathBench/deprecated_mathbench_2024_gen_de9ff9.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
single_choice_prompts = {
'single_choice_cn_with_reasoning': '以下是一道关于数学的单项选择题，请你一步一步推理，并在最后用“所以答案为选项X”给出答案，其中“X”为选项A，B，C，D中你认为正确的选项。下面是你要回答的问题\n{question}\n让我们一步一步思考：\n',
'single_choice_cn': '以下是一道关于数学的单项选择题，请你直接回答正确答案的选项序号。\n下面是你要回答的题目：\n{question}\n答案选项：',
'single_choice_en_with_reasoning': "Here is a multiple-choice question about mathematics. Please reason through it step by step, and at the end, provide your answer option with 'Therefore, the correct answer is option X', Where 'X' is the correct option you think from A，B，C，D. Here is the question you need to answer:\n{question}\nLet's think step by step:",
'single_choice_en': 'Here is a multiple-choice question about mathematics. Please provide the correct answer option directly.\nHere is the question you need to answer:\n{question}\nAnswer option:',
}
cloze_prompts = {
'cloze_cn': [
dict(role='HUMAN', prompt='Q: 林中有15棵树。林务工人员今天将在林中种植树木。完成后，将有21棵树。林务工人员今天种植了多少棵树？'),
dict(role='BOT', prompt='A: 我们从15棵树开始。后来有21棵树。差值必定是他们种植的树木数量。所以，他们必须种植了21 - 15 = 6棵树。答案是 6\n'),
dict(role='HUMAN', prompt='Q: 如果停车场有3辆车，又有2辆车进来，停车场里有多少辆车？'),
dict(role='BOT', prompt='A: 停车场已经有3辆车。又进来了2辆车。现在有3 + 2 = 5辆车。答案是 5\n'),
dict(role='HUMAN', prompt='Q: 黎恩有32块巧克力，她的妹妹有42块。如果他们吃了35块，他们总共剩下多少块？'),
dict(role='BOT', prompt='A: 黎恩有32块巧克力，Leah的妹妹有42块。这意味着原本有32 + 42 = 74块巧克力。被吃掉了35块。所以他们总共还剩下74 - 35 = 39块巧克力。答案是 39\n'),
dict(role='HUMAN', prompt='Q: 杰森有20个棒棒糖。他给丹妮一些棒棒糖。现在Jason只剩下12个棒棒糖。杰森给丹妮多少个棒棒糖？'),
dict(role='BOT', prompt='A: 杰森有20个棒棒糖。因为他现在只剩下12个，所以他必须把剩下的都给了丹妮。他给丹妮的棒棒糖数量必定是20 - 12 = 8个。答案是 8\n'),
dict(role='HUMAN', prompt='Q: 莎莎有五个玩具。在圣诞节，他从他的爸爸和妈妈那里各得到了两个玩具。现在他有多少个玩具？'),
dict(role='BOT', prompt='A: 她有5个玩具。他从妈妈那里得到了2个，所以之后他有5 + 2 = 7个玩具。然后他从爸爸那里得到了2个，所以总共他有7 + 2 = 9个玩具。答案是 9\n'),
dict(role='HUMAN', prompt='Q: 服务器房间里有九台电脑。从周一到周四每天增加五台电脑。现在服务器房里有多少台电脑？'),
dict(role='BOT', prompt='A: 从周一到周四有4天。每天增加5台电脑。这意味着总共增加了4 * 5 = 20台电脑。一开始有9台电脑，所以现在有9 + 20 = 29台电脑。答案是 29\n'),
dict(role='HUMAN', prompt='Q: 迈克尔有58个高尔夫球。星期二，他丢失了23个高尔夫球。星期三，他又丢失了2个。星期三结束时他还剩下多少个高尔夫球？'),
dict(role='BOT', prompt='A: 迈克尔一开始有58个球。星期二他丢失了23个，所以之后他还剩下58 - 23 = 35个球。星期三他又丢失了2个，所以现在他还剩下35 - 2 = 33个球。答案是 33\n'),
dict(role='HUMAN', prompt='Q: 奥利弗有23美元。她用每个3美元的价格买了五个百吉饼。她还剩下多少钱？'),
dict(role='BOT', prompt='A: 她以每个3美元的价格买了5个百吉饼。这意味着她在百吉饼上花费了5 * 3 = 15美元。她一开始有23美元，所以现在她还剩下23 - 15 = 8美元。答案是 8\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}'),
],
'cloze_en': [
dict(role='HUMAN', prompt='Q: There are 15 trees in the grove. Grove workers will plant trees in the grove today. After they are done, there will be 21 trees. How many trees did the grove workers plant today?'),
dict(role='BOT', prompt='A: We start with 15 trees. Later we have 21 trees. The difference must be the number of trees they planted. So, they must have planted 21 - 15 = 6 trees. The answer is 6.\n'),
dict(role='HUMAN', prompt='Q: If there are 3 cars in the parking lot and 2 more cars arrive, how many cars are in the parking lot?'),
dict(role='BOT', prompt='A: There are 3 cars in the parking lot already. 2 more arrive. Now there are 3 + 2 = 5 cars. The answer is 5.\n'),
dict(role='HUMAN', prompt='Q: Leah had 32 chocolates and her sister had 42. If they ate 35, how many pieces do they have left in total?'),
dict(role='BOT', prompt="A: Leah had 32 chocolates and Leah's sister had 42. That means there were originally 32 + 42 = 74 chocolates. 35 have been eaten. So in total they still have 74 - 35 = 39 chocolates. The answer is 39.\n"),
dict(role='HUMAN', prompt='Q: Jason had 20 lollipops. He gave Denny some lollipops. Now Jason has 12 lollipops. How many lollipops did Jason give to Denny?'),
dict(role='BOT', prompt='A: Jason had 20 lollipops. Since he only has 12 now, he must have given the rest to Denny. The number of lollipops he has given to Denny must have been 20 - 12 = 8 lollipops. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: Shawn has five toys. For Christmas, he got two toys each from his mom and dad. How many toys does he have now?'),
dict(role='BOT', prompt='A: He has 5 toys. He got 2 from mom, so after that he has 5 + 2 = 7 toys. Then he got 2 more from dad, so in total he has 7 + 2 = 9 toys. The answer is 9.\n'),
dict(role='HUMAN', prompt='Q: There were nine computers in the server room. Five more computers were installed each day, from monday to thursday. How many computers are now in the server room?'),
dict(role='BOT', prompt='A: There are 4 days from monday to thursday. 5 computers were added each day. That means in total 4 * 5 = 20 computers were added. There were 9 computers in the beginning, so now there are 9 + 20 = 29 computers. The answer is 29.\n'),
dict(role='HUMAN', prompt='Q: Michael had 58 golf balls. On tuesday, he lost 23 golf balls. On wednesday, he lost 2 more. How many golf balls did he have at the end of wednesday?'),
dict(role='BOT', prompt='A: Michael initially had 58 balls. He lost 23 on Tuesday, so after that he has 58 - 23 = 35 balls. On Wednesday he lost 2 more so now he has 35 - 2 = 33 balls. The answer is 33.\n'),
dict(role='HUMAN', prompt='Q: Olivia has $23. She bought five bagels for $3 each. How much money does she have left?'),
dict(role='BOT', prompt='A: She bought 5 bagels for $3 each. This means she spent 5 * $3 = $15 on the bagels. She had $23 in beginning, so now she has $23 - $15 = $8. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}\n'),
]}
mathbench_sets = {
'college': ['single_choice_cn', 'single_choice_en'],
'high': ['single_choice_cn', 'single_choice_en'],
'middle': ['single_choice_cn', 'single_choice_en'],
'primary': ['cloze_cn', 'cloze_en'],
'calculate': ['cloze_en'],
}
# Generate reasoning path or not, only for single choice
with_reasoning = True
# Use circular evaluation or not
with_circular_eval = True
mathbench_datasets = []
for _split in list(mathbench_sets.keys()):
for _name in mathbench_sets[_split]:
mathbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[
dict(
role='HUMAN',
prompt=single_choice_prompts[_name + '_with_reasoning'] if with_reasoning else single_choice_prompts[_name],
),
dict(role='BOT', prompt='{answer}')] if 'choice' in _name else cloze_prompts[_name],
),
),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=512),
)
mathbench_eval_cfg = dict(
evaluator=dict(type=CircularEvaluator if 'choice' in _name and with_circular_eval else AccEvaluator),
pred_postprocessor=dict(type=first_option_postprocess, options='ABCD') if 'single_choice' in _name else dict(type=mathbench_postprocess, name=_name))
mathbench_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'./data/mathbench/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=dict(
input_columns=['question'],
output_column='answer'
),
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
))
opencompass/configs/datasets/MathBench/deprecated_mathbench_agent_gen_48ec47.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import AgentInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
PROMPT_EN = {
'FEWSHOT_INSTRUCTION_CLOZE' : [
dict(role='HUMAN', prompt='Mark\'s basketball team scores 25 2 pointers, 8 3 pointers and 10 free throws. Their opponents score double the 2 pointers but half the 3 pointers and free throws. What\'s the total number of points scored by both teams added together?'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\ndef solution():\n mark_pointers_2 = 25 * 2\n mark_pointers_3 = 8 * 3\n mark_free_throws = 10 * 1\n mark_points_scored = mark_pointers_2 + mark_pointers_3 + mark_free_throws\n opponents_pointers_2 = mark_pointers_2 * 2\n opponents_pointers_3 = mark_pointers_3 / 2\n opponents_free_throws = mark_free_throws / 2\n opponents_points_scored = opponents_pointers_2 + opponents_pointers_3 + opponents_free_throws\n total_points_scored = mark_points_scored + opponents_points_scored\n result = total_points_scored\n return result\n```'),
dict(role='SYSTEM', prompt='Response:210'),
dict(role='BOT', prompt='Thought: According to the response, I got the answer\nFinalAnswer: 210'),
dict(role='HUMAN', prompt='Bella has two times as many marbles as frisbees. She also has 20 more frisbees than deck cards. If she buys 2/5 times more of each item, what would be the total number of the items she will have if she currently has 60 marbles?'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\ndef solution():\n marbles = 60\n num_increased_marbles = marbles * 2 / 5\n num_total_marbles = marbles + num_increased_marbles\n frisbees = marbles / 2\n num_increased_frisbees = frisbees * 2 / 5\n num_total_frisbees = frisbees + num_increased_frisbees\n deck_cards = frisbees - 20\n num_increased_deck_cards = deck_cards * 2 / 5\n num_total_deck_cards = deck_cards + num_increased_deck_cards\n num_total = num_total_marbles + num_total_frisbees + num_total_deck_cards\n result = num_total\n return result\n```'),
dict(role='SYSTEM', prompt='Response:140'),
dict(role='BOT', prompt='Thought: According to the response, I got the answer\nFinalAnswer: 140'),
dict(role='HUMAN', prompt='A group of 4 fruit baskets contains 9 apples, 15 oranges, and 14 bananas in the first three baskets and 2 less of each fruit in the fourth basket. How many fruits are there?'),
dict(role='BOT', prompt="""Tool:PythonInterpreter\nTool Input:```python\ndef solution():\n num_fruits_per_first_three_basket = 9 + 15 + 14\n num_fruits_first_three_basket = num_fruits_per_first_three_basket * 3\n num_apple_fourth_basket = 9 - 2\n num_orange_fourth_basket = 15 - 2\n num_banana_fourth_basket = 14 - 2\n num_fruits_fourth_basket = num_apple_fourth_basket + num_orange_fourth_basket + num_banana_fourth_basket\n num_fruits_total = num_fruits_first_three_basket + num_fruits_fourth_basket\n result = num_fruits_total\n return result\n```"""),
dict(role='SYSTEM', prompt='Response:146'),
dict(role='BOT', prompt='Thought: According to the response, I got the answer\nFinalAnswer: 146'),
dict(role='HUMAN', prompt='{question}'),
],
'FEWSHOT_INSTRUCTION_CHOICE' : [
dict(role='HUMAN', prompt='Given point P(-1,4) lies on the graph of the inverse proportionality function $y=\\frac{{k}}{{x}}$ (k≠0), what is the value of k? A. $-\\frac{{1}}{{4}}$ B. $\\frac{{1}}{{4}}$ C. $4$ D. $-4$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\nfrom sympy import solve, symbols, Ne, Eq\ndef solution():\n k = symbols(\'k\')\n result = solve([Eq(4, k / (-1)), Ne(k, 0)], k, dict=True)\n return result\n```'),
dict(role='SYSTEM', prompt='Response:{{Eq(k, -4)}}'),
dict(role='BOT', prompt='Thought: Substituting point P into the function yields the value of k as $-4$\nFinalAnswer: D'),
dict(role='HUMAN', prompt='The graph of the power function $y=(x)$ passes through the point$ (2, \\dfrac {{1}}{{4}}) $, what is the value of $f(-3)$? A. $\\frac{{1}}{{9}}$ B. $\\frac{{1}}{{8}})=196-x$ C. $\\frac{{2}}{{9}}$ D. $\\frac{{1}}{{4}}$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\nfrom sympy import *\ndef solution():\n x, y, k = symbols(\'x y k\')\n eq1 = Eq(2**k, Rational(1, 4))\n k_value = solve(eq1, k)[0]\n y = x**k_value\n result = y.subs(x, -3)\n return result\n```'),
dict(role='SYSTEM', prompt='Response:1/9'),
dict(role='BOT', prompt='Thought: The functional expression of the power function is $y=x^{{-2}}$. Substituting $x=-3$ yields $y=$\\frac{{1}}{{9}}$\nFinalAnswer: A'),
dict(role='HUMAN', prompt='If $3 x-y=12$, what is the value of $\\frac{8^{x}}{2^{y}} ?$\nA. The value cannot be determined from the information given.\nB. $2^{12}$\nC. 4\nD. $8^{2}$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\nfrom sympy import symbols, Eq, solve\n\ndef sloution():\n x, y = symbols(\'x y\')\n equation = Eq(3*x - y, 12)\n\n y_in_terms_of_x = solve(equation, y)[0]\n expression = 8**x / 2**y_in_terms_of_x\n result = expression.simplify()\n return result\n```'),
dict(role='SYSTEM', prompt='Response:2**12'),
dict(role='BOT', prompt='Thought: The value of $\\frac{8^{x}}{2^{y}}$ is $2^{12}$\nFinalAnswer: B'),
dict(role='HUMAN', prompt='{question}'),
]
}
PROMPT_CN = {
'FEWSHOT_INSTRUCTION_CLOZE' : [
dict(role='HUMAN', prompt='Mark的篮球队得到25个2分球，8个3分球和10个罚球。他们的对手得到2分球的两倍，但3分球和罚球的一半。两队得分的总和是多少？'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\ndef solution():\n mark_pointers_2 = 25 * 2\n mark_pointers_3 = 8 * 3\n mark_free_throws = 10 * 1\n mark_points_scored = mark_pointers_2 + mark_pointers_3 + mark_free_throws\n opponents_pointers_2 = mark_pointers_2 * 2\n opponents_pointers_3 = mark_pointers_3 / 2\n opponents_free_throws = mark_free_throws / 2\n opponents_points_scored = opponents_pointers_2 + opponents_pointers_3 + opponents_free_throws\n total_points_scored = mark_points_scored + opponents_points_scored\n result = total_points_scored\n return result\n```'),
dict(role='SYSTEM', prompt='Response:210'),
dict(role='BOT', prompt='Thought: 根据回答，我得到了答案\nFinalAnswer: 210'),
dict(role='HUMAN', prompt='Bella有两倍于飞盘的弹珠。她还比卡片多20个飞盘。如果她买每种物品多2/5，她会有多少总数的物品，如果她现在有60颗弹珠？'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\ndef solution():\n marbles = 60\n num_increased_marbles = marbles * 2 / 5\n num_total_marbles = marbles + num_increased_marbles\n frisbees = marbles / 2\n num_increased_frisbees = frisbees * 2 / 5\n num_total_frisbees = frisbees + num_increased_frisbees\n deck_cards = frisbees - 20\n num_increased_deck_cards = deck_cards * 2 / 5\n num_total_deck_cards = deck_cards + num_increased_deck_cards\n num_total = num_total_marbles + num_total_frisbees + num_total_deck_cards\n result = num_total\n return result\n```'),
dict(role='SYSTEM', prompt='Response:140'),
dict(role='BOT', prompt='Thought: 根据回答，我得到了答案\nFinalAnswer: 140'),
dict(role='HUMAN', prompt='一个有4个水果篮子，前三个篮子里有9个苹果、15个橙子和14个香蕉，第四个篮子里每种水果都少2个。总共有多少水果？'),
dict(role='BOT', prompt="""Tool:PythonInterpreter\nTool Input:```python\ndef solution():\n num_fruits_per_first_three_basket = 9 + 15 + 14\n num_fruits_first_three_basket = num_fruits_per_first_three_basket * 3\n num_apple_fourth_basket = 9 - 2\n num_orange_fourth_basket = 15 - 2\n num_banana_fourth_basket = 14 - 2\n num_fruits_fourth_basket = num_apple_fourth_basket + num_orange_fourth_basket + num_banana_fourth_basket\n num_fruits_total = num_fruits_first_three_basket + num_fruits_fourth_basket\n result = num_fruits_total\n return result\n```"""),
dict(role='SYSTEM', prompt='Response:146'),
dict(role='BOT', prompt='Thought: 根据回答，我得到了答案\nFinalAnswer: 146'),
dict(role='HUMAN', prompt='{question}'),
],
'FEWSHOT_INSTRUCTION_CHOICE' : [
dict(role='HUMAN', prompt='已知点P（-1，4）在反比例函数$y=\\frac{{k}}{{x}}$ (k≠0)的图象上，则k的值是____'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\nfrom sympy import solve, symbols, Ne, Eq\ndef solution():\n k = symbols(\'k\')\n result = solve([Eq(4, k / (-1)), Ne(k, 0)], k, dict=True)\n return result\n```'),
dict(role='SYSTEM', prompt='Response:{{Eq(k, -4)}}'),
dict(role='BOT', prompt='Thought: 将点 P 带入函数解出 k 的值为 $-4$\nFinalAnswer: D'),
dict(role='HUMAN', prompt='幂函数$ y=(x) $的图象经过点$ (2, \\dfrac {{1}}{{4}}) $，则$ f(-3) $的值为 ______ ．'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\nfrom sympy import *\ndef solution():\n x, y, k = symbols(\'x y k\')\n eq1 = Eq(2**k, Rational(1, 4))\n k_value = solve(eq1, k)[0]\n y = x**k_value\n result = y.subs(x, -3)\n return result\n```'),
dict(role='SYSTEM', prompt='Response:1/9'),
dict(role='BOT', prompt='Thought: 求出幂函数的函数表达式为 $y=x^{{-2}}$，代入 $x=-3$ 得到 $y=$\\frac{{1}}{{9}}$\nFinalAnswer: A'),
dict(role='HUMAN', prompt='如果$3 x-y=12$，则$\\frac{8^{x}}{2^{y}}$的值是多少？\nA. 无法从给定的信息中确定值。\nB. $2^{12}$\nC. 4\nD. $8^{2}$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:```python\nfrom sympy import symbols, Eq, solve\n\ndef sloution():\n x, y = symbols(\'x y\')\n equation = Eq(3*x - y, 12)\n\n y_in_terms_of_x = solve(equation, y)[0]\n expression = 8**x / 2**y_in_terms_of_x\n result = expression.simplify()\n return result\n```'),
dict(role='SYSTEM', prompt='Response:2**12'),
dict(role='BOT', prompt='Thought: $\\frac{8^{x}}{2^{y}}$的值是$2^{12}$\nFinalAnswer: B'),
dict(role='HUMAN', prompt='{question}'),
]
}
mathbench_sets = {
'college': ['single_choice_cn', 'cloze_en'],
'high': ['single_choice_cn', 'single_choice_en'],
'middle': ['single_choice_cn'],
'primary': ['cloze_cn'],
'primary_refine': ['refine_cloze_cn']
}
# Use circular evaluation or not
with_circular_eval = True
mathbench_agent_datasets = []
for _split in list(mathbench_sets.keys()):
for _name in mathbench_sets[_split]:
prompt_example = PROMPT_CN if '_cn' in _name else PROMPT_EN
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate,
template=dict(
round = prompt_example['FEWSHOT_INSTRUCTION_CLOZE'] if 'cloze' in _name else prompt_example['FEWSHOT_INSTRUCTION_CHOICE'])),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=AgentInferencer)
)
mathbench_eval_cfg = dict(
evaluator=dict(type=CircularEvaluator if 'choice' in _name and with_circular_eval else AccEvaluator),
pred_postprocessor=dict(type=first_option_postprocess, options='ABCD') if 'single_choice' in _name else dict(type=mathbench_postprocess, name=_name))
mathbench_agent_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name + '-agent',
type=MathBenchDataset,
path=f'./data/mathbench/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=dict(
input_columns=['question'],
output_column='answer'
),
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
))
opencompass/configs/datasets/MathBench/deprecated_mathbench_agent_gen_fbe13b.py 0 → 100644
View file @ be3dfa50
# GONNA BE DEPRECATED, DON'T USE IT
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import AgentInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
PROMPT_EN = {
'FEWSHOT_INSTRUCTION_CLOZE' : [
dict(role='HUMAN', prompt='Mark\'s basketball team scores 25 2 pointers, 8 3 pointers and 10 free throws. Their opponents score double the 2 pointers but half the 3 pointers and free throws. What\'s the total number of points scored by both teams added together?'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:def solution():\n mark_pointers_2 = 25 * 2\n mark_pointers_3 = 8 * 3\n mark_free_throws = 10 * 1\n mark_points_scored = mark_pointers_2 + mark_pointers_3 + mark_free_throws\n opponents_pointers_2 = mark_pointers_2 * 2\n opponents_pointers_3 = mark_pointers_3 / 2\n opponents_free_throws = mark_free_throws / 2\n opponents_points_scored = opponents_pointers_2 + opponents_pointers_3 + opponents_free_throws\n total_points_scored = mark_points_scored + opponents_points_scored\n result = total_points_scored\n return result'),
dict(role='SYSTEM', prompt='Response:210'),
dict(role='BOT', prompt='Thought: According to the response, I got the answer\nFinalAnswer: 210'),
dict(role='HUMAN', prompt='Bella has two times as many marbles as frisbees. She also has 20 more frisbees than deck cards. If she buys 2/5 times more of each item, what would be the total number of the items she will have if she currently has 60 marbles?'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:def solution():\n marbles = 60\n num_increased_marbles = marbles * 2 / 5\n num_total_marbles = marbles + num_increased_marbles\n frisbees = marbles / 2\n num_increased_frisbees = frisbees * 2 / 5\n num_total_frisbees = frisbees + num_increased_frisbees\n deck_cards = frisbees - 20\n num_increased_deck_cards = deck_cards * 2 / 5\n num_total_deck_cards = deck_cards + num_increased_deck_cards\n num_total = num_total_marbles + num_total_frisbees + num_total_deck_cards\n result = num_total\n return result'),
dict(role='SYSTEM', prompt='Response:140'),
dict(role='BOT', prompt='Thought: According to the response, I got the answer\nFinalAnswer: 140'),
dict(role='HUMAN', prompt='A group of 4 fruit baskets contains 9 apples, 15 oranges, and 14 bananas in the first three baskets and 2 less of each fruit in the fourth basket. How many fruits are there?'),
dict(role='BOT', prompt="""Tool:PythonInterpreter\nTool Input:def solution():\n num_fruits_per_first_three_basket = 9 + 15 + 14\n num_fruits_first_three_basket = num_fruits_per_first_three_basket * 3\n num_apple_fourth_basket = 9 - 2\n num_orange_fourth_basket = 15 - 2\n num_banana_fourth_basket = 14 - 2\n num_fruits_fourth_basket = num_apple_fourth_basket + num_orange_fourth_basket + num_banana_fourth_basket\n num_fruits_total = num_fruits_first_three_basket + num_fruits_fourth_basket\n result = num_fruits_total\n return result"""),
dict(role='SYSTEM', prompt='Response:146'),
dict(role='BOT', prompt='Thought: According to the response, I got the answer\nFinalAnswer: 146'),
dict(role='HUMAN', prompt='{question}'),
],
'FEWSHOT_INSTRUCTION_CHOICE' : [
dict(role='HUMAN', prompt='Given point P(-1,4) lies on the graph of the inverse proportionality function $y=\\frac{{k}}{{x}}$ (k≠0), what is the value of k? A. $-\\frac{{1}}{{4}}$ B. $\\frac{{1}}{{4}}$ C. $4$ D. $-4$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:from sympy import solve, symbols, Ne, Eq\ndef solution():\n k = symbols(\'k\')\n result = solve([Eq(4, k / (-1)), Ne(k, 0)], k, dict=True)\n return result'),
dict(role='SYSTEM', prompt='Response:{{Eq(k, -4)}}'),
dict(role='BOT', prompt='Thought: Substituting point P into the function yields the value of k as $-4$\nFinalAnswer: D'),
dict(role='HUMAN', prompt='The graph of the power function $y=(x)$ passes through the point$ (2, \\dfrac {{1}}{{4}}) $, what is the value of $f(-3)$? A. $\\frac{{1}}{{9}}$ B. $\\frac{{1}}{{8}})=196-x$ C. $\\frac{{2}}{{9}}$ D. $\\frac{{1}}{{4}}$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:from sympy import *\ndef solution():\n x, y, k = symbols(\'x y k\')\n eq1 = Eq(2**k, Rational(1, 4))\n k_value = solve(eq1, k)[0]\n y = x**k_value\n result = y.subs(x, -3)\n return result'),
dict(role='SYSTEM', prompt='Response:1/9'),
dict(role='BOT', prompt='Thought: The functional expression of the power function is $y=x^{{-2}}$. Substituting $x=-3$ yields $y=$\\frac{{1}}{{9}}$\nFinalAnswer: A'),
dict(role='HUMAN', prompt='If $3 x-y=12$, what is the value of $\\frac{8^{x}}{2^{y}} ?$\nA. The value cannot be determined from the information given.\nB. $2^{12}$\nC. 4\nD. $8^{2}$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:from sympy import symbols, Eq, solve\n\ndef sloution():\n x, y = symbols(\'x y\')\n equation = Eq(3*x - y, 12)\n\n y_in_terms_of_x = solve(equation, y)[0]\n expression = 8**x / 2**y_in_terms_of_x\n result = expression.simplify()\n return result'),
dict(role='SYSTEM', prompt='Response:2**12'),
dict(role='BOT', prompt='Thought: The value of $\\frac{8^{x}}{2^{y}}$ is $2^{12}$\nFinalAnswer: B'),
dict(role='HUMAN', prompt='{question}'),
]
}
PROMPT_CN = {
'FEWSHOT_INSTRUCTION_CLOZE' : [
dict(role='HUMAN', prompt='Mark的篮球队得到25个2分球，8个3分球和10个罚球。他们的对手得到2分球的两倍，但3分球和罚球的一半。两队得分的总和是多少？'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:def solution():\n mark_pointers_2 = 25 * 2\n mark_pointers_3 = 8 * 3\n mark_free_throws = 10 * 1\n mark_points_scored = mark_pointers_2 + mark_pointers_3 + mark_free_throws\n opponents_pointers_2 = mark_pointers_2 * 2\n opponents_pointers_3 = mark_pointers_3 / 2\n opponents_free_throws = mark_free_throws / 2\n opponents_points_scored = opponents_pointers_2 + opponents_pointers_3 + opponents_free_throws\n total_points_scored = mark_points_scored + opponents_points_scored\n result = total_points_scored\n return result'),
dict(role='SYSTEM', prompt='Response:210'),
dict(role='BOT', prompt='Thought: 根据回答，我得到了答案\nFinalAnswer: 210'),
dict(role='HUMAN', prompt='Bella有两倍于飞盘的弹珠。她还比卡片多20个飞盘。如果她买每种物品多2/5，她会有多少总数的物品，如果她现在有60颗弹珠？'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:def solution():\n marbles = 60\n num_increased_marbles = marbles * 2 / 5\n num_total_marbles = marbles + num_increased_marbles\n frisbees = marbles / 2\n num_increased_frisbees = frisbees * 2 / 5\n num_total_frisbees = frisbees + num_increased_frisbees\n deck_cards = frisbees - 20\n num_increased_deck_cards = deck_cards * 2 / 5\n num_total_deck_cards = deck_cards + num_increased_deck_cards\n num_total = num_total_marbles + num_total_frisbees + num_total_deck_cards\n result = num_total\n return result'),
dict(role='SYSTEM', prompt='Response:140'),
dict(role='BOT', prompt='Thought: 根据回答，我得到了答案\nFinalAnswer: 140'),
dict(role='HUMAN', prompt='一个有4个水果篮子，前三个篮子里有9个苹果、15个橙子和14个香蕉，第四个篮子里每种水果都少2个。总共有多少水果？'),
dict(role='BOT', prompt="""Tool:PythonInterpreter\nTool Input:def solution():\n num_fruits_per_first_three_basket = 9 + 15 + 14\n num_fruits_first_three_basket = num_fruits_per_first_three_basket * 3\n num_apple_fourth_basket = 9 - 2\n num_orange_fourth_basket = 15 - 2\n num_banana_fourth_basket = 14 - 2\n num_fruits_fourth_basket = num_apple_fourth_basket + num_orange_fourth_basket + num_banana_fourth_basket\n num_fruits_total = num_fruits_first_three_basket + num_fruits_fourth_basket\n result = num_fruits_total\n return result"""),
dict(role='SYSTEM', prompt='Response:146'),
dict(role='BOT', prompt='Thought: 根据回答，我得到了答案\nFinalAnswer: 146'),
dict(role='HUMAN', prompt='{question}'),
],
'FEWSHOT_INSTRUCTION_CHOICE' : [
dict(role='HUMAN', prompt='已知点P（-1，4）在反比例函数$y=\\frac{{k}}{{x}}$ (k≠0)的图象上，则k的值是____'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:from sympy import solve, symbols, Ne, Eq\ndef solution():\n k = symbols(\'k\')\n result = solve([Eq(4, k / (-1)), Ne(k, 0)], k, dict=True)\n return result'),
dict(role='SYSTEM', prompt='Response:{{Eq(k, -4)}}'),
dict(role='BOT', prompt='Thought: 将点 P 带入函数解出 k 的值为 $-4$\nFinalAnswer: D'),
dict(role='HUMAN', prompt='幂函数$ y=(x) $的图象经过点$ (2, \\dfrac {{1}}{{4}}) $，则$ f(-3) $的值为 ______ ．'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:from sympy import *\ndef solution():\n x, y, k = symbols(\'x y k\')\n eq1 = Eq(2**k, Rational(1, 4))\n k_value = solve(eq1, k)[0]\n y = x**k_value\n result = y.subs(x, -3)\n return result'),
dict(role='SYSTEM', prompt='Response:1/9'),
dict(role='BOT', prompt='Thought: 求出幂函数的函数表达式为 $y=x^{{-2}}$，代入 $x=-3$ 得到 $y=$\\frac{{1}}{{9}}$\nFinalAnswer: A'),
dict(role='HUMAN', prompt='如果$3 x-y=12$，则$\\frac{8^{x}}{2^{y}}$的值是多少？\nA. 无法从给定的信息中确定值。\nB. $2^{12}$\nC. 4\nD. $8^{2}$'),
dict(role='BOT', prompt='Tool:PythonInterpreter\nTool Input:from sympy import symbols, Eq, solve\n\ndef sloution():\n x, y = symbols(\'x y\')\n equation = Eq(3*x - y, 12)\n\n y_in_terms_of_x = solve(equation, y)[0]\n expression = 8**x / 2**y_in_terms_of_x\n result = expression.simplify()\n return result'),
dict(role='SYSTEM', prompt='Response:2**12'),
dict(role='BOT', prompt='Thought: $\\frac{8^{x}}{2^{y}}$的值是$2^{12}$\nFinalAnswer: B'),
dict(role='HUMAN', prompt='{question}'),
]
}
mathbench_sets = {
'college': ['single_choice_cn', 'cloze_en'],
'high': ['single_choice_cn', 'single_choice_en'],
'middle': ['single_choice_cn'],
'primary': ['cloze_cn'],
'primary_refine': ['refine_cloze_cn']
}
# Use circular evaluation or not
with_circular_eval = True
mathbench_agent_datasets = []
for _split in list(mathbench_sets.keys()):
for _name in mathbench_sets[_split]:
prompt_example = PROMPT_CN if '_cn' in _name else PROMPT_EN
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate,
template=dict(
round = prompt_example['FEWSHOT_INSTRUCTION_CLOZE'] if 'cloze' in _name else prompt_example['FEWSHOT_INSTRUCTION_CHOICE'])),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=AgentInferencer)
)
mathbench_eval_cfg = dict(
evaluator=dict(type=CircularEvaluator if 'choice' in _name and with_circular_eval else AccEvaluator),
pred_postprocessor=dict(type=first_option_postprocess, options='ABCD') if 'single_choice' in _name else dict(type=mathbench_postprocess, name=_name))
mathbench_agent_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name + '-agent',
type=MathBenchDataset,
path=f'./data/mathbench/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=dict(
input_columns=['question'],
output_column='answer'
),
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
))
opencompass/configs/datasets/MathBench/deprecated_mathbench_arith_gen_ccd638.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer
from opencompass.openicl.icl_evaluator import AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
cloze_prompts ={
'cloze_arith_en': [
dict(role='HUMAN', prompt='Q: Calculate (341/11)/(9/(-6)*(-2)/3).'),
dict(role='BOT', prompt='A: First, (9/(-6)*(-2)/3) can be simplified by : 9/(-6) = -1.5, -1.5 * (-2) = 3, 3 / 3 = 1. So, (9/(-6)*(-2)/3) is equal to 1. Now, we have `(341/11)/1` equals `341/11`. Finally, calculate `341/11 = 31`. The answer is 31.\n'),
dict(role='HUMAN', prompt='Q: In base 14, what is 5 - 638d8d?'),
dict(role='BOT', prompt='A: 5 - 638d8d = -638d88. The answer is -638d88.\n'),
dict(role='HUMAN', prompt='Q: What is -491354 times -0.34?'),
dict(role='BOT', prompt='A: The product of -491354 and -0.34 is 167060.36. The answer is 167060.36.\n'),
dict(role='HUMAN', prompt='Q: What is the value of (-55)/(6930/(-382)) + (0 - 3)?.'),
dict(role='BOT', prompt='A: First, (-55)/(6930/(-382)) = (-55)/(-(6930/382)) = 55*382/6930 = 21010/6930 = 2101/693. Then, 2101/693 + (0 - 3) = 2101/693 - 3 = 2101/693 - 3*693/693 = (2101-2079)/693 = 22/693 = 2/63. The answer is 2/63.\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}\n'),
]
}
mathbench_sets = {
'arithmetic': ['cloze_arith_en'],
}
mathbench_datasets = []
for _split in list(mathbench_sets.keys()):
for _name in mathbench_sets[_split]:
mathbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=cloze_prompts[_name],
),
),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=512),
)
mathbench_eval_cfg = dict(
evaluator=dict(type=AccEvaluator),
pred_postprocessor=dict(type=mathbench_postprocess, name=_name))
mathbench_datasets.append(
dict(
type=MathBenchDataset,
path=f'./data/mathbench/{_split}',
name=_name,
with_circular=False,
abbr='mathbench-arithmetic' + _split + '-' + _name,
reader_cfg=dict(
input_columns=['question'],
output_column='answer'
),
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
))
opencompass/configs/datasets/MathBench/deprecated_mathbench_cot_gen_66f329.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
single_choice_prompts = {
'single_choice_cn_with_reasoning': '以下是一道关于数学的单项选择题，请你一步一步推理，并在最后用“所以答案为选项X”给出答案，其中“X”为选项A，B，C，D中你认为正确的选项。下面是你要回答的问题\n{question}\n让我们一步一步思考：\n',
'single_choice_cn': '以下是一道关于数学的单项选择题，请你直接回答正确答案的选项序号。\n下面是你要回答的题目：\n{question}\n答案选项：',
'single_choice_en_with_reasoning': "Here is a multiple-choice question about mathematics. Please reason through it step by step, and at the end, provide your answer option with 'Therefore, the correct answer is option X', Where 'X' is the correct option you think from A，B，C，D. Here is the question you need to answer:\n{question}\nLet's think step by step:",
'single_choice_en': 'Here is a multiple-choice question about mathematics. Please provide the correct answer option directly.\nHere is the question you need to answer:\n{question}\nAnswer option:',
}
cloze_prompts = {
'cloze_cn': [
dict(role='HUMAN', prompt='Q: 林中有15棵树。林务工人员今天将在林中种植树木。完成后，将有21棵树。林务工人员今天种植了多少棵树？'),
dict(role='BOT', prompt='A: 我们从15棵树开始。后来有21棵树。差值必定是他们种植的树木数量。所以，他们必须种植了21 - 15 = 6棵树。答案是 6\n'),
dict(role='HUMAN', prompt='Q: 如果停车场有3辆车，又有2辆车进来，停车场里有多少辆车？'),
dict(role='BOT', prompt='A: 停车场已经有3辆车。又进来了2辆车。现在有3 + 2 = 5辆车。答案是 5\n'),
dict(role='HUMAN', prompt='Q: 黎恩有32块巧克力，她的妹妹有42块。如果他们吃了35块，他们总共剩下多少块？'),
dict(role='BOT', prompt='A: 黎恩有32块巧克力，Leah的妹妹有42块。这意味着原本有32 + 42 = 74块巧克力。被吃掉了35块。所以他们总共还剩下74 - 35 = 39块巧克力。答案是 39\n'),
dict(role='HUMAN', prompt='Q: 杰森有20个棒棒糖。他给丹妮一些棒棒糖。现在Jason只剩下12个棒棒糖。杰森给丹妮多少个棒棒糖？'),
dict(role='BOT', prompt='A: 杰森有20个棒棒糖。因为他现在只剩下12个，所以他必须把剩下的都给了丹妮。他给丹妮的棒棒糖数量必定是20 - 12 = 8个。答案是 8\n'),
dict(role='HUMAN', prompt='Q: 莎莎有五个玩具。在圣诞节，他从他的爸爸和妈妈那里各得到了两个玩具。现在他有多少个玩具？'),
dict(role='BOT', prompt='A: 她有5个玩具。他从妈妈那里得到了2个，所以之后他有5 + 2 = 7个玩具。然后他从爸爸那里得到了2个，所以总共他有7 + 2 = 9个玩具。答案是 9\n'),
dict(role='HUMAN', prompt='Q: 服务器房间里有九台电脑。从周一到周四每天增加五台电脑。现在服务器房里有多少台电脑？'),
dict(role='BOT', prompt='A: 从周一到周四有4天。每天增加5台电脑。这意味着总共增加了4 * 5 = 20台电脑。一开始有9台电脑，所以现在有9 + 20 = 29台电脑。答案是 29\n'),
dict(role='HUMAN', prompt='Q: 迈克尔有58个高尔夫球。星期二，他丢失了23个高尔夫球。星期三，他又丢失了2个。星期三结束时他还剩下多少个高尔夫球？'),
dict(role='BOT', prompt='A: 迈克尔一开始有58个球。星期二他丢失了23个，所以之后他还剩下58 - 23 = 35个球。星期三他又丢失了2个，所以现在他还剩下35 - 2 = 33个球。答案是 33\n'),
dict(role='HUMAN', prompt='Q: 奥利弗有23美元。她用每个3美元的价格买了五个百吉饼。她还剩下多少钱？'),
dict(role='BOT', prompt='A: 她以每个3美元的价格买了5个百吉饼。这意味着她在百吉饼上花费了5 * 3 = 15美元。她一开始有23美元，所以现在她还剩下23 - 15 = 8美元。答案是 8\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}'),
],
'cloze_en': [
dict(role='HUMAN', prompt='Q: There are 15 trees in the grove. Grove workers will plant trees in the grove today. After they are done, there will be 21 trees. How many trees did the grove workers plant today?'),
dict(role='BOT', prompt='A: We start with 15 trees. Later we have 21 trees. The difference must be the number of trees they planted. So, they must have planted 21 - 15 = 6 trees. The answer is 6.\n'),
dict(role='HUMAN', prompt='Q: If there are 3 cars in the parking lot and 2 more cars arrive, how many cars are in the parking lot?'),
dict(role='BOT', prompt='A: There are 3 cars in the parking lot already. 2 more arrive. Now there are 3 + 2 = 5 cars. The answer is 5.\n'),
dict(role='HUMAN', prompt='Q: Leah had 32 chocolates and her sister had 42. If they ate 35, how many pieces do they have left in total?'),
dict(role='BOT', prompt="A: Leah had 32 chocolates and Leah's sister had 42. That means there were originally 32 + 42 = 74 chocolates. 35 have been eaten. So in total they still have 74 - 35 = 39 chocolates. The answer is 39.\n"),
dict(role='HUMAN', prompt='Q: Jason had 20 lollipops. He gave Denny some lollipops. Now Jason has 12 lollipops. How many lollipops did Jason give to Denny?'),
dict(role='BOT', prompt='A: Jason had 20 lollipops. Since he only has 12 now, he must have given the rest to Denny. The number of lollipops he has given to Denny must have been 20 - 12 = 8 lollipops. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: Shawn has five toys. For Christmas, he got two toys each from his mom and dad. How many toys does he have now?'),
dict(role='BOT', prompt='A: He has 5 toys. He got 2 from mom, so after that he has 5 + 2 = 7 toys. Then he got 2 more from dad, so in total he has 7 + 2 = 9 toys. The answer is 9.\n'),
dict(role='HUMAN', prompt='Q: There were nine computers in the server room. Five more computers were installed each day, from monday to thursday. How many computers are now in the server room?'),
dict(role='BOT', prompt='A: There are 4 days from monday to thursday. 5 computers were added each day. That means in total 4 * 5 = 20 computers were added. There were 9 computers in the beginning, so now there are 9 + 20 = 29 computers. The answer is 29.\n'),
dict(role='HUMAN', prompt='Q: Michael had 58 golf balls. On tuesday, he lost 23 golf balls. On wednesday, he lost 2 more. How many golf balls did he have at the end of wednesday?'),
dict(role='BOT', prompt='A: Michael initially had 58 balls. He lost 23 on Tuesday, so after that he has 58 - 23 = 35 balls. On Wednesday he lost 2 more so now he has 35 - 2 = 33 balls. The answer is 33.\n'),
dict(role='HUMAN', prompt='Q: Olivia has $23. She bought five bagels for $3 each. How much money does she have left?'),
dict(role='BOT', prompt='A: She bought 5 bagels for $3 each. This means she spent 5 * $3 = $15 on the bagels. She had $23 in beginning, so now she has $23 - $15 = $8. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}\n'),
]}
cloze_prompts['refine_cloze_cn'] = cloze_prompts['cloze_cn']
mathbench_sets = {
'college': ['single_choice_cn', 'cloze_en'],
'high': ['single_choice_cn', 'single_choice_en'],
'middle': ['single_choice_cn'],
'primary': ['cloze_cn'],
'primary_refine': ['refine_cloze_cn']
}
# Generate reasoning path or not, only for single choice
with_reasoning = True
# Use circular evaluation or not
with_circular_eval = True
mathbench_datasets = []
for _split in list(mathbench_sets.keys()):
for _name in mathbench_sets[_split]:
mathbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[
dict(
role='HUMAN',
prompt=single_choice_prompts[_name + '_with_reasoning'] if with_reasoning else single_choice_prompts[_name],
),
dict(role='BOT', prompt='{answer}')] if 'choice' in _name else cloze_prompts[_name],
),
),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=512),
)
mathbench_eval_cfg = dict(
evaluator=dict(type=CircularEvaluator if 'choice' in _name and with_circular_eval else AccEvaluator),
pred_postprocessor=dict(type=first_option_postprocess, options='ABCD') if 'single_choice' in _name else dict(type=mathbench_postprocess, name=_name))
mathbench_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'./data/mathbench/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=dict(
input_columns=['question'],
output_column='answer'
),
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
))
opencompass/configs/datasets/MathBench/deprecated_mathbench_gen_7b734b.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
single_choice_prompts = {
'single_choice_cn_with_reasoning': '以下是一道关于数学的单项选择题，请你一步一步推理，并在最后用“所以答案为选项X”给出答案，其中“X”为选项A，B，C，D中你认为正确的选项。下面是你要回答的问题\n{question}\n让我们一步一步思考：\n',
'single_choice_cn': '以下是一道关于数学的单项选择题，请你直接回答正确答案的选项序号。\n下面是你要回答的题目：\n{question}\n答案选项：',
'single_choice_en_with_reasoning': "Here is a multiple-choice question about mathematics. Please reason through it step by step, and at the end, provide your answer option with 'Therefore, the correct answer is option X', Where 'X' is the correct option you think from A，B，C，D. Here is the question you need to answer:\n{question}\nLet's think step by step:",
'single_choice_en': 'Here is a multiple-choice question about mathematics. Please provide the correct answer option directly.\nHere is the question you need to answer:\n{question}\nAnswer option:',
}
cloze_prompts = {
'cloze_cn': [
dict(role='HUMAN', prompt='Q: 林中有15棵树。林务工人员今天将在林中种植树木。完成后，将有21棵树。林务工人员今天种植了多少棵树？'),
dict(role='BOT', prompt='A: 我们从15棵树开始。后来有21棵树。差值必定是他们种植的树木数量。所以，他们必须种植了21 - 15 = 6棵树。答案是 6\n'),
dict(role='HUMAN', prompt='Q: 如果停车场有3辆车，又有2辆车进来，停车场里有多少辆车？'),
dict(role='BOT', prompt='A: 停车场已经有3辆车。又进来了2辆车。现在有3 + 2 = 5辆车。答案是 5\n'),
dict(role='HUMAN', prompt='Q: 黎恩有32块巧克力，她的妹妹有42块。如果他们吃了35块，他们总共剩下多少块？'),
dict(role='BOT', prompt='A: 黎恩有32块巧克力，Leah的妹妹有42块。这意味着原本有32 + 42 = 74块巧克力。被吃掉了35块。所以他们总共还剩下74 - 35 = 39块巧克力。答案是 39\n'),
dict(role='HUMAN', prompt='Q: 杰森有20个棒棒糖。他给丹妮一些棒棒糖。现在Jason只剩下12个棒棒糖。杰森给丹妮多少个棒棒糖？'),
dict(role='BOT', prompt='A: 杰森有20个棒棒糖。因为他现在只剩下12个，所以他必须把剩下的都给了丹妮。他给丹妮的棒棒糖数量必定是20 - 12 = 8个。答案是 8\n'),
dict(role='HUMAN', prompt='Q: 莎莎有五个玩具。在圣诞节，他从他的爸爸和妈妈那里各得到了两个玩具。现在他有多少个玩具？'),
dict(role='BOT', prompt='A: 她有5个玩具。他从妈妈那里得到了2个，所以之后他有5 + 2 = 7个玩具。然后他从爸爸那里得到了2个，所以总共他有7 + 2 = 9个玩具。答案是 9\n'),
dict(role='HUMAN', prompt='Q: 服务器房间里有九台电脑。从周一到周四每天增加五台电脑。现在服务器房里有多少台电脑？'),
dict(role='BOT', prompt='A: 从周一到周四有4天。每天增加5台电脑。这意味着总共增加了4 * 5 = 20台电脑。一开始有9台电脑，所以现在有9 + 20 = 29台电脑。答案是 29\n'),
dict(role='HUMAN', prompt='Q: 迈克尔有58个高尔夫球。星期二，他丢失了23个高尔夫球。星期三，他又丢失了2个。星期三结束时他还剩下多少个高尔夫球？'),
dict(role='BOT', prompt='A: 迈克尔一开始有58个球。星期二他丢失了23个，所以之后他还剩下58 - 23 = 35个球。星期三他又丢失了2个，所以现在他还剩下35 - 2 = 33个球。答案是 33\n'),
dict(role='HUMAN', prompt='Q: 奥利弗有23美元。她用每个3美元的价格买了五个百吉饼。她还剩下多少钱？'),
dict(role='BOT', prompt='A: 她以每个3美元的价格买了5个百吉饼。这意味着她在百吉饼上花费了5 * 3 = 15美元。她一开始有23美元，所以现在她还剩下23 - 15 = 8美元。答案是 8\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}'),
],
'cloze_en': [
dict(role='HUMAN', prompt='Q: There are 15 trees in the grove. Grove workers will plant trees in the grove today. After they are done, there will be 21 trees. How many trees did the grove workers plant today?'),
dict(role='BOT', prompt='A: We start with 15 trees. Later we have 21 trees. The difference must be the number of trees they planted. So, they must have planted 21 - 15 = 6 trees. The answer is 6.\n'),
dict(role='HUMAN', prompt='Q: If there are 3 cars in the parking lot and 2 more cars arrive, how many cars are in the parking lot?'),
dict(role='BOT', prompt='A: There are 3 cars in the parking lot already. 2 more arrive. Now there are 3 + 2 = 5 cars. The answer is 5.\n'),
dict(role='HUMAN', prompt='Q: Leah had 32 chocolates and her sister had 42. If they ate 35, how many pieces do they have left in total?'),
dict(role='BOT', prompt="A: Leah had 32 chocolates and Leah's sister had 42. That means there were originally 32 + 42 = 74 chocolates. 35 have been eaten. So in total they still have 74 - 35 = 39 chocolates. The answer is 39.\n"),
dict(role='HUMAN', prompt='Q: Jason had 20 lollipops. He gave Denny some lollipops. Now Jason has 12 lollipops. How many lollipops did Jason give to Denny?'),
dict(role='BOT', prompt='A: Jason had 20 lollipops. Since he only has 12 now, he must have given the rest to Denny. The number of lollipops he has given to Denny must have been 20 - 12 = 8 lollipops. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: Shawn has five toys. For Christmas, he got two toys each from his mom and dad. How many toys does he have now?'),
dict(role='BOT', prompt='A: He has 5 toys. He got 2 from mom, so after that he has 5 + 2 = 7 toys. Then he got 2 more from dad, so in total he has 7 + 2 = 9 toys. The answer is 9.\n'),
dict(role='HUMAN', prompt='Q: There were nine computers in the server room. Five more computers were installed each day, from monday to thursday. How many computers are now in the server room?'),
dict(role='BOT', prompt='A: There are 4 days from monday to thursday. 5 computers were added each day. That means in total 4 * 5 = 20 computers were added. There were 9 computers in the beginning, so now there are 9 + 20 = 29 computers. The answer is 29.\n'),
dict(role='HUMAN', prompt='Q: Michael had 58 golf balls. On tuesday, he lost 23 golf balls. On wednesday, he lost 2 more. How many golf balls did he have at the end of wednesday?'),
dict(role='BOT', prompt='A: Michael initially had 58 balls. He lost 23 on Tuesday, so after that he has 58 - 23 = 35 balls. On Wednesday he lost 2 more so now he has 35 - 2 = 33 balls. The answer is 33.\n'),
dict(role='HUMAN', prompt='Q: Olivia has $23. She bought five bagels for $3 each. How much money does she have left?'),
dict(role='BOT', prompt='A: She bought 5 bagels for $3 each. This means she spent 5 * $3 = $15 on the bagels. She had $23 in beginning, so now she has $23 - $15 = $8. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}\n'),
]}
cloze_prompts['refine_cloze_cn'] = cloze_prompts['cloze_cn']
mathbench_sets = {
'college': ['single_choice_cn', 'cloze_en'],
'high': ['single_choice_cn', 'single_choice_en'],
'middle': ['single_choice_cn'],
'primary': ['cloze_cn'],
'primary_refine': ['refine_cloze_cn']
}
# Generate reasoning path or not, only for single choice
with_reasoning = False
# Use circular evaluation or not
with_circular_eval = True
mathbench_datasets = []
for _split in list(mathbench_sets.keys()):
for _name in mathbench_sets[_split]:
mathbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[
dict(
role='HUMAN',
prompt=single_choice_prompts[_name + '_with_reasoning'] if with_reasoning else single_choice_prompts[_name],
),
dict(role='BOT', prompt='{answer}')] if 'choice' in _name else cloze_prompts[_name],
),
),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=512),
)
mathbench_eval_cfg = dict(
evaluator=dict(type=CircularEvaluator if 'choice' in _name and with_circular_eval else AccEvaluator),
pred_postprocessor=dict(type=first_option_postprocess, options='ABCD') if 'single_choice' in _name else dict(type=mathbench_postprocess, name=_name))
mathbench_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'./data/mathbench/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=dict(
input_columns=['question'],
output_column='answer'
),
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
))
opencompass/configs/datasets/MathBench/mathbench_2024_few_shot_mixed_4a3fd4.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
from copy import deepcopy
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer, PPLInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
with read_base():
from .mathbench_prompt import zero_shot_prompts, few_shot_prompts, mathbench_sets
# Max for this dataset is 4
num_shot = 4
# Generate reasoning path or not, only for single choice
with_reasoning = True
# Use circular evaluation or not
with_circular_eval = True
# Use PPL mode in single choice test or not
use_ppl_single_choice = True
assert 0 <= num_shot <= 4
if num_shot == 0:
prompts = zero_shot_prompts
else:
prompts = {name: p[- 2 * num_shot - 2:] for name, p in few_shot_prompts.items()}
mathbench_datasets = []
for _split in mathbench_sets:
for _name in mathbench_sets[_split]:
if 'single_choice' in _name:
if with_reasoning and not use_ppl_single_choice:
template_round = prompts[_name + '_with_reasoning']
else:
template_round = prompts[_name]
else:
template_round = prompts[_name]
if 'single_choice' in _name:
pred_postprocessor = dict(type=first_option_postprocess, options='ABCD')
else:
pred_postprocessor = dict(type=mathbench_postprocess, name=_name)
if 'single_choice' in _name and with_circular_eval:
evaluator = dict(type=CircularEvaluator)
else:
evaluator = dict(type=AccEvaluator)
# assemble the final config
mathbench_reader_cfg = dict(input_columns=['question'], output_column='answer')
if use_ppl_single_choice and 'single_choice' in _name:
template = {}
for answer in ['A', 'B', 'C', 'D']:
one_template_round = deepcopy(template_round)
one_template_round[-1]['prompt'] = one_template_round[-1]['prompt'].format(answer=answer)
template[answer] = dict(round=one_template_round)
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=template),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=PPLInferencer),
)
else:
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=dict(round=template_round)),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=2048, stopping_criteria=['Question:']),
)
mathbench_eval_cfg = dict(evaluator=evaluator, pred_postprocessor=pred_postprocessor)
mathbench_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'data/mathbench_v1/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=mathbench_reader_cfg,
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
)
)
opencompass/configs/datasets/MathBench/mathbench_2024_gen_19e486.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
single_choice_prompts = {
'single_choice_cn_with_reasoning': '以下是一道关于数学的单项选择题，请你一步一步推理，并在最后用“所以答案为选项X”给出答案，其中“X”为选项A，B，C，D中你认为正确的选项。下面是你要回答的问题\n{question}\n让我们一步一步思考：\n',
'single_choice_cn': '以下是一道关于数学的单项选择题，请你直接回答正确答案的选项序号。\n下面是你要回答的题目：\n{question}\n答案选项：',
'single_choice_en_with_reasoning': "Here is a multiple-choice question about mathematics. Please reason through it step by step, and at the end, provide your answer option with 'Therefore, the correct answer is option X', Where 'X' is the correct option you think from A，B，C，D. Here is the question you need to answer:\n{question}\nLet's think step by step:",
'single_choice_en': 'Here is a multiple-choice question about mathematics. Please provide the correct answer option directly.\nHere is the question you need to answer:\n{question}\nAnswer option:',
}
cloze_prompts = {
'cloze_cn': [
dict(role='HUMAN', prompt='Q: 林中有15棵树。林务工人员今天将在林中种植树木。完成后，将有21棵树。林务工人员今天种植了多少棵树？'),
dict(role='BOT', prompt='A: 我们从15棵树开始。后来有21棵树。差值必定是他们种植的树木数量。所以，他们必须种植了21 - 15 = 6棵树。答案是 6\n'),
dict(role='HUMAN', prompt='Q: 如果停车场有3辆车，又有2辆车进来，停车场里有多少辆车？'),
dict(role='BOT', prompt='A: 停车场已经有3辆车。又进来了2辆车。现在有3 + 2 = 5辆车。答案是 5\n'),
dict(role='HUMAN', prompt='Q: 黎恩有32块巧克力，她的妹妹有42块。如果他们吃了35块，他们总共剩下多少块？'),
dict(role='BOT', prompt='A: 黎恩有32块巧克力，Leah的妹妹有42块。这意味着原本有32 + 42 = 74块巧克力。被吃掉了35块。所以他们总共还剩下74 - 35 = 39块巧克力。答案是 39\n'),
dict(role='HUMAN', prompt='Q: 杰森有20个棒棒糖。他给丹妮一些棒棒糖。现在Jason只剩下12个棒棒糖。杰森给丹妮多少个棒棒糖？'),
dict(role='BOT', prompt='A: 杰森有20个棒棒糖。因为他现在只剩下12个，所以他必须把剩下的都给了丹妮。他给丹妮的棒棒糖数量必定是20 - 12 = 8个。答案是 8\n'),
dict(role='HUMAN', prompt='Q: 莎莎有五个玩具。在圣诞节，他从他的爸爸和妈妈那里各得到了两个玩具。现在他有多少个玩具？'),
dict(role='BOT', prompt='A: 她有5个玩具。他从妈妈那里得到了2个，所以之后他有5 + 2 = 7个玩具。然后他从爸爸那里得到了2个，所以总共他有7 + 2 = 9个玩具。答案是 9\n'),
dict(role='HUMAN', prompt='Q: 服务器房间里有九台电脑。从周一到周四每天增加五台电脑。现在服务器房里有多少台电脑？'),
dict(role='BOT', prompt='A: 从周一到周四有4天。每天增加5台电脑。这意味着总共增加了4 * 5 = 20台电脑。一开始有9台电脑，所以现在有9 + 20 = 29台电脑。答案是 29\n'),
dict(role='HUMAN', prompt='Q: 迈克尔有58个高尔夫球。星期二，他丢失了23个高尔夫球。星期三，他又丢失了2个。星期三结束时他还剩下多少个高尔夫球？'),
dict(role='BOT', prompt='A: 迈克尔一开始有58个球。星期二他丢失了23个，所以之后他还剩下58 - 23 = 35个球。星期三他又丢失了2个，所以现在他还剩下35 - 2 = 33个球。答案是 33\n'),
dict(role='HUMAN', prompt='Q: 奥利弗有23美元。她用每个3美元的价格买了五个百吉饼。她还剩下多少钱？'),
dict(role='BOT', prompt='A: 她以每个3美元的价格买了5个百吉饼。这意味着她在百吉饼上花费了5 * 3 = 15美元。她一开始有23美元，所以现在她还剩下23 - 15 = 8美元。答案是 8\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}'),
],
'cloze_en': [
dict(role='HUMAN', prompt='Q: There are 15 trees in the grove. Grove workers will plant trees in the grove today. After they are done, there will be 21 trees. How many trees did the grove workers plant today?'),
dict(role='BOT', prompt='A: We start with 15 trees. Later we have 21 trees. The difference must be the number of trees they planted. So, they must have planted 21 - 15 = 6 trees. The answer is 6.\n'),
dict(role='HUMAN', prompt='Q: If there are 3 cars in the parking lot and 2 more cars arrive, how many cars are in the parking lot?'),
dict(role='BOT', prompt='A: There are 3 cars in the parking lot already. 2 more arrive. Now there are 3 + 2 = 5 cars. The answer is 5.\n'),
dict(role='HUMAN', prompt='Q: Leah had 32 chocolates and her sister had 42. If they ate 35, how many pieces do they have left in total?'),
dict(role='BOT', prompt="A: Leah had 32 chocolates and Leah's sister had 42. That means there were originally 32 + 42 = 74 chocolates. 35 have been eaten. So in total they still have 74 - 35 = 39 chocolates. The answer is 39.\n"),
dict(role='HUMAN', prompt='Q: Jason had 20 lollipops. He gave Denny some lollipops. Now Jason has 12 lollipops. How many lollipops did Jason give to Denny?'),
dict(role='BOT', prompt='A: Jason had 20 lollipops. Since he only has 12 now, he must have given the rest to Denny. The number of lollipops he has given to Denny must have been 20 - 12 = 8 lollipops. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: Shawn has five toys. For Christmas, he got two toys each from his mom and dad. How many toys does he have now?'),
dict(role='BOT', prompt='A: He has 5 toys. He got 2 from mom, so after that he has 5 + 2 = 7 toys. Then he got 2 more from dad, so in total he has 7 + 2 = 9 toys. The answer is 9.\n'),
dict(role='HUMAN', prompt='Q: There were nine computers in the server room. Five more computers were installed each day, from monday to thursday. How many computers are now in the server room?'),
dict(role='BOT', prompt='A: There are 4 days from monday to thursday. 5 computers were added each day. That means in total 4 * 5 = 20 computers were added. There were 9 computers in the beginning, so now there are 9 + 20 = 29 computers. The answer is 29.\n'),
dict(role='HUMAN', prompt='Q: Michael had 58 golf balls. On tuesday, he lost 23 golf balls. On wednesday, he lost 2 more. How many golf balls did he have at the end of wednesday?'),
dict(role='BOT', prompt='A: Michael initially had 58 balls. He lost 23 on Tuesday, so after that he has 58 - 23 = 35 balls. On Wednesday he lost 2 more so now he has 35 - 2 = 33 balls. The answer is 33.\n'),
dict(role='HUMAN', prompt='Q: Olivia has $23. She bought five bagels for $3 each. How much money does she have left?'),
dict(role='BOT', prompt='A: She bought 5 bagels for $3 each. This means she spent 5 * $3 = $15 on the bagels. She had $23 in beginning, so now she has $23 - $15 = $8. The answer is 8.\n'),
dict(role='HUMAN', prompt='Q: {question}'),
dict(role='BOT', prompt='A: {answer}\n'),
]}
mathbench_sets = {
# Practice Part
'college': ['single_choice_cn', 'single_choice_en'],
'high': ['single_choice_cn', 'single_choice_en'],
'middle': ['single_choice_cn', 'single_choice_en'],
'primary': ['cloze_cn', 'cloze_en'],
'arithmetic': ['cloze_en'],
# Theory part
'college_knowledge': ['single_choice_cn','single_choice_en'],
'high_knowledge': ['single_choice_cn','single_choice_en'],
'middle_knowledge': ['single_choice_cn','single_choice_en'],
'primary_knowledge': ['single_choice_cn','single_choice_en'],
}
# Generate reasoning path or not, only for single choice
with_reasoning = True
# Use circular evaluation or not
with_circular_eval = True
mathbench_datasets = []
for _split in list(mathbench_sets.keys()):
for _name in mathbench_sets[_split]:
mathbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[
dict(
role='HUMAN',
prompt=single_choice_prompts[_name + '_with_reasoning'] if with_reasoning else single_choice_prompts[_name],
),
dict(role='BOT', prompt='{answer}')] if 'choice' in _name else cloze_prompts[_name],
),
),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=2048),
)
mathbench_eval_cfg = dict(
evaluator=dict(type=CircularEvaluator if 'choice' in _name and with_circular_eval else AccEvaluator),
pred_postprocessor=dict(type=first_option_postprocess, options='ABCD') if 'single_choice' in _name else dict(type=mathbench_postprocess, name=_name))
mathbench_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'./data/mathbench_v1/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=dict(
input_columns=['question'],
output_column='answer'
),
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
))
opencompass/configs/datasets/MathBench/mathbench_2024_gen_1dc21d.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
from copy import deepcopy
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer, PPLInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
with read_base():
from .mathbench_prompt import zero_shot_prompts, few_shot_prompts, mathbench_sets
# Max for this dataset is 4
num_shot = 4
# Generate reasoning path or not, only for single choice
with_reasoning = False
# Use circular evaluation or not
with_circular_eval = True
# Use PPL mode in single choice test or not
use_ppl_single_choice = False
assert 0 <= num_shot <= 4
if num_shot == 0:
prompts = zero_shot_prompts
else:
prompts = {name: p[- 2 * num_shot - 2:] for name, p in few_shot_prompts.items()}
mathbench_datasets = []
for _split in mathbench_sets:
for _name in mathbench_sets[_split]:
if 'single_choice' in _name:
if with_reasoning:
template_round = prompts[_name + '_with_reasoning']
else:
template_round = prompts[_name]
else:
template_round = prompts[_name]
if 'single_choice' in _name:
pred_postprocessor = dict(type=first_option_postprocess, options='ABCD')
else:
pred_postprocessor = dict(type=mathbench_postprocess, name=_name)
if 'single_choice' in _name and with_circular_eval:
evaluator = dict(type=CircularEvaluator)
else:
evaluator = dict(type=AccEvaluator)
# assemble the final config
mathbench_reader_cfg = dict(input_columns=['question'], output_column='answer')
if use_ppl_single_choice and 'single_choice' in _name and not with_reasoning:
template = {}
for answer in ['A', 'B', 'C', 'D']:
one_template_round = deepcopy(template_round)
one_template_round['round'][-1]['prompt'] = one_template_round['round'][-1]['prompt'].format(answer=answer)
template[answer] = dict(round=one_template_round)
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=template),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=PPLInferencer),
)
else:
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=dict(round=template_round)),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=2048),
)
mathbench_eval_cfg = dict(evaluator=evaluator, pred_postprocessor=pred_postprocessor)
mathbench_datasets.append(
dict(
abbr='mathbench-no_cot-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'data/mathbench_v1/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=mathbench_reader_cfg,
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
)
)
opencompass/configs/datasets/MathBench/mathbench_2024_gen_50a320.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
from copy import deepcopy
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer, PPLInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, math_postprocess_v2
from opencompass.utils.text_postprocessors import first_option_postprocess
with read_base():
from .mathbench_prompt import zero_shot_prompts, few_shot_prompts, mathbench_sets
# Max for this dataset is 4
num_shot = 0
# Generate reasoning path or not, only for single choice
with_reasoning = True
# Use circular evaluation or not
with_circular_eval = True
# Use PPL mode in single choice test or not
use_ppl_single_choice = False
assert 0 <= num_shot <= 4
if num_shot == 0:
prompts = zero_shot_prompts
else:
prompts = {name: p[- 2 * num_shot - 2:] for name, p in few_shot_prompts.items()}
mathbench_datasets = []
for _split in mathbench_sets:
for _name in mathbench_sets[_split]:
if 'single_choice' in _name:
if with_reasoning:
template_round = prompts[_name + '_with_reasoning']
else:
template_round = prompts[_name]
else:
template_round = prompts[_name]
if 'single_choice' in _name:
pred_postprocessor = dict(type=first_option_postprocess, options='ABCD')
else:
pred_postprocessor = dict(type=math_postprocess_v2)
if 'single_choice' in _name and with_circular_eval:
evaluator = dict(type=CircularEvaluator)
else:
evaluator = dict(type=AccEvaluator)
# assemble the final config
mathbench_reader_cfg = dict(input_columns=['question'], output_column='answer')
if use_ppl_single_choice and 'single_choice' in _name and not with_reasoning:
template = {}
for answer in ['A', 'B', 'C', 'D']:
one_template_round = deepcopy(template_round)
one_template_round['round'][-1]['prompt'] = one_template_round['round'][-1]['prompt'].format(answer=answer)
template[answer] = dict(round=one_template_round)
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=template),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=PPLInferencer),
)
else:
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=dict(round=template_round)),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=2048),
)
mathbench_eval_cfg = dict(evaluator=evaluator, pred_postprocessor=pred_postprocessor)
mathbench_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'data/mathbench_v1/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=mathbench_reader_cfg,
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
)
)
opencompass/configs/datasets/MathBench/mathbench_2024_gen_fc2a24.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
from copy import deepcopy
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer, PPLInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
with read_base():
from .mathbench_prompt import zero_shot_prompts, few_shot_prompts, mathbench_sets
# Max for this dataset is 4
num_shot = 4
# Generate reasoning path or not, only for single choice
with_reasoning = True
# Use circular evaluation or not
with_circular_eval = True
# Use PPL mode in single choice test or not
use_ppl_single_choice = False
assert 0 <= num_shot <= 4
if num_shot == 0:
prompts = zero_shot_prompts
else:
prompts = {name: p[- 2 * num_shot - 2:] for name, p in few_shot_prompts.items()}
mathbench_datasets = []
for _split in mathbench_sets:
for _name in mathbench_sets[_split]:
if 'single_choice' in _name:
if with_reasoning:
template_round = prompts[_name + '_with_reasoning']
else:
template_round = prompts[_name]
else:
template_round = prompts[_name]
if 'single_choice' in _name:
pred_postprocessor = dict(type=first_option_postprocess, options='ABCD')
else:
pred_postprocessor = dict(type=mathbench_postprocess, name=_name)
if 'single_choice' in _name and with_circular_eval:
evaluator = dict(type=CircularEvaluator)
else:
evaluator = dict(type=AccEvaluator)
# assemble the final config
mathbench_reader_cfg = dict(input_columns=['question'], output_column='answer')
if use_ppl_single_choice and 'single_choice' in _name and not with_reasoning:
template = {}
for answer in ['A', 'B', 'C', 'D']:
one_template_round = deepcopy(template_round)
one_template_round['round'][-1]['prompt'] = one_template_round['round'][-1]['prompt'].format(answer=answer)
template[answer] = dict(round=one_template_round)
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=template),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=PPLInferencer),
)
else:
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=dict(round=template_round)),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=2048),
)
mathbench_eval_cfg = dict(evaluator=evaluator, pred_postprocessor=pred_postprocessor)
mathbench_datasets.append(
dict(
abbr='mathbench-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'data/mathbench_v1/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=mathbench_reader_cfg,
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
)
)
opencompass/configs/datasets/MathBench/mathbench_2024_wocircular_gen_1dc21d.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
from copy import deepcopy
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer, PPLInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
with read_base():
from .mathbench_prompt import zero_shot_prompts, few_shot_prompts, mathbench_sets
# Max for this dataset is 4
num_shot = 4
# Generate reasoning path or not, only for single choice
with_reasoning = False
# Use circular evaluation or not
with_circular_eval = False
# Use PPL mode in single choice test or not
use_ppl_single_choice = False
assert 0 <= num_shot <= 4
if num_shot == 0:
prompts = zero_shot_prompts
else:
prompts = {name: p[- 2 * num_shot - 2:] for name, p in few_shot_prompts.items()}
mathbench_datasets = []
for _split in mathbench_sets:
for _name in mathbench_sets[_split]:
if 'single_choice' in _name:
if with_reasoning:
template_round = prompts[_name + '_with_reasoning']
else:
template_round = prompts[_name]
else:
template_round = prompts[_name]
if 'single_choice' in _name:
pred_postprocessor = dict(type=first_option_postprocess, options='ABCD')
else:
pred_postprocessor = dict(type=mathbench_postprocess, name=_name)
if 'single_choice' in _name and with_circular_eval:
evaluator = dict(type=CircularEvaluator)
else:
evaluator = dict(type=AccEvaluator)
# assemble the final config
mathbench_reader_cfg = dict(input_columns=['question'], output_column='answer')
if use_ppl_single_choice and 'single_choice' in _name and not with_reasoning:
template = {}
for answer in ['A', 'B', 'C', 'D']:
one_template_round = deepcopy(template_round)
one_template_round['round'][-1]['prompt'] = one_template_round['round'][-1]['prompt'].format(answer=answer)
template[answer] = dict(round=one_template_round)
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=template),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=PPLInferencer),
)
else:
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=dict(round=template_round)),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=2048),
)
mathbench_eval_cfg = dict(evaluator=evaluator, pred_postprocessor=pred_postprocessor)
mathbench_datasets.append(
dict(
abbr='mathbench-wocircular-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'data/mathbench_v1/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=mathbench_reader_cfg,
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
)
)
opencompass/configs/datasets/MathBench/mathbench_2024_wocircular_mixed_8eb12b.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
from copy import deepcopy
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer, PPLInferencer
from opencompass.openicl.icl_evaluator import CircularEvaluator, AccEvaluator
from opencompass.datasets import MathBenchDataset, mathbench_postprocess
from opencompass.utils.text_postprocessors import first_option_postprocess
with read_base():
from .mathbench_prompt import zero_shot_prompts, few_shot_prompts, mathbench_sets
# Max for this dataset is 4
num_shot = 4
# Generate reasoning path or not, only for single choice
with_reasoning = False
# Use circular evaluation or not
with_circular_eval = False
# Use PPL mode in single choice test or not
use_ppl_single_choice = True
assert 0 <= num_shot <= 4
if num_shot == 0:
prompts = zero_shot_prompts
else:
prompts = {name: p[- 2 * num_shot - 2:] for name, p in few_shot_prompts.items()}
mathbench_datasets = []
for _split in mathbench_sets:
for _name in mathbench_sets[_split]:
if 'single_choice' in _name:
if with_reasoning:
template_round = prompts[_name + '_with_reasoning']
else:
template_round = prompts[_name]
else:
template_round = prompts[_name]
if 'single_choice' in _name:
pred_postprocessor = dict(type=first_option_postprocess, options='ABCD')
else:
pred_postprocessor = dict(type=mathbench_postprocess, name=_name)
if 'single_choice' in _name and with_circular_eval:
evaluator = dict(type=CircularEvaluator)
else:
evaluator = dict(type=AccEvaluator)
# assemble the final config
mathbench_reader_cfg = dict(input_columns=['question'], output_column='answer')
if use_ppl_single_choice and 'single_choice' in _name and not with_reasoning:
template = {}
for answer in ['A', 'B', 'C', 'D']:
one_template_round = deepcopy(template_round)
one_template_round[-1]['prompt'] = one_template_round[-1]['prompt'].format(answer=answer)
template[answer] = dict(round=one_template_round)
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=template),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=PPLInferencer),
)
else:
mathbench_infer_cfg = dict(
prompt_template=dict(type=PromptTemplate, template=dict(round=template_round)),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer, max_out_len=2048),
)
mathbench_eval_cfg = dict(evaluator=evaluator, pred_postprocessor=pred_postprocessor)
mathbench_datasets.append(
dict(
abbr='mathbench-wocircular-' + _split + '-' + _name,
type=MathBenchDataset,
path=f'data/mathbench_v1/{_split}',
name=_name,
with_circular=with_circular_eval,
reader_cfg=mathbench_reader_cfg,
infer_cfg=mathbench_infer_cfg,
eval_cfg=mathbench_eval_cfg,
)
)
opencompass/configs/datasets/MathBench/mathbench_gen.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
with read_base():
from .mathbench_2024_gen_19e486 import mathbench_datasets # noqa: F401, F403
opencompass/configs/datasets/MathBench/mathbench_prompt.py 0 → 100644
View file @ be3dfa50
zero_shot_prompts = {
'single_choice_cn_with_reasoning': [
dict(role='HUMAN', prompt='问题: 以下是一道关于数学的单项选择题，请你一步一步推理，并在最后用“所以答案为选项X”给出答案，其中“X”为选项A，B，C，D中你认为正确的选项。下面是你要回答的问题\n{question}\n让我们一步一步思考：'),
],
'single_choice_cn': [
dict(role='HUMAN', prompt='问题: 以下是一道关于数学的单项选择题，请你直接回答正确答案的选项序号。\n下面是你要回答的题目：\n{question}\n所以答案是：'),
],
'single_choice_en_with_reasoning': [
dict(role='HUMAN', prompt='Question: Here is a multiple-choice question about mathematics. Please reason through it step by step, and at the end, provide your answer option with "Therefore, the correct answer is option X", Where "X" is the correct option you think from A，B，C，D. Here is the question you need to answer:\n{question}\nLet\'s think step by step:'),
],
'single_choice_en': [
dict(role='HUMAN', prompt='Question: Here is a multiple-choice question about mathematics. Please provide the correct answer option directly.\nHere is the question you need to answer:\n{question}\nThe answer is:'),
],
'cloze_en': [
dict(role='HUMAN', prompt='{question}\nPlease reason step by step, and put your final answer within \\boxed{}.'),
],
'cloze_cn': [
dict(role='HUMAN', prompt='{question}\n请一步一步推理，并在最后用\\boxed{}给出你的答案。'),
]
}
few_shot_prompts = {
'single_choice_cn': [
dict(role='HUMAN', prompt='问题: 已知i是虚数单位，z为复数，$2+\\frac{1}{i}=z(3+i)$，则在复平面内z对应的点位于____?\nA. 第一象限\nB. 第二象限\nC. 第三象限\nD. 第四象限'),
dict(role='BOT', prompt='回答: D'),
dict(role='HUMAN', prompt='问题: 将函数$y=\\tan(\\omega x-1)(\\omega>0)$的图像向左平移2个单位长度后,与函数$y=\\tan(\\omega x+3)$的图象重合,则的最小值等于____\nA. $2-\\frac{\\pi}{2}$\nB. 1\nC. $\\pi - 2$\nD. 2'),
dict(role='BOT', prompt='回答: D'),
dict(role='HUMAN', prompt='问题: 设$（1+2i）a+b=2i$，其中a,b为实数，则（ ）\nA. $a=1,b=-1$\nB. $a=1,b=1$\nC. $a=-1,b=1$\nD. $a=-1,b=-1$'),
dict(role='BOT', prompt='回答: A'),
dict(role='HUMAN', prompt='问题: 圆锥的底面半径为2，高为4.一个圆柱的下底面在圆锥的底面上，上底面的圆周在圆锥的侧面上，当圆柱侧面积为$4 \\pi$时，该圆柱的体积为____\nA. $\\pi$\nB. $2\\pi$\nC. $3\\pi$\nD. $4\\pi$'),
dict(role='BOT', prompt='回答: B'),
dict(role='HUMAN', prompt='问题: {question}'),
dict(role='BOT', prompt='回答: {answer}'),
],
'single_choice_cn_with_reasoning' : [
dict(role='HUMAN', prompt='问题: 已知i是虚数单位，z为复数，$2+\\frac{1}{i}=z(3+i)$，则在复平面内z对应的点位于____\nA. 第一象限\nB. 第二象限\nC. 第三象限\nD. 第四象限'),
dict(role='BOT', prompt='回答: 因为首先，我们将等式两边同时乘以$(3 + i)$的共轭复数$(3 - i)$，以便消去分母中的虚数部分：$z = \\frac{2 - i}{3 + i} \\cdot \\frac{3 - i}{3 - i}$，这样做的目的是利用复数乘法的性质，从而消去分母中的虚数部分。我们进行乘法运算：$z = \\frac{(2 - i)(3 - i)}{(3 + i)(3 - i)}$，$z = \\frac{6 - 2i - 3i + i^2}{9 - 3i + 3i - i^2}$，由于$i^2 = -1$，我们可以进一步简化：$z = \\frac{6 - 5i - 1}{9 + 1}$ $z = \\frac{{1}{2} - \\frac{1}{2}i$，在复平面上，这个点位于第四象限，因为它的实部是正的（$\\frac{1}{2}$），而虚部是负的（$-\\frac{1}{2}$）\n答案选项：D'),
dict(role='HUMAN', prompt='问题: 将函数$y=\\tan(\\omega x-1)(\\omega>0)$的图像向左平移2个单位长度后,与函数$y=\\tan(\\omega x+3)$的图象重合,则的最小值等于____\nA. $2-\\frac{\\pi}{2}$\nB. 1\nC. $\\pi - 2$\nD. 2'),
dict(role='BOT', prompt='回答: 将函数$y=\\tan(\\omega x-1)$（其中$\\omega>0$）的图像向左平移2个单位长度后，得到的新函数为$y=\\tan(\\omega(x+2)-1)$。为了使这个新函数的图像与$y=\\tan(\\omega x+3)$的图像重合，我们需要满足以下条件：\n$\\tan(\\omega(x+2)-1) = \\tan(\\omega x+3)$\n由于正切函数的周期性，我们知道$\\tan(\\theta) = \\tan(\\theta + k\\pi)$，其中$k$是任意整数。因此，我们可以写出：\n$\\omega(x+2)-1 = \\omega x+3 + k\\pi$\n解这个方程，我们得到：$\\omega x + 2\\omega - 1 = \\omega x + 3 + k\\pi$\n$2\\omega - 1 = 3 + k\\pi$\n$\\omega = 2 + \\frac{k\\pi}{2}$\n由于$\\omega>0$，我们需要找到满足条件的最小正整数$k$。当$k=0$时，我们得到$\\omega=2$，这是满足条件的最小正值。因此，最小值等于2。答案选项：D'),
dict(role='HUMAN', prompt='问题: 设$（1+2i）a+b=2i$，其中a,b为实数，则（ ）\nA. $a=1,b=-1$\nB. $a=1,b=1$\nC. $a=-1,b=1$\nD. $a=-1,b=-1$'),
dict(role='BOT', prompt='回答: 我们可以将这个方程写成：$a + 2ai + b = 2i$，然后将实部和虚部分开：\n$a + b = 0$\n$2a = 2$\n由于$a + b = 0$，我们可以得到$a = -b$。将这个值代入$2a = 2$，我们得到$2(-b) = 2$，因此$b = -1$。因此，$a = 1$。答案选项：A'),
dict(role='HUMAN', prompt='问题: 一直角三角形的两边长分别为3和4．则第三边的长为____\nA. 5\nB. $\\sqrt 7$\nC. $\\sqrt 5$\nD. 5或者$\\sqrt 7$'),
dict(role='BOT', prompt='回答: 我们可以使用勾股定理来解决这个问题。勾股定理告诉我们，如果一个三角形的两条边长分别为$a$和$b$，那么第三条边的长度为$c = \\sqrt{a^2 + b^2}$。在这个问题中，$a = 3$，$b = 4$，因此$c = \\sqrt{3^2 + 4^2} = \\sqrt{9 + 16} = \\sqrt{25} = 5$。答案选项：A'),
dict(role='HUMAN', prompt='问题: {question}'),
dict(role='BOT', prompt='回答: {answer}'),
],
'single_choice_en': [
dict(role='HUMAN', prompt='Question: Given that $i$ is the imaginary unit and $z$ is a complex number, $2 + \\frac{1}{i} = z(3 + i)$, the point corresponding to $z$ in the complex plane is located in the ______\nA. First quadrant\nB. Second quadrant\nC. Third quadrant\nD. Fourth quadrant'),
dict(role='BOT', prompt='Response: D'),
dict(role='HUMAN', prompt='Question: The graph of the function $y = \\tan(\\omega x - 1)$ is shifted 2 units to the left and coincides with the graph of the function $y = \\tan(\\omega x + 3)$. The minimum value of $\\omega$ is ______\nA. $2 - \\frac{\\pi}{2}$\nB. 1\nC. $\\pi - 2$\nD. 2'),
dict(role='BOT', prompt='Response: D'),
dict(role='HUMAN', prompt='Question: If $(1 + 2i)a + b = 2i$, where $a$ and $b$ are real numbers, then ______\nA. $a = 1, b = -1$\nB. $a = 1, b = 1$\nC. $a = -1, b = 1$\nD. $a = -1, b = -1$'),
dict(role='BOT', prompt='Response: A'),
dict(role='HUMAN', prompt='Question: The radius of a cone is 2 and its height is 4. A cylinder has its lower base on the base of the cone and its upper base on the lateral surface of the cone. When the lateral surface area of the cylinder is $4\\pi$, the volume of the cylinder is ______\nA. $\\pi$\nB. $2\\pi$\nC. $3\\pi$\nD. $4\\pi$'),
dict(role='BOT', prompt='Response: B'),
dict(role='HUMAN', prompt='Question: {question}'),
dict(role='BOT', prompt='Response: {answer}'),
],
'single_choice_en_with_reasoning': [
dict(role='HUMAN', prompt='Question: Given that $i$ is the imaginary unit and $z$ is a complex number, $2 + \\frac{1}{i} = z(3 + i)$, the point corresponding to $z$ in the complex plane is located in the ______\nA. First quadrant\nB. Second quadrant\nC. Third quadrant\nD. Fourth quadrant'),
dict(role='BOT', prompt='Response: First, we multiply both sides of the equation by the conjugate of $(3 + i)$: $z = \\frac{2 - i}{3 + i} \\cdot \\frac{3 - i}{3 - i}$. We perform the multiplication: $z = \\frac{(2 - i)(3 - i)}{(3 + i)(3 - i)}$, $z = \\frac{6 - 2i - 3i + i^2}{9 - 3i + 3i - i^2}$. Since $i^2 = -1$, we can simplify this further: $z = \\frac{6 - 5i - 1}{9 + 1}$ $z = \\frac{1}{2} - \\frac{1}{2}i$. In the complex plane, this point is located in the fourth quadrant, because its real part is positive ($\\frac{1}{2}$) and its imaginary part is negative ($-\\frac{1}{2}$)\nAnswer option: D'),
dict(role='HUMAN', prompt='Question: The graph of the function $y = \\tan(\\omega x - 1)$ is shifted 2 units to the left and coincides with the graph of the function $y = \\tan(\\omega x + 3)$. The minimum value of $\\omega$ is ______\nA. $2 - \\frac{\\pi}{2}$\nB. 1\nC. $\\pi - 2$\nD. 2'),
dict(role='BOT', prompt='Response: In order for the graph of this new function to coincide with the graph of $y = \\tan(\\omega x + 3)$, we need to satisfy the following condition: $\\tan(\\omega(x + 2) - 1) = \\tan(\\omega x + 3)$. Therefore, we can write: $\\omega(x + 2) - 1 = \\omega x + 3 + k\\pi$. Solving this equation, we get: $\\omega x + 2\\omega - 1 = \\omega x + 3 + k\\pi$. $2\\omega - 1 = 3 + k\\pi$. $\\omega = 2 + \\frac{k\\pi}{2}$. Since $\\omega > 0$, we need to find the smallest positive integer $k$ that satisfies the condition. When $k = 0$, we get $\\omega = 2$, which is the smallest positive value that satisfies the condition. Therefore, the minimum value is 2. Answer option: D'),
dict(role='HUMAN', prompt='Question: If $(1 + 2i)a + b = 2i$, where $a$ and $b$ are real numbers, then ______\nA. $a = 1, b = -1$\nB. $a = 1, b = 1$\nC. $a = -1, b = 1$\nD. $a = -1, b = -1$'),
dict(role='BOT', prompt='Response: We can write this equation as: $a + 2ai + b = 2i$, and then separate the real and imaginary parts: $a + b = 0$. $2a = 2$. Since $a + b = 0$, we can get $a = -b$. Substituting this value into $2a = 2$, we get $2(-b) = 2$, so $b = -1$. Therefore, $a = 1$. Answer option: A'),
dict(role='HUMAN', prompt='Question: The radius of a cone is 2 and its height is 4. A cylinder has its lower base on the base of the cone and its upper base on the lateral surface of the cone. When the lateral surface area of the cylinder is $4\\pi$, the volume of the cylinder is ______\nA. $\\pi$\nB. $2\\pi$\nC. $3\\pi$\nD. $4\\pi$'),
dict(role='BOT', prompt='Response: We can use the Pythagorean theorem to solve this problem. The Pythagorean theorem tells us that if the two sides of a triangle are $a$ and $b$, then the length of the third side is $c = \\sqrt{a^2 + b^2}$. In this problem, $a = 3$ and $b = 4$, so $c = \\sqrt{3^2 + 4^2} = \\sqrt{9 + 16} = \\sqrt{25} = 5$. Answer option: A'),
dict(role='HUMAN', prompt='Question: {question}'),
dict(role='BOT', prompt='Response: {answer}'),
],
'cloze_cn': [
dict(role='HUMAN', prompt='问题: 莎莎有五个玩具。在圣诞节，他从他的爸爸和妈妈那里各得到了两个玩具。现在他有多少个玩具？'),
dict(role='BOT', prompt='回答: 她有5个玩具。他从妈妈那里得到了2个，所以之后他有5 + 2 = 7个玩具。然后他从爸爸那里得到了2个，所以总共他有7 + 2 = 9个玩具。答案是 9'),
dict(role='HUMAN', prompt='问题: 服务器房间里有九台电脑。从周一到周四每天增加五台电脑。现在服务器房里有多少台电脑？'),
dict(role='BOT', prompt='回答: 从周一到周四有4天。每天增加5台电脑。这意味着总共增加了4 * 5 = 20台电脑。一开始有9台电脑，所以现在有9 + 20 = 29台电脑。答案是 29'),
dict(role='HUMAN', prompt='问题: 迈克尔有58个高尔夫球。星期二，他丢失了23个高尔夫球。星期三，他又丢失了2个。星期三结束时他还剩下多少个高尔夫球？'),
dict(role='BOT', prompt='回答: 迈克尔一开始有58个球。星期二他丢失了23个，所以之后他还剩下58 - 23 = 35个球。星期三他又丢失了2个，所以现在他还剩下35 - 2 = 33个球。答案是 33'),
dict(role='HUMAN', prompt='问题: 奥利弗有23美元。她用每个3美元的价格买了五个百吉饼。她还剩下多少钱？'),
dict(role='BOT', prompt='回答: 她以每个3美元的价格买了5个百吉饼。这意味着她在百吉饼上花费了5 * 3 = 15美元。她一开始有23美元，所以现在她还剩下23 - 15 = 8美元。答案是 8'),
dict(role='HUMAN', prompt='问题: {question}'),
dict(role='BOT', prompt='回答: {answer}'),
],
'cloze_en': [
dict(role='HUMAN', prompt='Question: Shawn has five toys. For Christmas, he got two toys each from his mom and dad. How many toys does he have now?'),
dict(role='BOT', prompt='Response: He has 5 toys. He got 2 from mom, so after that he has 5 + 2 = 7 toys. Then he got 2 more from dad, so in total he has 7 + 2 = 9 toys. The answer is 9.'),
dict(role='HUMAN', prompt='Question: There were nine computers in the server room. Five more computers were installed each day, from monday to thursday. How many computers are now in the server room?'),
dict(role='BOT', prompt='Response: There are 4 days from monday to thursday. 5 computers were added each day. That means in total 4 * 5 = 20 computers were added. There were 9 computers in the beginning, so now there are 9 + 20 = 29 computers. The answer is 29.'),
dict(role='HUMAN', prompt='Question: Michael had 58 golf balls. On tuesday, he lost 23 golf balls. On wednesday, he lost 2 more. How many golf balls did he have at the end of wednesday?'),
dict(role='BOT', prompt='Response: Michael initially had 58 balls. He lost 23 on Tuesday, so after that he has 58 - 23 = 35 balls. On Wednesday he lost 2 more so now he has 35 - 2 = 33 balls. The answer is 33.'),
dict(role='HUMAN', prompt='Question: Olivia has $23. She bought five bagels for $3 each. How much money does she have left?'),
dict(role='BOT', prompt='Response: She bought 5 bagels for $3 each. This means she spent 5 * $3 = $15 on the bagels. She had $23 in beginning, so now she has $23 - $15 = $8. The answer is 8.'),
dict(role='HUMAN', prompt='Question: {question}'),
dict(role='BOT', prompt='Response: {answer}'),
],
}
mathbench_sets = {
# Practice Part
'college': ['single_choice_cn', 'single_choice_en'],
'high': ['single_choice_cn', 'single_choice_en'],
'middle': ['single_choice_cn', 'single_choice_en'],
'primary': ['cloze_cn', 'cloze_en'],
'arithmetic': ['cloze_en'],
# Theory part
'college_knowledge': ['single_choice_cn', 'single_choice_en'],
'high_knowledge': ['single_choice_cn', 'single_choice_en'],
'middle_knowledge': ['single_choice_cn', 'single_choice_en'],
'primary_knowledge': ['single_choice_cn', 'single_choice_en'],
}
opencompass/configs/datasets/MedBench/medbench_gen.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
with read_base():
from .medbench_gen_d44f24 import medbench_datasets # noqa: F401, F403
opencompass/configs/datasets/MedBench/medbench_gen_0b4fff.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer
from opencompass.openicl.icl_evaluator import AccEvaluator
from opencompass.datasets import MedBenchDataset, MedBenchEvaluator, MedBenchEvaluator_Cloze, MedBenchEvaluator_CMeEE, MedBenchEvaluator_CMeIE, MedBenchEvaluator_CHIP_CDEE, MedBenchEvaluator_CHIP_CDN, MedBenchEvaluator_CHIP_CTC, MedBenchEvaluator_NLG, MedBenchEvaluator_TF, MedBenchEvaluator_DBMHG, MedBenchEvaluator_SMDoc, MedBenchEvaluator_IMCS_V2_MRG
from opencompass.utils.text_postprocessors import first_capital_postprocess
medbench_reader_cfg = dict(
input_columns=['problem_input'], output_column='label')
medbench_multiple_choices_sets = ['Med-Exam', 'DDx-basic', 'DDx-advanced', 'MedSafety'] # 选择题，用acc判断
medbench_qa_sets = ['MedHC', 'MedMC', 'MedDG', 'MedSpeQA', 'MedTreat', 'CMB-Clin'] # 开放式QA，有标答
medbench_cloze_sets = ['MedHG'] # 限定域QA，有标答
medbench_single_choice_sets = ['DrugCA'] # 正确与否判断，有标答
medbench_ie_sets = ['DBMHG', 'CMeEE', 'CMeIE', 'CHIP-CDEE', 'CHIP-CDN', 'CHIP-CTC', 'SMDoc', 'IMCS-V2-MRG'] # 判断识别的实体是否一致，用F1评价
medbench_datasets = []
for name in medbench_single_choice_sets + medbench_multiple_choices_sets:
medbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[dict(role='HUMAN', prompt='{problem_input}')])),
retriever=dict(type=ZeroRetriever
), # retriver 不起作用，以输入参数为准 (zero-shot / few-shot)
inferencer=dict(type=GenInferencer))
medbench_eval_cfg = dict(
evaluator=dict(type=MedBenchEvaluator), pred_role='BOT')
medbench_datasets.append(
dict(
type=MedBenchDataset,
path='./data/MedBench/' + name,
name=name,
abbr='medbench-' + name,
setting_name='zero-shot',
reader_cfg=medbench_reader_cfg,
infer_cfg=medbench_infer_cfg.copy(),
eval_cfg=medbench_eval_cfg.copy()))
for name in medbench_qa_sets:
medbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[dict(role='HUMAN', prompt='{problem_input}')])),
retriever=dict(type=ZeroRetriever
), # retriver 不起作用，以输入参数为准 (zero-shot / few-shot)
inferencer=dict(type=GenInferencer))
medbench_eval_cfg = dict(
evaluator=dict(type=MedBenchEvaluator_NLG), pred_role='BOT')
medbench_datasets.append(
dict(
type=MedBenchDataset,
path='./data/MedBench/' + name,
name=name,
abbr='medbench-' + name,
setting_name='zero-shot',
reader_cfg=medbench_reader_cfg,
infer_cfg=medbench_infer_cfg.copy(),
eval_cfg=medbench_eval_cfg.copy()))
for name in medbench_cloze_sets:
medbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[dict(role='HUMAN', prompt='{problem_input}')])),
retriever=dict(type=ZeroRetriever
), # retriver 不起作用，以输入参数为准 (zero-shot / few-shot)
inferencer=dict(type=GenInferencer))
medbench_eval_cfg = dict(
evaluator=dict(type=MedBenchEvaluator_Cloze), pred_role='BOT')
medbench_datasets.append(
dict(
type=MedBenchDataset,
path='./data/MedBench/' + name,
name=name,
abbr='medbench-' + name,
setting_name='zero-shot',
reader_cfg=medbench_reader_cfg,
infer_cfg=medbench_infer_cfg.copy(),
eval_cfg=medbench_eval_cfg.copy()))
for name in medbench_ie_sets:
medbench_infer_cfg = dict(
prompt_template=dict(
type=PromptTemplate,
template=dict(
round=[dict(role='HUMAN', prompt='{problem_input}')])),
retriever=dict(type=ZeroRetriever
), # retriver 不起作用，以输入参数为准 (zero-shot / few-shot)
inferencer=dict(type=GenInferencer))
medbench_eval_cfg = dict(
evaluator=dict(type=eval('MedBenchEvaluator_'+name.replace('-', '_'))), pred_role='BOT')
medbench_datasets.append(
dict(
type=MedBenchDataset,
path='./data/MedBench/' + name,
name=name,
abbr='medbench-' + name,
setting_name='zero-shot',
reader_cfg=medbench_reader_cfg,
infer_cfg=medbench_infer_cfg.copy(),
eval_cfg=medbench_eval_cfg.copy()))
del name, medbench_infer_cfg, medbench_eval_cfg
opencompass/configs/datasets/NPHardEval/NPHardEval_gen.py 0 → 100644
View file @ be3dfa50
from mmengine.config import read_base
with read_base():
from .NPHardEval_gen_22aac5 import NPHardEval_datasets # noqa: F401, F403
opencompass/configs/datasets/NPHardEval/NPHardEval_gen_22aac5.py 0 → 100644
View file @ be3dfa50
from opencompass.openicl.icl_prompt_template import PromptTemplate
from opencompass.openicl.icl_retriever import ZeroRetriever
from opencompass.openicl.icl_inferencer import GenInferencer
from opencompass.datasets.NPHardEval import (
HardGCPDataset, HardGCPEvaluator,
Hard_TSP_Dataset, Hard_TSP_Evaluator,
Hard_MSP_Dataset, Hard_MSP_Evaluator,
CMP_GCP_D_Dataset, CMP_GCP_D_Evaluator,
CMP_TSP_D_Dataset, CMP_TSP_D_Evaluator,
CMP_KSP_Dataset, CMP_KSP_Evaluator,
P_BSP_Dataset, P_BSP_Evaluator,
P_EDP_Dataset, P_EDP_Evaluator,
P_SPP_Dataset, P_SPP_Evaluator,
)
NPHardEval_tasks = [
['hard_GCP', 'GCP', HardGCPDataset, HardGCPEvaluator],
['hard_TSP', 'TSP', Hard_TSP_Dataset, Hard_TSP_Evaluator],
['hard_MSP', 'MSP', Hard_MSP_Dataset, Hard_MSP_Evaluator],
['cmp_GCP_D', 'GCP_Decision', CMP_GCP_D_Dataset, CMP_GCP_D_Evaluator],
['cmp_TSP_D', 'TSP_Decision', CMP_TSP_D_Dataset, CMP_TSP_D_Evaluator],
['cmp_KSP', 'KSP', CMP_KSP_Dataset, CMP_KSP_Evaluator],
['p_BSP', 'BSP', P_BSP_Dataset, P_BSP_Evaluator],
['p_EDP', 'EDP', P_EDP_Dataset, P_EDP_Evaluator],
['p_SPP', 'SPP', P_SPP_Dataset, P_SPP_Evaluator],
]
NPHardEval_datasets = []
for name, path_name, dataset, evaluator in NPHardEval_tasks:
NPHardEval_reader_cfg = dict(input_columns=['prompt', 'level'], output_column='q')
NPHardEval_infer_cfg = dict(
ice_template=dict(
type=PromptTemplate,
template=dict(
begin='</E>',
round=[
dict(role='HUMAN', prompt='</E>{prompt}'),
dict(role='BOT', prompt=''),
],
),
ice_token='</E>',
),
retriever=dict(type=ZeroRetriever),
inferencer=dict(type=GenInferencer),
)
NPHardEval_eval_cfg = dict(evaluator=dict(type=evaluator), pred_role='BOT')
NPHardEval_datasets.append(
dict(
type=dataset,
abbr=name,
path=f'./data/NPHardEval/{path_name}/',
reader_cfg=NPHardEval_reader_cfg,
infer_cfg=NPHardEval_infer_cfg,
eval_cfg=NPHardEval_eval_cfg,
)
)
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