refactor: optimize overlap removal logic in remove_overlaps_min_blocks function

mineru/utils/model_utils.py

@@ -209,22 +209,34 @@ def remove_overlaps_min_blocks(res_list):
     # 重叠block，小的不能直接删除，需要和大的那个合并成一个更大的。
     # 删除重叠blocks中较小的那些
     need_remove = []
-    for res1 in res_list:
-        for res2 in res_list:
-            if res1 != res2:
+    for i in range(len(res_list)):
+        # 如果当前元素已在需要移除列表中，则跳过
+        if res_list[i] in need_remove:
+            continue
+
+        for j in range(i + 1, len(res_list)):
+            # 如果比较对象已在需要移除列表中，则跳过
+            if res_list[j] in need_remove:
+                continue
+
             overlap_box = get_minbox_if_overlap_by_ratio(
-                    res1['bbox'], res2['bbox'], 0.8
+                res_list[i]['bbox'], res_list[j]['bbox'], 0.8
             )
+
             if overlap_box is not None:
-                    res_to_remove = next(
-                        (res for res in res_list if res['bbox'] == overlap_box),
-                        None,
-                    )
-                    if (
-                        res_to_remove is not None
-                        and res_to_remove not in need_remove
-                    ):
-                        large_res = res1 if res1 != res_to_remove else res2
+                res_to_remove = None
+                large_res = None
+
+                # 确定哪个是小块（要移除的）
+                if overlap_box == res_list[i]['bbox']:
+                    res_to_remove = res_list[i]
+                    large_res = res_list[j]
+                elif overlap_box == res_list[j]['bbox']:
+                    res_to_remove = res_list[j]
+                    large_res = res_list[i]
+
+                if res_to_remove is not None and res_to_remove not in need_remove:
+                    # 更新大块的边界为两者的并集
                     x1, y1, x2, y2 = large_res['bbox']
                     sx1, sy1, sx2, sy2 = res_to_remove['bbox']
                     x1 = min(x1, sx1)
@@ -234,7 +246,7 @@ def remove_overlaps_min_blocks(res_list):
                     large_res['bbox'] = [x1, y1, x2, y2]
                     need_remove.append(res_to_remove)
 
-    if len(need_remove) > 0:
+    # 从列表中移除标记的元素
     for res in need_remove:
         res_list.remove(res)