Fix https://github.com/pytorch/vision/issues/2354 (#2355)

torchvision/transforms/functional_tensor.py

@@ -288,22 +288,35 @@ def _blend(img1, img2, ratio):
 def _rgb2hsv(img):
     r, g, b = img.unbind(0)
 
-    maxc, _ = torch.max(img, dim=0)
-    minc, _ = torch.min(img, dim=0)
+    maxc = torch.max(img, dim=0).values
+    minc = torch.min(img, dim=0).values
+
+    # The algorithm erases S and H channel where `maxc = minc`. This avoids NaN
+    # from happening in the results, because
+    #   + S channel has division by `maxc`, which is zero only if `maxc = minc`
+    #   + H channel has division by `(maxc - minc)`.
+    #
+    # Instead of overwriting NaN afterwards, we just prevent it from occuring so
+    # we don't need to deal with it in case we save the NaN in a buffer in
+    # backprop, if it is ever supported, but it doesn't hurt to do so.
+    eqc = maxc == minc
 
     cr = maxc - minc
-    s = cr / maxc
-    rc = (maxc - r) / cr
-    gc = (maxc - g) / cr
-    bc = (maxc - b) / cr
+    # Since `eqc => cr = 0`, replacing denominator with 1 when `eqc` is fine.
+    s = cr / torch.where(eqc, maxc.new_ones(()), maxc)
+    # Note that `eqc => maxc = minc = r = g = b`. So the following calculation
+    # of `h` would reduce to `bc - gc + 2 + rc - bc + 4 + rc - bc = 6` so it
+    # would not matter what values `rc`, `gc`, and `bc` have here, and thus
+    # replacing denominator with 1 when `eqc` is fine.
+    cr_divisor = torch.where(eqc, maxc.new_ones(()), cr)
+    rc = (maxc - r) / cr_divisor
+    gc = (maxc - g) / cr_divisor
+    bc = (maxc - b) / cr_divisor
 
-    t = (maxc != minc)
-    s = t * s
     hr = (maxc == r) * (bc - gc)
     hg = ((maxc == g) & (maxc != r)) * (2.0 + rc - bc)
     hb = ((maxc != g) & (maxc != r)) * (4.0 + gc - rc)
     h = (hr + hg + hb)
-    h = t * h
     h = torch.fmod((h / 6.0 + 1.0), 1.0)
     return torch.stack((h, s, maxc))