Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Fix F_t rgb2hsv #2355

Merged
merged 1 commit into from
Jun 29, 2020
Merged
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Jump to
Jump to file
Failed to load files.
Diff view
Diff view
31 changes: 22 additions & 9 deletions torchvision/transforms/functional_tensor.py
Original file line number Diff line number Diff line change
Expand Up @@ -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))

Expand Down