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Replies: 11 comments
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The easiest way to do that is to use the Note that to get the ProPhoto RGB colourspace class instance, you don't have to do what you did with your first line of code. We do that in the
Thus to convert from ProPhoto RGB to sRGB you would just do that: colour.RGB_to_RGB(
my_array,
colour.RGB_COLOURSPACES['ProPhoto RGB'],
colour.RGB_COLOURSPACES['sRGB']) Cheers, Thomas |
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Hi @KelSolaar , |
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Hi @KelSolaar , Best, |
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It is entirely possible that ProPhoto RGB colours are out of sRGB Gamut, the former is much larger than the latter. |
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Hi @KelSolaar Best, |
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Unfortunately no, but I have started to look at that seriously here: https://github.com/colour-science/colour/tree/feature/gamut_mapping, and related issues are #154 and #311. |
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Something about that plot is fishy; the chromaticities shouldn’t be clustered along the edge of the locus. Something seems odd there? |
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@sobotka , as KelSolaar mentioned earlier, they are investigating about the gamut mapping. |
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It’s unrelated to gamut mapping. The chromaticities in the source to destination will remain in relative positions. The issue is the collapsing of the chromaticities along that right edge of the spectral locus. Is it possible that the ProPhoto image was clipped to ProPhoto? |
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Hi @sobotka, I think this process may have some effects on color chromaticities. Best |
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I think it may be related to the use of unlinearised data. |
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The easiest way to do that is to use the
colour.RGB_to_RGB
definition, it will compute a transformation matrix with chromatic adaptation if needed (and required by the user).Note that to get the ProPhoto RGB colourspace class instance, you don't have to do what you did with your first line of code. We do that in the
colour.plotting
sub-package because the filtering definitions accept regular expressions but in that case you colour access it this way directly:colour.RGB_COLOURSPACES['ProPhoto RGB']
colour.models.PROPHOTO_RGB_COLOURSPACE
Thus to convert from ProPhoto RGB to sRGB you would just do that: