Thanks, Jerome!

I should have waited a bit longer to respond. I definitely want to encourage 
others to jump in with help when they know the answers to people's problems.



> On Feb 16, 2022, at 9:28 AM, Jerome Raim <jerome.r...@gmail.com> wrote:
> 
> Hi Tony,
> See here:
> https://openimageio.readthedocs.io/en/latest/pythonbindings.html#image-transformations-and-data-movement
>  
> <https://openimageio.readthedocs.io/en/latest/pythonbindings.html#image-transformations-and-data-movement>
> 
> So instead of
> new_img = ImageBufAlgo.channels(img, channel_ids)
> 
> You need to add the newchannelnames. I think this should do it:
> new_img = ImageBufAlgo.channels(img, channel_ids, ('R','G','B'))
> 
> 
> 
> On Wed, Feb 16, 2022 at 12:22 PM Tony DUGARD <tony.dug...@gmail.com 
> <mailto:tony.dug...@gmail.com>> wrote:
> Hi,
> 
> I'm currently writing a tool for splitting vray rendered exr into separate 
> channels :
> For exemple, I have 6 channels :
> Channel 1 : "R"
> Channel 2 : "G"
> Channel 3 :  "B"
> Channel 4 : 'VRaySpecular.R'
> Channel 5 : "VRaySpecular.G"
> Channel 6 : "VRaySpecular.B"
> 
> I've successful grouped the channel into a dict :
> channel_dict = {"" : (1, 2, 3), "VRaySpecular": (4, 5, 6)}
> 
> With the following code :
> img = ImageBuf(path)
> for key, channel_ids in channel_dict.tems():
>     new_img = ImageBufAlgo.channels(img, channel_ids)
>     filename, filetype = os.path.splitext(path)
>     out_path = filename + "." + key + filetype
>     new_img.write(out_path)
> 
> The problem i got is that the channel isn't shuffled into the new_img R, G, B 
> channel but instead stay in the VRaySpecular.R, VRaySpecular.G, VRaySpecular.B
> 
> I don't know how to shuffle the original channel to the RGB channel.
> 
> Thanks.
> Tony
> 
> 
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--
Larry Gritz
l...@larrygritz.com




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