On Thu, 16 Oct 2008 9:34:48 am Frédéric wrote: > Ok, I see. Is there any gain (speed ?) to do in-place transform, instead of > doing : > > rgbColours = myTransform.doTransform(rgbColours)
Well, of course. Don't know how big it is, but it is present.
Generally the doTransform() method calls internally the lcms cmsDoTransform()
function, which *requires* an input and an output buffer, however both may be
the same (for an in-place transform).
The doTransform() method is implemented in a way that it allocates a buffer of
adequate size, shape and type (using a numpy array) to hold the
transformation results if you do not pass in an outputBuffer. So you will
definitely have the overhead of the object creation (numpy.zeros() is
called).
If you do not want to perform an in-place transformation, but still don't want
it to slow down, you can also create a suitable buffer yourself and use that
one. To modify the in-place example to use a self created buffer just go as
this:
destinationColours = numpy.zeros(rgbColours.shape,
dtype=numpy.uint8)
myTransform.doTransform(rgbColours, destinationBuffer=destinationColours)
This way they should be equally fast (without verification, but I doubt that
cmsDoTransform() differs between two buffer and in-place transformations).
Maybe I should package my stuff for others to give it a try.
Guy
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Guy K. Kloss
Institute of Information and Mathematical Sciences
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