This is probably much slower than the point (or eval) operation, but it works.

> >> image = Image.new('I', (10, 10))
> >> img = Image.eval(image, math.sqrt)

> >> for x in xrange(img.size[0]):
...     for y in xrange(img.size[1]):
...             img.putpixel((x,y),math.sqrt(image.getpixel((x,y))))
...

BTW I'm wondering what's the reason for this special form lambda in the point
operator

Hope that helps.

-- 
regards
   Bartek Wilczyñski


Quoting wmcclain@salamander.com:

>  On 13 Jul 2003 20:10:17 -0400
>  "Alan Grosskurth" <alan.grosskurth@[...].ca> wrote:
> 
>  > That works fine for 8-bit images, but I'm
>  > still have trouble with mode 'I' images (32-bit integer):
>  >
>  >     image = Image.new('I', (10, 10))
>  >     out = Image.eval(image, math.sqrt)
> 
>  This gets beyond my knowledge. That code calls point(), and according to
>  the docs:
> 
>      If the image has mode "I" (integer) or "F" (floating point), you
>      must use a function, and it must have the following format: argument
>      * scale+ offset
> 
>      Example: Map floating point images
> 
>          out = im.point(lambda i: i * 1.2 + 10)
> 
>      You can leave out either the scale or the offset.
> 
>  Using the example lambda works in your eval(), but I don't see how to
>  get in a sqrt(). Perhaps someone with better Python knowledge can help?
> 
>  -Bill
>  --
>  Sattre Press                                The King in Yellow
>  http://sattre-press.com/                 by Robert W. Chambers
>  info@[...].com             http://kiy.sattre-press.com/
>  



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