I did mean to subtract -lo. I used y because the initial definition used y as the lowest point. For your second comment, I don't get the difference. you still have 256 separate gradations.
On Thu, Feb 9, 2012 at 5:48 PM, Randolph Bentson <bent...@holmsjoen.com>wrote: > On Thu, Feb 09, 2012 at 12:40:49PM -0500, Chris Mitchell wrote: > > For a linear equation, you would want do use this: > > > > outJ = outI.point(lambda i:(i-y)*x) > > I think this should be outJ = outI.point(lambda i:(i-lo)*x), but the > point conversion method doesn't deal with these lambda definitions. > My definition of y should have been "y = -lo*x" which brings my > code to an algebraic equivalent to what I think Chris intends. > > This point conversion is only appropriate for use of the show > method of the new image. Although outJ.show() creates a properly > formed Netpbm PGM "rawbits" image, if outJ is saved, the resulting > grayscale png file is black because it's a grayscale image which > has been scaled down a lot. > > The single statement > outI.point(lambda i:i*x+y).show() > avoids the temptation to save the rescaled image. :-) > > ----------------------------------------------------- > > Although this design gives a non-white image, it is flawed in that > the displayed image has greater contrast than the original. This > is because the image under discussion doesn't use the full 16 bit > "I" mode range of 0 to 65535. My code maps a lesser source range > to 0 to 255 instead of 32 to 236 which would preserve the contrast. > > So I propose a revised solution > scale = 256.0/2**16 > outI.point(lambda i:i*scale+0).show() > > This relies on the "I" mode image having 16 bit pixels. I couldn't > find a way to determine the source pixel size in PIL. The description > of "I" mode in pil-handbook.pdf says it's 32 bit, which suggests > other grayscale files may use other pixel sizes. > > -- > Randolph Bentson > bent...@holmsjoen.com >
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