Re: [matplotlib-devel] Quantization of normalized float to uint8

[Christoph G. <cg...@uc...>]

On 9/19/2011 2:23 AM, Christoph Gohlke wrote:
>
>
> On 9/18/2011 2:30 PM, Eric Firing wrote:
>> On 09/18/2011 09:30 AM, Christoph Gohlke wrote:
>>> Hello,
>>>
>>> matplotlib uses int(x*255) or np.array(x*255, np.uint8) to quantize
>>> normalized floating point numbers x in the range [0.0 to 1.0] to
>>> integers in the range [0 to 255]. This way only 1.0 is mapped to 255,
>>> not for example 0.999. Is this really intended or would not the largest
>>> floating point number below 256.0 be a better scale factor than 255? The
>>> exact factor depends on the floating point precision (~255.999992 for
>>> np.float32, ~255.93 for np.float16).
>>>
>>> Christoph
>>
>> Christoph,
>>
>> It's a reasonable question; but do you have use cases in mind where it
>> actually makes a difference?
>>
>> The simple scaling with truncation is used in many places, both in the
>> python and the c++ code.
>>
>> Eric
>>
>
> Hi Eric,
>
> visually it will be hardly noticeable in most cases. However, I'd expect
> the histogram of normalized intensity data to be the same as the
> histogram of a linear grayscale image of that data (neglecting gamma
> correction, image scaling/interpolation for now). Consider this code for
> example:
>
> import numpy as np
> a = np.random.rand(1024*1024)
> a[0], a[-1] = 0.0, 1.0
> h0 = np.histogram(a, bins=256, range=(0, 1))[0]
> h1 = np.bincount(np.uint8(a * 255))
> h2 = np.bincount(np.uint8(a * 255.9999999999999))
> print (h0 - h1)
> print (h0 - h2)
>
> Christoph
>


To make this work with any float type one could use:

np.uint8(a * np.nextafter(a.dtype.type(256), a.dtype.type(0)))

Christoph