On Wed, Apr 8, 2009 at 20:39, Ian Mallett <geometr...@gmail.com> wrote:
> Hello,
>
> I want to make an array of size sqrt(n) by sqrt(n) by 3, filled with special
> values.
>
> The values range from 0.0 to 3.0, starting with 0.0 at one corner and ending
> at 3.0 in the opposite, increasing going row by row. The value is to be
> encoded in each color. Because this is somewhat abstract, here's a small
> example (n=25), generated using the attached code (it also multiplies the
> number by 255 to obtain a RGB color and not messy floats) to show the
> concept. The real version should be done by NumPy. This is where I need
> help; I have no idea how to even approach the problem.
>
> [[ 0, 0, 0],[ 32, 0, 0],[ 64, 0, 0],[ 96, 0, 0],[128, 0, 0],
> [159, 0, 0],[191, 0, 0],[223, 0, 0],[255, 0, 0],[255, 32, 0],
> [255, 64, 0],[255, 96, 0],[255,128, 0],[255,159, 0],[255,191, 0],
> [255,223, 0],[255,255, 0],[255,255, 32],[255,255, 64],[255,255, 96],
> [255,255,128],[255,255,159],[255,255,191],[255,255,223],[255,255,255]]
>
> Arrays like this need to be generated quite quickly, so the per-pixel method
> I presented will not work. How should I do it with NumPy?
In [1]: from numpy import *
In [2]: sqrtn = 5
In [3]: n = sqrtn**2
In [4]: x = linspace(0.0, 3.0, n)
In [5]: y = column_stack([x, x-1, x-2]).clip(0, 1).reshape([sqrtn, sqrtn, 3])
In [6]: z = (y * 255).round().astype(uint8)
In [7]: z
Out[7]:
array([[[ 0, 0, 0],
[ 32, 0, 0],
[ 64, 0, 0],
[ 96, 0, 0],
[128, 0, 0]],
[[159, 0, 0],
[191, 0, 0],
[223, 0, 0],
[255, 0, 0],
[255, 32, 0]],
[[255, 64, 0],
[255, 96, 0],
[255, 128, 0],
[255, 159, 0],
[255, 191, 0]],
[[255, 223, 0],
[255, 255, 0],
[255, 255, 32],
[255, 255, 64],
[255, 255, 96]],
[[255, 255, 128],
[255, 255, 159],
[255, 255, 191],
[255, 255, 223],
[255, 255, 255]]], dtype=uint8)
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it had an underlying truth."
-- Umberto Eco
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