[Numpy-discussion] Geometrically defined masking arrays; how to optimize?

[Julian Taylor]

meshgrid also has the sparse keyword argument which archives the same.

On 12.02.2014 20:04, Chris Barker wrote:
> An extra note here:
> 
> One of the great things about numpy (as opposed, to say, MATLAB), is
> array broadcasting  Thus you generally don't need meshgrid -- why carry
> all that extra repetitive data around. So here's a version that uses
> broadcasting instead
> 
> 
> radius = 5
> 
> # Center at 0, 0
> 
> x = np.linspace(-5,5,11)
> 
> # y is re-shaped to be a column vector...
> y = np.linspace(-8,8,17).reshape((-1,1))
> 
> # when you add x and y you get the 2-d array for the results...
> 
> out_circle = (x * x + y * y < radius**2)
> 
> print out_circle
> 
> 
> 
> 
> On Wed, Feb 12, 2014 at 2:43 AM, Wolfgang Draxinger
> <Wolfgang.Draxinger at physik.uni-muenchen.de
> <mailto:Wolfgang.Draxinger at physik.uni-muenchen.de>> wrote:
> 
>     On Tue, 11 Feb 2014 22:16:46 +0100
>     Daπid <davidmenhur at gmail.com <mailto:davidmenhur at gmail.com>> wrote:
> 
>     > Here it is an example:
>     >
>     > import numpy as np
>     > import pylab as plt
>     >
>     > N = 100
>     > M = 200
>     > x, y = np.meshgrid(np.arange(N), np.arange(M))
>     >
>     > # Center at 40, 70, radius 20
>     > x -= 40
>     > y -= 70
>     > out_circle = (x * x + y * y < 20**2)
> 
>     Neat.
> 
>     > Note that I have avoided taking the costly square root of each
>     > element by just taking the square of the radius. It can also be
>     > generalised to ellipses or rectangles, if you need it.
> 
>     Oops, why did I take the "long" tour anyway?
> 
>     > Also, don't use 0 as a False value, and don't force it to be a 0.
>     > Instead, use "if not ring:"
> 
>     'ring' is not a boolean, but a numeric index and the test is if the
>     index is nonzero. I could have also written 'if ring > 0:' and swapped
>     the clauses. Yes I know that for a value 0 zero it will test as not,
>     but in this case I wanted to have it written down, that this tests for
>     a certain numerical value.
> 
> 
>     Thanks,
> 
>     Wolfgang
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> 
> 
> 
> -- 
> 
> Christopher Barker, Ph.D.
> Oceanographer
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