Hi,
Given a (possibly masked) 2d array x, is there a fast(er) way in Numpy to obtain
the same result as the following few lines?
d = 1 # neighbourhood 'radius'
Nrow = x.shape[0]
Ncol = x.shape[1]
y = array([[x[i-d:i+d+1,j-d:j+d+1].ravel() for j in range(d,Ncol-d)]
A Divendres 23 Febrer 2007 17:38, [EMAIL PROTECTED] escrigué:
> Hi,
>
> Given a (possibly masked) 2d array x, is there a fast(er) way in Numpy to
> obtain the same result as the following few lines?
>
> d = 1 # neighbourhood 'radius'
> Nrow = x.shape[0]
> Ncol = x.s
On Fri, 2007-02-23 at 17:38 +0100, [EMAIL PROTECTED] wrote:
> Hi,
>
> Given a (possibly masked) 2d array x, is there a fast(er) way in Numpy to
> obtain
> the same result as the following few lines?
>
> d = 1 # neighbourhood 'radius'
> Nrow = x.shape[0]
> Ncol =
Scipy's ndimage module has a function that takes a generic callback
and calls it with the values of each neighborhood (of a given size,
and optionally with a particular "mask" footprint) centered on each
array element. That function handles boundary conditions, etc nicely.
Unfortunately, I'm
On Friday 23 February 2007 14:53:05 Zachary Pincus wrote:
> Scipy's ndimage module has a function that takes a generic callback
> and calls it with the values of each neighborhood (of a given size,
> and optionally with a particular "mask" footprint) centered on each
> array element. That function
Thanks for all the useful comments. Some feedback about the improved version of
my code snippet. For a 40x40 matrix and d=1 the new version is 44 times faster,
and for d=2 it's 27 times faster. For my astronomical images (typical
2000x2000)
the new version saves my day.
# improved version
d = 1
N
