Re: [Numpy-discussion] help on fast slicing on a grid

On Thu, Jan 29, 2009 at 00:09, frank wang wrote:

Here is the for loop that I am think about. Also, I do not know whether the where commands can handle the complicated logic. The where command basically find the data in the square around the point cnstl[j].

cnstl is a 2D array from your previous description.

Let the data array is qam with size N

I don't see qam anywhere. Did you mean X?

Out = X error = X

Don't you want something like zeros_like(X) for these?

for i in arange(N): for j in arange(L): aa = np.where((real(X)real(cnstl[j])-1) & (imag(X)imag(cnstl[j]-1)) Out[aa]=cnstl[j] error[aa]=abs(X)**2 - abs(cnstl[j])**2

I'm still confused. Can you show me a complete, working script with possibly fake data? -- 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