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

frank wang f.yw@hotmail....
Thu Jan 29 00:09:01 CST 2009


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].
 
Let the data array is qam with size N
 
Out = X
error = X
for i in arange(N):
    for j in arange(L):
         aa = np.where((real(X)<real(cnstl[j])+1) & (real(X)>real(cnstl[j])-1) & (imag(X)<imag(cnstl[j])+1) & (imag(X)>imag(cnstl[j]-1))         Out[aa]=cnstl[j]
         error[aa]=abs(X)**2 - abs(cnstl[j])**2
 
 
Thanks
 
Frank
> Date: Wed, 28 Jan 2009 23:57:16 -0600> From: robert.kern@gmail.com> To: numpy-discussion@scipy.org> Subject: Re: [Numpy-discussion] help on fast slicing on a grid> > On Wed, Jan 28, 2009 at 23:52, frank wang <f.yw@hotmail.com> wrote:> >> > Hi,> >> > I have to buidl a grid with 256 point by the command:> > a = arange(-15,16,2)> > L = len(a)> > cnstl = a.reshape(L,1)+1j*a> >> > My problem is that I have a big data array that contains the data round the> > points in cnstl. I want to slice the point to the closest cnstl point and> > also compute the error. The condition is in the middle of the two point in x> > and y axis. I can do it in a for loop. Since Python and numpy have a lot of> > magic, I want to find an efficient way to do. This problem arise from QAM> > 256 modulation.> > Can you show us the for loop? I'm not really sure what you want to compute.> > -- > 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> _______________________________________________> Numpy-discussion mailing list> Numpy-discussion@scipy.org> http://projects.scipy.org/mailman/listinfo/numpy-discussion
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