无法将MATLAB转换为repmat和对称性的Python代码

Question

MATLAB code:

n = 2048;
d = 1;
order = 2048;
nn = [-(n/2):(n/2-1)]';
h = zeros(size(nn),'single');
h(n/2+1) = 1 / 4;
odd = mod(nn,2) == 1;
h(odd) = -1 ./ (pi * nn(odd)).^2;
f_kernel = abs(fft(h))*2;
filt = f_kernel(1:order/2+1)';
w = 2*pi*(0:size(filt,2)-1)/order;
filt(w>pi*d) = 0;                     
filt = [filt , filt(end-1:-1:2)]; 
filt = repmat(filt',[1 1024]);

Python Code :

import numpy as np
import numpy.matlib
from numpy.matlib import repmat
d = 1
filt_length = 2048
nn = np.linspace(-1024,1023,2048)
nn = np.transpose(nn)
h = np.zeros((2048))
h[1024] = 0.25
odd = (nn%2)
for i in range(0,2048) :
    if odd[i] == 1 :
    h[i] = -1/((np.pi*nn[i])**2)       

f_kernel = abs(fft(h))*2
filt = np.transpose(f_kernel[0:1024])
w = (np.pi)*np.linspace(0,1,1025)

However I have been unable to convert the last 3 lines of the MATLAB code to Python. Any suggestions? The second last step of MATLAB code creates a ramp filter of size 2048 (goes from 0 to 1 in steps of 1024 and 1 to 0 in another 1024 steps). The last repmat makes the size of filt as (2048,1024).

Answer 1

If I were on my computer, I'd start Octave and IPython sessions, and start replicating the code line by line. I'd use smaller dimensions to easily watch the results. I'd pay special attention to shapes. Since my Matlab is rusty it is easier to do it this way than in my head. And more reliable.

np.arange is easier to use than linspace . transpose is unnecessary with 1d arrays. np.repeat or np.tile will probably perform repmat's job. I don't recall exactly what repmat does. Initially I would copy loops, but I'd try replace them when code is working. fft may require scipy .