CODEWITHSUNDEEP
pythonmatlab
MasterWizard
MasterWizard

Reputation: 877

Converting this operation from matlab to python

I have this line in some matlab script that Im trying to convert to python. So, m=20, and n=20. The dimensions of I_true equals [400,1].

I want to convert following Matlab code:

A=zeros((2*m*n),(2*m*n)+2);
A(1:m*n,(2*m*n)+1)=-I_true(:);

Am I converting it right?

Converted code in Python:

for i in range(0,m*n):
    for j in range((2*m*n)+1):
        A[i][j] = I_true[i]

Upvotes: 0

Views: 34

Answers (2)

unutbu
unutbu

Reputation: 880887

Let's look at a small example, with n = 2, m = 2: In Octave (and presumably Matlab): 

octave:50> m = 2; n = 2;
octave:51> I_true = [1;2;3;4];
octave:52> A = zeros((2*m*n),(2*m*n)+2);
octave:53> A(1:m*n,(2*m*n)+1)=-I_true(:)
A =

   0   0   0   0   0   0   0   0  -1   0
   0   0   0   0   0   0   0   0  -2   0
   0   0   0   0   0   0   0   0  -3   0
   0   0   0   0   0   0   0   0  -4   0
   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0

The equivalent in Python (with n = 20, m = 20) would be

import numpy as np
n, m = 20, 20
I_true = np.arange(1, n*m+1)  # just as an example
A = np.zeros((2*m*n, 2*(n*m+1)), dtype=I.dtype)
A[:m*n, 2*m*n] = -I_true

The reason why the last line uses A[:m*n, 2*m*n] and not A[1:m*n, (2*m*n)+1] is because Python uses 0-based indexing whereas Matlab uses 1-based indexing.

Upvotes: 1

sheshkovsky
sheshkovsky

Reputation: 1340

Check this so question as well.

You can define a matrix with 2*m*n rows and 2*m*n+2 columns in python like this:

m = 20
n = 20
a = [[0 for i in range(2*m*n)] for j in range((2*m*n)+2)]

Now you have your matrix you can assign values to its elements using different ways. One example would be using for loops to assign values from another matrix with same size:

for i in range(2*m*n):
    for j in range((2*m*n)+2):
        a[i][j] = I_true[i][j]

I hope it helps.

Upvotes: 0

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