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]
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.
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.
