The following octave code shows a sample 3D matrix using Octave/Matlab
octave:1> A=zeros(3,3,3);
octave:2>
octave:2> A(:,:,1)= [[1 2 3];[4 5 6];[7
MATLAB and Python index differently. To investigate this, lets create a linear array of number 1 to 8 and then reshape the result to be a 2-by-2-by-2 matrix in each language:
MATLAB:
M_flat = 1:8
M = reshape(M_flat, [2,2,2])
which returns
M =
ans(:,:,1) =
1 3
2 4
ans(:,:,2) =
5 7
6 8
Python:
import numpy as np
P_flat = np.array(range(1,9))
P = np.reshape(P, [2,2,2])
which returns
array([[[1, 2],
[3, 4]],
[[5, 6],
[7, 8]]])
The first thing you should notice is that the first two dimensions have switched. This is because MATLAB uses column-major indexing which means we count down the columns first whereas Python use row-major indexing and hence it counts across the rows first.
Now let's try indexing them. So let's try slicing along the different dimensions. In MATLAB, I know to get a slice out of the third dimension I can do
M(:,:,1)
ans =
1 3
2 4
Now let's try the same in Python
P[:,:,0]
array([[1, 3],
[5, 7]])
So that's completely different. To get the MATLAB 'equivalent' we need to go
P[0,:,:]
array([[1, 2],
[3, 4]])
Now this returns the transpose of the MATLAB version which is to be expected due the the row-major vs column-major difference.
So what does this mean for indexing? It looks like Python puts the major index at the end which is the reverse of MALTAB.
Let's say I index as follows in MATLAB
M(1,2,2)
ans =
7
now to get the 7 from Python we should go
P(1,1,0)
which is the MATLAB syntax reversed. Note that is is reversed because we created the Python matrix with a row-major ordering in mind. If you create it as you did in your code you would have to swap the last 2 indices so rather create the matrix correctly in the first place as Ander has suggested in the comments.