I'd like to know how to remove the first and last rows and columns from a 2D array in numpy. For example, say we have a (N+1) x (N+1) matrix called H then in MATLAB/Octave, the code I'd use would be: 

Hsub = H(2:N,2:N);

What's the equivalent code in Numpy? I thought that np.reshape might do what I want but I'm not sure how to get it to remove just the target rows as I think if I reshape to a (N-1) x (N-1) matrix, it'll remove the last two rows and columns. 

Answer: 

How about this? 

Hsub = H[1:-1, 1:-1]

The 1:-1 range means that we access elements from the second index, or 1, and we go up to the second last index, as indicated by the -1 for a dimension. We do this for both dimensions independently. When you do this independently for both dimensions, the result is the intersection of how you're accessing each dimension, which is essentially chopping off the first row, first column, last row and last column. 

Remember, the ending index is exclusive, so if we did 0:3 for example, we only get the first three elements of a dimension, not four. 

Also, negative indices mean that we access the array from the end. -1 is the last value to access in a particular dimension, but because of the exclusivity, we are getting up to the second last element, not the last element. Essentially, this is the same as doing: 

Hsub = H[1:H.shape[0]-1, 1:H.shape[1]-1]

... but using negative indices is much more elegant. You also don't have to use the number of rows and columns to extract out what you need. The above syntax is dimension agnostic. However, you need to make sure that the matrix is at least 3 x 3, or you'll get an error. 

How about this?

Hsub = H[1:-1, 1:-1]

The 1:-1 range means that we access elements from the second index, or 1, and we go up to the second last index, as indicated by the -1 for a dimension. We do this for both dimensions independently. When you do this independently for both dimensions, the result is the intersection of how you're accessing each dimension, which is essentially chopping off the first row, first column, last row and last column.

Remember, the ending index is exclusive, so if we did 0:3 for example, we only get the first three elements of a dimension, not four.

Also, negative indices mean that we access the array from the end. -1 is the last value to access in a particular dimension, but because of the exclusivity, we are getting up to the second last element, not the last element. Essentially, this is the same as doing:

Hsub = H[1:H.shape[0]-1, 1:H.shape[1]-1]

... but using negative indices is much more elegant. You also don't have to use the number of rows and columns to extract out what you need. The above syntax is dimension agnostic. However, you need to make sure that the matrix is at least 3 x 3, or you'll get an error.

Small bonus 

In MATLAB / Octave, you can achieve the same thing without using the dimensions by: 

Hsub = H(2:end-1, 2:end-1);

The end keyword with regards to indexing means to get the last element for a particular dimension. 

Example use 

Here's an example (using IPython): 

In [1]: import numpy as np

In [2]: H = np.meshgrid(np.arange(5), np.arange(5))[0]

In [3]: H
Out[3]: 
array([[0, 1, 2, 3, 4],
       [0, 1, 2, 3, 4],
       [0, 1, 2, 3, 4],
       [0, 1, 2, 3, 4],
       [0, 1, 2, 3, 4]])

In [4]: Hsub = H[1:-1,1:-1]

In [5]: Hsub
Out[5]: 
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])