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.
Now to remove the first element from the array create a Boolean array with length same as the array. In at all the elements are True except for the first element. Then pass this Boolean array as index to the given array. This will give an array with first element removed.
Now to remove the last element from the array create a Boolean array with length same as the array. All the elements in this Boolean array are True except for the last element. Then pass this Boolean array as index to the original NumPy Array. This will give an array with last element removed.
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.
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.
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]])
As you can see, the first row, first column, last row and last column have been removed from the source matrix H
and the remainder has been placed in the output matrix Hsub
.
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