Say I have an array like this:
x = [1, 2, 3]
[4, 5, 6]
[7, 8, 9]
And I want to delete both the ith row and column. So if i=1, I'd create (with 0-indexing):
[1, 3]
[7, 9]
Is there an easy way of doing this with a one-liner? I know I can call np.delete() twice, but that seems a little unclean.
It'd be exactly equivalent to np.delete(np.delete(x, idx, 0), idx, 1), where idx is the index of the row/column pair to delete - it'd just look cleaner.
This can be easily achieved by numpy's delete function. It would be:
arr = np.delete(arr, index, 0) # deletes the desired row
arr = np.delete(arr, index, 1) # deletes the desired column at index
The third argument is the axis.
In [196]: x = np.arange(1,10).reshape(3,3)
If you look at np.delete code, you'll see that it's python (not compiled) and takes different approaches depending on how the delete values are specified. One is to make a res array of right size, and copy two slices to it.
Another is to make a boolean mask. For example:
In [197]: mask = np.ones(x.shape[0], bool)
In [198]: mask[1] = 0
In [199]: mask
Out[199]: array([ True, False, True])
Since you are deleting the same row and column, use this indexing:
In [200]: x[mask,:][:,mask]
Out[200]:
array([[1, 3],
[7, 9]])
A 1d boolean mask like this can't be 'broadcasted' in the same ways a integer array can.
We can do the 2d advanced indexing with:
In [201]: idx = np.nonzero(mask)[0]
In [202]: idx
Out[202]: array([0, 2])
In [203]: np.ix_(idx,idx)
Out[203]:
(array([[0],
[2]]),
array([[0, 2]]))
In [204]: x[np.ix_(idx,idx)]
Out[204]:
array([[1, 3],
[7, 9]])
Actually ix_ can work directly from the boolean array(s):
In [207]: np.ix_(mask,mask)
Out[207]:
(array([[0],
[2]]),
array([[0, 2]]))
This isn't a one-liner, but it probably is faster than the double delete, since it strips off all the extra baggage that the more general function requires.
np.delete(np.delete(x, idx, 0), idx, 1)is an easy solution, I just feel like there should be a cleaner way.