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.