Numpy K-th Diagonal Indices

I'd like to do arithmetics with k-th diagonal of a numpy.array. I need those indices. For example, something like: >>> a = numpy.eye(2) >>> a[numpy.diag_indices(a

Solution 1:

A bit late, but this version also works for k = 0 (and does not alter the arrays, so does not need to make a copy).

defkth_diag_indices(a, k):
    rows, cols = np.diag_indices_from(a)
    if k < 0:
        return rows[-k:], cols[:k]
    elif k > 0:
        return rows[:-k], cols[k:]
    else:return rows, cols

Solution 2:

Here's a way:

1. Create index value arrays.
2. Get the daigonal index values you want.
3. Thats it! :)

Like this:

>>>import numpy as np>>>rows, cols = np.indices((3,3))>>>row_vals = np.diag(rows, k=-1)>>>col_vals = np.diag(cols, k=-1)>>>z = np.zeros((3,3))>>>z[row_vals, col_vals]=1>>>z
array([[ 0.,  0.,  0.],
       [ 1.,  0.,  0.],
       [ 0.,  1.,  0.]])

Solution 3:

The indices of the k'th diagonal of a can be computed with

defkth_diag_indices(a, k):
    rowidx, colidx = np.diag_indices_from(a)
    colidx = colidx.copy()  # rowidx and colidx share the same bufferif k > 0:
        colidx += k
    else:
        rowidx -= k
    k = np.abs(k)

    return rowidx[:-k], colidx[:-k]

Demo:

>>> a
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24]])
>>> a[kth_diag_indices(a, 1)]
array([ 1,  7, 13, 19])
>>> a[kth_diag_indices(a, 2)]
array([ 2,  8, 14])
>>> a[kth_diag_indices(a, -1)]
array([ 5, 11, 17, 23])

Solution 4:

So since np.diag_indices() doesn't have the same functionality as np.triu_indices() does to get the kth diagonals/triangles, another approach is to just use np.eye(n,k) to construct an nxn matrix with 1's on the kth diagonal, and then use np.where to extract a tuple of indices of where the 1's are located.

So we can do this with just:

T = np.where(np.eye(5,k=-1) == 1)

This is an extra allocation of the eye matrix that may be excessive in some cases, but its an easy one-liner.

Solution 5:

There is another solution. Create a matrix E with np.eye. You can just change the main diagonal like the following. Then, create two more matrizes with the k parameter in eye. In the end just add all matrizes together.

E = np.eye(5)
E = E*2
F = -1*np.eye(len(E),k=1)
G = -1*np.eye(len(E),k=-1)

E = E+F+G
print(E)