numpy.mask_indices

numpy.mask_indices(n, mask_func, k=0) [source]

Return the indices to access (n, n) arrays, given a masking function.

Assume mask_func is a function that, for a square array a of size (n, n) with a possible offset argument k, when called as mask_func(a, k) returns a new array with zeros in certain locations (functions like triu or tril do precisely this). Then this function returns the indices where the non-zero values would be located.

Parameters:

n : int The returned indices will be valid to access arrays of shape (n, n). mask_func : callable A function whose call signature is similar to that of triu, tril. That is, mask_func(x, k) returns a boolean array, shaped like x. k is an optional argument to the function. k : scalar An optional argument which is passed through to mask_func. Functions like triu, tril take a second argument that is interpreted as an offset.

Returns:

indices : tuple of arrays. The n arrays of indices corresponding to the locations where mask_func(np.ones((n, n)), k) is True.

See also

triu, tril, triu_indices, tril_indices

Notes

New in version 1.4.0.

Examples

These are the indices that would allow you to access the upper triangular part of any 3x3 array:

>>> iu = np.mask_indices(3, np.triu)

For example, if a is a 3x3 array:

>>> a = np.arange(9).reshape(3, 3)
>>> a
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])
>>> a[iu]
array([0, 1, 2, 4, 5, 8])

An offset can be passed also to the masking function. This gets us the indices starting on the first diagonal right of the main one:

>>> iu1 = np.mask_indices(3, np.triu, 1)

with which we now extract only three elements:

>>> a[iu1]
array([1, 2, 5])