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:
nint

The returned indices will be valid to access arrays of shape (n, n).

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.

kscalar

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:
indicestuple of arrays.

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

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])
```