This seems simple (and is trivial to write a three-line loop for), but how can I use numpy slicing make a list of the index locations of the upper diagonal of a numpy array? I.e.

Given a 4x4 array, I'd like the index locations at the X's:

`[ X X X X ] [ 0 X X X ] [ 0 0 X X ] [ 0 0 0 X ] `

Giving:

`[ (0,0), (0,1), (0,2), (0,3), (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) ] `

-------------Problems Reply------------

carnieri beat me to the `numpy.triu_indices`

answer, but there is also `numpy.triu_indices_from`

which takes an array as input rather than the dimensions.

Though the format of index locations is different, it seems like you want the function `numpy.triu_indices`

.

`scipy.linalg.triu`

and `nonzero`

If you're running Ubuntu and don't want to upgrade Numpy just for this, you can use the following function:

`from itertools import chain`

triu_indices = lambda x, y=0: zip(*list(chain(*[[(i, j) for j in range(i + y, x)] for i in range(x - y)])))

Example:

`In [26]: triu_indices = lambda x, y=0: zip(*list(chain(*[[(i, j) for j in range(i + y, x)] for i in range(x - y)])))`

```
```In [27]: triu_indices(4)

Out[27]: [(0, 0, 0, 0, 1, 1, 1, 2, 2, 3), (0, 1, 2, 3, 1, 2, 3, 2, 3, 3)]

`In [28]: zip(*triu_indices(4))`

Out[28]:

[(0, 0),

(0, 1),

(0, 2),

(0, 3),

(1, 1),

(1, 2),

(1, 3),

(2, 2),

(2, 3),

(3, 3)]