Find indexes of repeated elements in an array (Python, NumPy)

Question

Assume, I have a NumPy-array of integers, as:

[34,2,3,22,22,22,22,22,22,18,90,5,-55,-19,22,6,6,6,6,6,6,6,6,23,53,1,5,-42,82]

I want to find

Answer 1

Here is a solution using Python's native itertools.

Code

import itertools as it


def find_ranges(lst, n=2):
    """Return ranges for `n` or more repeated values."""
    groups = ((k, tuple(g)) for k, g in it.groupby(enumerate(lst), lambda x: x[-1]))
    repeated = (idx_g for k, idx_g in groups if len(idx_g) >=n)
    return ((sub[0][0], sub[-1][0]) for sub in repeated)

lst = [34,2,3,22,22,22,22,22,22,18,90,5,-55,-19,22,6,6,6,6,6,6,6,6,23,53,1,5,-42,82]    
list(find_ranges(lst, 5))
# [(3, 8), (15, 22)]

Tests

import nose.tools as nt


def test_ranges(f):
    """Verify list results identifying ranges."""
    nt.eq_(list(f([])), [])
    nt.eq_(list(f([0, 1,1,1,1,1,1, 2], 5)), [(1, 6)])
    nt.eq_(list(f([1,1,1,1,1,1, 2,2, 1, 3, 1,1,1,1,1,1], 5)), [(0, 5), (10, 15)])
    nt.eq_(list(f([1,1, 2, 1,1,1,1, 2, 1,1,1], 3)), [(3, 6), (8, 10)])    
    nt.eq_(list(f([1,1,1,1, 2, 1,1,1, 2, 1,1,1,1], 3)), [(0, 3), (5, 7), (9, 12)])

test_ranges(find_ranges)

This example captures (index, element) pairs in lst, and then groups them by element. Only repeated pairs are retained. Finally, first and last pairs are sliced, yielding (start, end) indices from each repeated group.

See also this post for finding ranges of indices using itertools.groupby.