itertools.combinations()

itertools.combinations(iterable, r)

Return r length subsequences of elements from the input iterable.

Combinations are emitted in lexicographic sort order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.

Elements are treated as unique based on their position, not on their value. So if the input elements are unique, there will be no repeat values in each combination.

Roughly equivalent to:

def combinations(iterable, r):
    # combinations('ABCD', 2) --> AB AC AD BC BD CD
    # combinations(range(4), 3) --> 012 013 023 123
    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = list(range(r))
    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)

The code for combinations() can be also expressed as a subsequence of permutations() after filtering entries where the elements are not in sorted order (according to their position in the input pool):

def combinations(iterable, r):
    pool = tuple(iterable)
    n = len(pool)
    for indices in permutations(range(n), r):
        if sorted(indices) == list(indices):
            yield tuple(pool[i] for i in indices)

The number of items returned is n! / r! / (n-r)! when 0 <= r <= n or zero when r > n.

doc_python
2016-10-07 17:35:47
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