ID EN
Itertools

permutations

Python 3.11 🇮🇩 Bahasa Indonesia

Mengembalikan permutasi panjang r elemen berturut-turut dari iterable.

Syntax

PYTHON
itertools.permutations(iterable, r=None)

Contoh

Example 1
PYTHON
def permutations(iterable, r=None):
     permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
     permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = list(range(n))
    cycles = list(range(n, n-r, -1))
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return
Example 2
PYTHON
def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

See Also

permutations() product()