Answers for "python permutation list"

3

python permutation list

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
print ( list(permutations([1, 2, 3], 2)) )
# [(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]
Posted by: Guest on May-15-2021
2

python permutation

import itertools

a = [1, 2, 3]
n = 3

perm_iterator = itertools.permutations(a, n)

for item in perm_iterator:
    print(item)
Posted by: Guest on October-30-2020

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