# How to generate all permutations of a list?

How do you generate all the permutations of a list in Python, independently of the type of elements in that list?

For example:

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

Starting with Python 2.6 (and if you’re on Python 3) you have a standard-library tool for this: `itertools.permutations`.

``````import itertools
list(itertools.permutations([1, 2, 3]))
``````

If you’re using an older Python (<2.6) for some reason or are just curious to know how it works, here’s one nice approach, taken from http://code.activestate.com/recipes/252178/:

``````def all_perms(elements):
if len(elements) <=1:
yield elements
else:
for perm in all_perms(elements[1:]):
for i in range(len(elements)):
# nb elements[0:1] works in both string and list contexts
yield perm[:i] + elements[0:1] + perm[i:]
``````

A couple of alternative approaches are listed in the documentation of `itertools.permutations`. Here’s one:

``````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 = range(n)
cycles = 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
``````

And another, based on `itertools.product`:

``````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)
``````
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