从Python 2.6（如果你使用的是Python 3）开始，你可以使用标准库工具：itertools.permutations。
import itertoolslist(itertools.permutations([1, 2, 3]))
如果你出于某种原因使用旧版Python（
<2.6），或者只是想知道它的工作原理，那么这是一种不错的方法，取自 http://pre.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:]
的文档中列出了几种替代方法
itertools.permutations。这是一个：
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另一个基于
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)
