如何生成列表的所有排列?

问题描述

如何在Python中生成列表的所有排列，而与列表中元素的类型无关?

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

例如:

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]

推荐答案

从Python 2.6开始(如果您使用的是Python 3)，则有一个 standard-library >为此的工具: itertools.permutations .

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]))

---

如果您出于某种原因使用旧版Python(，或者只是想知道它的工作原理，那么这是一种很好的方法，取材于 http://code.activestate.com/recipes/252178/:

---

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:]

itertools.permutations的文档中列出了几种替代方法.这是一个:

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

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)

这篇关于如何生成列表的所有排列?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！