如何生成列表的所有排列?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]

当前回答

我看到在这些递归函数中进行了很多迭代,而不是纯粹的递归。。。

所以对于那些连一个循环都不能遵守的人来说,这里有一个粗略的、完全不必要的完全递归的解决方案

def all_insert(x, e, i=0):
    return [x[0:i]+[e]+x[i:]] + all_insert(x,e,i+1) if i<len(x)+1 else []

def for_each(X, e):
    return all_insert(X[0], e) + for_each(X[1:],e) if X else []

def permute(x):
    return [x] if len(x) < 2 else for_each( permute(x[1:]) , x[0])


perms = permute([1,2,3])

其他回答

def permutations(head, tail=''):
    if len(head) == 0:
        print(tail)
    else:
        for i in range(len(head)):
            permutations(head[:i] + head[i+1:], tail + head[i])

称为:

permutations('abc')

对于Python 2.6及以上版本:

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

这将作为生成器返回。使用列表(排列(xs))作为列表返回。

为了节省您可能的搜索和实验时间,下面是Python中的非递归置换解决方案,它也适用于Numba(从0.41版开始):

@numba.njit()
def permutations(A, k):
    r = [[i for i in range(0)]]
    for i in range(k):
        r = [[a] + b for a in A for b in r if (a in b)==False]
    return r
permutations([1,2,3],3)
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

要给人留下绩效印象:

%timeit permutations(np.arange(5),5)

243 µs ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
time: 406 ms

%timeit list(itertools.permutations(np.arange(5),5))
15.9 µs ± 8.61 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
time: 12.9 s

因此,只有在必须从njit函数调用它时才使用此版本,否则更倾向于itertools实现。

如果有人喜欢这个丑陋的单行线(虽然只适用于字符串):

def p(a):
    return a if len(a) == 1 else [[a[i], *j] for i in range(len(a)) for j in p(a[:i] + a[i + 1:])]

生成所有可能的排列

我正在使用python3.4:

def calcperm(arr, size):
    result = set([()])
    for dummy_idx in range(size):
        temp = set()
        for dummy_lst in result:
            for dummy_outcome in arr:
                if dummy_outcome not in dummy_lst:
                    new_seq = list(dummy_lst)
                    new_seq.append(dummy_outcome)
                    temp.add(tuple(new_seq))
        result = temp
    return result

测试用例:

lst = [1, 2, 3, 4]
#lst = ["yellow", "magenta", "white", "blue"]
seq = 2
final = calcperm(lst, seq)
print(len(final))
print(final)