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]
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
对于性能,一个由Knuth启发的numpy解决方案(第22页):
from numpy import empty, uint8
from math import factorial
def perms(n):
f = 1
p = empty((2*n-1, factorial(n)), uint8)
for i in range(n):
p[i, :f] = i
p[i+1:2*i+1, :f] = p[:i, :f] # constitution de blocs
for j in range(i):
p[:i+1, f*(j+1):f*(j+2)] = p[j+1:j+i+2, :f] # copie de blocs
f = f*(i+1)
return p[:n, :]
复制大量内存可节省时间-它比列表(itertools.permutations(range(n))快20倍:
In [1]: %timeit -n10 list(permutations(range(10)))
10 loops, best of 3: 815 ms per loop
In [2]: %timeit -n100 perms(10)
100 loops, best of 3: 40 ms per loop
其他回答
def permutate(l):
for i, x in enumerate(l):
for y in l[i + 1:]:
yield x, y
if __name__ == '__main__':
print(list(permutate(list('abcd'))))
print(list(permutate([1, 2, 3, 4])))
#[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
#[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
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。
此解决方案实现了一个生成器,以避免在内存中保留所有排列:
def permutations (orig_list):
if not isinstance(orig_list, list):
orig_list = list(orig_list)
yield orig_list
if len(orig_list) == 1:
return
for n in sorted(orig_list):
new_list = orig_list[:]
pos = new_list.index(n)
del(new_list[pos])
new_list.insert(0, n)
for resto in permutations(new_list[1:]):
if new_list[:1] + resto <> orig_list:
yield new_list[:1] + resto
如果不想使用内置方法,例如:
import itertools
list(itertools.permutations([1, 2, 3]))
你可以自己实现permute函数
from collections.abc import Iterable
def permute(iterable: Iterable[str]) -> set[str]:
perms = set()
if len(iterable) == 1:
return {*iterable}
for index, char in enumerate(iterable):
perms.update([char + perm for perm in permute(iterable[:index] + iterable[index + 1:])])
return perms
if __name__ == '__main__':
print(permute('abc'))
# {'bca', 'abc', 'cab', 'acb', 'cba', 'bac'}
print(permute(['1', '2', '3']))
# {'123', '312', '132', '321', '213', '231'}
另一种方法(无libs)
def permutation(input):
if len(input) == 1:
return input if isinstance(input, list) else [input]
result = []
for i in range(len(input)):
first = input[i]
rest = input[:i] + input[i + 1:]
rest_permutation = permutation(rest)
for p in rest_permutation:
result.append(first + p)
return result
输入可以是字符串或列表
print(permutation('abcd'))
print(permutation(['a', 'b', 'c', 'd']))