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
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')
其他回答
我使用了一种基于阶乘数系统的算法——对于长度为n的列表,您可以逐项组装每个排列,从每个阶段留下的项目中进行选择。第一项有n个选项,第二项有n-1个选项,最后一项只有一个选项,因此可以使用阶乘数系统中数字的数字作为索引。这是数字0到n-1对应于词典顺序中的所有可能的排列。
from math import factorial
def permutations(l):
permutations=[]
length=len(l)
for x in xrange(factorial(length)):
available=list(l)
newPermutation=[]
for radix in xrange(length, 0, -1):
placeValue=factorial(radix-1)
index=x/placeValue
newPermutation.append(available.pop(index))
x-=index*placeValue
permutations.append(newPermutation)
return permutations
permutations(range(3))
输出:
[[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]
此方法是非递归的,但在我的计算机上速度稍慢,xrange在n!太大,无法转换为C长整数(我的n=13)。当我需要它的时候,它已经足够了,但它远没有itertools.permutations。
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。
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)
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)]
在我看来,一个很明显的方式可能是:
def permutList(l):
if not l:
return [[]]
res = []
for e in l:
temp = l[:]
temp.remove(e)
res.extend([[e] + r for r in permutList(temp)])
return res