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
无论如何,我们可以使用sympy库,也支持多集合排列
import sympy
from sympy.utilities.iterables import multiset_permutations
t = [1,2,3]
p = list(multiset_permutations(t))
print(p)
# [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
答案的灵感来自获取numpy数组的所有排列
其他回答
另一种解决方案:
def permutation(flag, k =1 ):
N = len(flag)
for i in xrange(0, N):
if flag[i] != 0:
continue
flag[i] = k
if k == N:
print flag
permutation(flag, k+1)
flag[i] = 0
permutation([0, 0, 0])
def pzip(c, seq):
result = []
for item in seq:
for i in range(len(item)+1):
result.append(item[i:]+c+item[:i])
return result
def perm(line):
seq = [c for c in line]
if len(seq) <=1 :
return seq
else:
return pzip(seq[0], perm(seq[1:]))
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']))
免责声明:无耻的插件由包作者。:)
trotter包与大多数实现的不同之处在于,它生成的伪列表实际上不包含排列,而是描述排列与排序中各个位置之间的映射,从而可以处理非常大的排列“列表”,如本演示所示,它在一个包含字母表中所有字母排列的伪列表中执行相当即时的操作和查找,而不使用比典型网页更多的内存或处理。
在任何情况下,要生成排列列表,我们可以执行以下操作。
import trotter
my_permutations = trotter.Permutations(3, [1, 2, 3])
print(my_permutations)
for p in my_permutations:
print(p)
输出:
A pseudo-list containing 6 3-permutations of [1, 2, 3]. [1, 2, 3] [1, 3, 2] [3, 1, 2] [3, 2, 1] [2, 3, 1] [2, 1, 3]
def permuteArray (arr):
arraySize = len(arr)
permutedList = []
if arraySize == 1:
return [arr]
i = 0
for item in arr:
for elem in permuteArray(arr[:i] + arr[i + 1:]):
permutedList.append([item] + elem)
i = i + 1
return permutedList
我不打算在一个新的行中穷尽所有的可能性,以使它有点独特。