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
from __future__ import print_function
def perm(n):
p = []
for i in range(0,n+1):
p.append(i)
while True:
for i in range(1,n+1):
print(p[i], end=' ')
print("")
i = n - 1
found = 0
while (not found and i>0):
if p[i]<p[i+1]:
found = 1
else:
i = i - 1
k = n
while p[i]>p[k]:
k = k - 1
aux = p[i]
p[i] = p[k]
p[k] = aux
for j in range(1,(n-i)/2+1):
aux = p[i+j]
p[i+j] = p[n-j+1]
p[n-j+1] = aux
if not found:
break
perm(5)
其他回答
注意,该算法具有n个阶乘时间复杂度,其中n是输入列表的长度
打印跑步结果:
global result
result = []
def permutation(li):
if li == [] or li == None:
return
if len(li) == 1:
result.append(li[0])
print result
result.pop()
return
for i in range(0,len(li)):
result.append(li[i])
permutation(li[:i] + li[i+1:])
result.pop()
例子:
permutation([1,2,3])
输出:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
首先,导入itertools:
import itertools
排列(顺序重要):
print(list(itertools.permutations([1,2,3,4], 2)))
[(1, 2), (1, 3), (1, 4),
(2, 1), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 4),
(4, 1), (4, 2), (4, 3)]
组合(顺序无关紧要):
print(list(itertools.combinations('123', 2)))
[('1', '2'), ('1', '3'), ('2', '3')]
笛卡尔积(具有多个可迭代项):
print(list(itertools.product([1,2,3], [4,5,6])))
[(1, 4), (1, 5), (1, 6),
(2, 4), (2, 5), (2, 6),
(3, 4), (3, 5), (3, 6)]
笛卡尔积(具有一个可迭代的和自身):
print(list(itertools.product([1,2], repeat=3)))
[(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2),
(2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)]
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])
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
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]