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
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
其他回答
这里有一个算法,它在不创建新的中间列表的情况下处理列表,类似于Ber在https://stackoverflow.com/a/108651/184528.
def permute(xs, low=0):
if low + 1 >= len(xs):
yield xs
else:
for p in permute(xs, low + 1):
yield p
for i in range(low + 1, len(xs)):
xs[low], xs[i] = xs[i], xs[low]
for p in permute(xs, low + 1):
yield p
xs[low], xs[i] = xs[i], xs[low]
for p in permute([1, 2, 3, 4]):
print p
您可以在这里亲自尝试代码:http://repl.it/J9v
功能性风格
def addperm(x,l):
return [ l[0:i] + [x] + l[i:] for i in range(len(l)+1) ]
def perm(l):
if len(l) == 0:
return [[]]
return [x for y in perm(l[1:]) for x in addperm(l[0],y) ]
print perm([ i for i in range(3)])
结果:
[[0, 1, 2], [1, 0, 2], [1, 2, 0], [0, 2, 1], [2, 0, 1], [2, 1, 0]]
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]
这是受Haskell实现使用列表理解的启发:
def permutation(list):
if len(list) == 0:
return [[]]
else:
return [[x] + ys for x in list for ys in permutation(delete(list, x))]
def delete(list, item):
lc = list[:]
lc.remove(item)
return lc
递归之美:
>>> import copy
>>> def perm(prefix,rest):
... for e in rest:
... new_rest=copy.copy(rest)
... new_prefix=copy.copy(prefix)
... new_prefix.append(e)
... new_rest.remove(e)
... if len(new_rest) == 0:
... print new_prefix + new_rest
... continue
... perm(new_prefix,new_rest)
...
>>> perm([],['a','b','c','d'])
['a', 'b', 'c', 'd']
['a', 'b', 'd', 'c']
['a', 'c', 'b', 'd']
['a', 'c', 'd', 'b']
['a', 'd', 'b', 'c']
['a', 'd', 'c', 'b']
['b', 'a', 'c', 'd']
['b', 'a', 'd', 'c']
['b', 'c', 'a', 'd']
['b', 'c', 'd', 'a']
['b', 'd', 'a', 'c']
['b', 'd', 'c', 'a']
['c', 'a', 'b', 'd']
['c', 'a', 'd', 'b']
['c', 'b', 'a', 'd']
['c', 'b', 'd', 'a']
['c', 'd', 'a', 'b']
['c', 'd', 'b', 'a']
['d', 'a', 'b', 'c']
['d', 'a', 'c', 'b']
['d', 'b', 'a', 'c']
['d', 'b', 'c', 'a']
['d', 'c', 'a', 'b']
['d', 'c', 'b', 'a']