如何生成列表的所有排列组合?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组合?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]
使用标准库中的itertools.permutations
:
import itertools
list(itertools.permutations([1, 2, 3]))
以下是演示如何实现itertools.permutations
的来源代码:
def permutations(elements):
if len(elements) <= 1:
yield elements
return
for perm in permutations(elements[1:]):
for i in range(len(elements)):
# nb elements[0:1] works in both string and list contexts
yield perm[:i] + elements[0:1] + perm[i:]
在 itertools.permutations
的文档中列出了几种备选方法,这里介绍其中一种:
def permutations(iterable, r=None):
# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
# permutations(range(3)) --> 012 021 102 120 201 210
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
if r > n:
return
indices = range(n)
cycles = range(n, n-r, -1)
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[i:] = indices[i+1:] + indices[i:i+1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return
另一个基于itertools.product
的例子:
def permutations(iterable, r=None):
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
for indices in product(range(n), repeat=r):
if len(set(indices)) == r:
yield tuple(pool[i] for i in indices)
从 Python 2.6 开始:
import itertools
itertools.permutations([1, 2, 3])
这将返回一个生成器。使用list(permutations(xs))
可以将其转换为列表。
r
参数,例如 itertools.permutations([1,2,3], r=2)
,将生成选取 2 个元素的所有可能排列:[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]
。 - toto_tico首先,导入itertools
:
import itertools
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)]
print(list(itertools.combinations('123', 2)))
[('1', '2'), ('1', '3'), ('2', '3')]
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)]
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)]
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])
被称为:
permutations('abc')
perms=[]
参数来轻松完成,在每个print
上附加它,并有一个最终的return perms
。 - Alex Moore-Niemi#!/usr/bin/env python
def perm(a, k=0):
if k == len(a):
print a
else:
for i in xrange(k, len(a)):
a[k], a[i] = a[i] ,a[k]
perm(a, k+1)
a[k], a[i] = a[i], a[k]
perm([1,2,3])
输出:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 2, 1]
[3, 1, 2]
因为我在交换列表内容时需要可变序列类型作为输入。例如,perm(list("ball"))
可以工作,而perm("ball")
则不行,因为您无法更改字符串。
这个Python实现受到了Horowitz、Sahni和Rajasekeran的书《计算机算法》中所介绍的算法的启发。
这个解决方案实现了一个生成器,以避免将所有排列都保存在内存中:
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
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]]
普通实现(无yield-将在内存中完成所有操作):
def getPermutations(array):
if len(array) == 1:
return [array]
permutations = []
for i in range(len(array)):
# get all perm's of subarray w/o current item
perms = getPermutations(array[:i] + array[i+1:])
for p in perms:
permutations.append([array[i], *p])
return permutations
产量实现:
def getPermutations(array):
if len(array) == 1:
yield array
else:
for i in range(len(array)):
perms = getPermutations(array[:i] + array[i+1:])
for p in perms:
yield [array[i], *p]
基本想法是对于第一个位置遍历数组中的所有元素,然后在第二个位置上遍历除了第一个选择的元素之外的所有其他元素,以此类推。您可以使用递归来做到这一点,其中停止条件是到达一个只有1个元素的数组 - 在这种情况下,您将返回该数组。
perms = getPermutations(array[:i] + array[i+1:])
。 - RK1numpy
数组 _> getPermutations(np.array([1, 2, 3]))
,我发现它对于列表有效,只是因为函数参数是 array
而感到困惑 :) - RK1我认为一个相当明显的方法也可能是:
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
def permute_in_place(a):
a.sort()
yield list(a)
if len(a) <= 1:
return
first = 0
last = len(a)
while 1:
i = last - 1
while 1:
i = i - 1
if a[i] < a[i+1]:
j = last - 1
while not (a[i] < a[j]):
j = j - 1
a[i], a[j] = a[j], a[i] # swap the values
r = a[i+1:last]
r.reverse()
a[i+1:last] = r
yield list(a)
break
if i == first:
a.reverse()
return
if __name__ == '__main__':
for n in range(5):
for a in permute_in_place(range(1, n+1)):
print a
print
for a in permute_in_place([0, 0, 1, 1, 1]):
print a
print
for i in range(len(elements))
代替for i in range(len(elements)+1)
。实际上,被单独选出的元素elements[0:1]
可以在结果中出现len(elements)
次,而不是len(elements)+1
次。 - Eric O. Lebigot