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
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)]
其他回答
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)]
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
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])
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)
#!/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]
当我交换列表的内容时,需要一个可变的序列类型作为输入。例如,烫发(list(“ball”)会起作用,而烫发(“ball”)不会起作用,因为你不能更改字符串。
这种Python实现的灵感来自Horowitz、Sahni和Rajasekeran在《计算机算法》一书中提出的算法。