python中itertools里的product和permutation
平时经常碰到全排列或者在n个数组中每个数组选一个值组成的所有序列等等问题,可以用permutation和product解决,很方便,所以在此mark一下吧
直接上代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
from itertools import *
if __name__ == '__main__':
for j in permutations([2,5,6]):
print(j)
'''
(2, 5, 6)
(2, 6, 5)
(5, 2, 6)
(5, 6, 2)
(6, 2, 5)
(6, 5, 2)
'''
list1 = [1, 2, 3]
list2 = [4, 5, 6]
list3 = [7, 8, 9]
for i in product(list1,list2,list3):
print(i)
'''
(1, 4, 7)
(1, 4, 8)
(1, 4, 9)
(1, 5, 7)
(1, 5, 8)
(1, 5, 9)
(1, 6, 7)
(1, 6, 8)
(1, 6, 9)
(2, 4, 7)
(2, 4, 8)
(2, 4, 9)
(2, 5, 7)
(2, 5, 8)
(2, 5, 9)
(2, 6, 7)
(2, 6, 8)
(2, 6, 9)
(3, 4, 7)
(3, 4, 8)
(3, 4, 9)
(3, 5, 7)
(3, 5, 8)
(3, 5, 9)
(3, 6, 7)
(3, 6, 8)
(3, 6, 9)
'''
#[list2]*3表示[list2,list2,list2]
#最前面的*号表示将[list2,list2,list2]列表解析成独立的参数
#也就是相当于入参是(list2,list2,list2)
for i in product(*[list2]*3):
print(i)
'''
(4, 4, 4)
(4, 4, 5)
(4, 4, 6)
(4, 5, 4)
(4, 5, 5)
(4, 5, 6)
(4, 6, 4)
(4, 6, 5)
(4, 6, 6)
(5, 4, 4)
(5, 4, 5)
(5, 4, 6)
(5, 5, 4)
(5, 5, 5)
(5, 5, 6)
(5, 6, 4)
(5, 6, 5)
(5, 6, 6)
(6, 4, 4)
(6, 4, 5)
(6, 4, 6)
(6, 5, 4)
(6, 5, 5)
(6, 5, 6)
(6, 6, 4)
(6, 6, 5)
(6, 6, 6)