python中itertools里的product和permutation

转载自博客园xqn2017
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

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)
    '''