python中itertools里的product和permutation，,python中ite

python中itertools里的product和permutation，,python中ite

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123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081from 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)

python中itertools里的product和permutation