python中itertools裏的product和permutation

平時經常碰到全排列或者在n個數組中每個數組選一個值組成的所有序列等等問題，可以用permutation和product解決，很方便，所以在此mark一下吧

直接上代碼

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)
    '''
    print('*'*20)
    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)
    '''

解釋一下第三個product的參數，*[list2]*3，也就是*[[4,5,6]]*3，後面的乘以3就是複製三份，參數變爲*[[4,5,6],[4,5,6],[4,5,6]]，最前面的*號是將參數分解爲三個獨立的參數，也即通常所說的解包，如果是字典的話需要兩個星號**，這裏的一個*號也即代表了product([4,5,6],[4,5,6],[4,5,6])