python中itertools里的product和permutation
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)