【385】itertools 的 product 和 chain 和 accumulate

参考：itertools模块

product

相当于返回两个集合中数据的所有组合可能

Examples from Eric Martin

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

from itertools import product   print(list(product([0, 1], 'abc'))) print() print(list(product(['A', 'B'], ('a', 'b'), range(2)))) print() print(list(product([0, 1], repeat = 2))) print() print(list(product('ab', repeat = 4)))   output: [(0, 'a'), (0, 'b'), (0, 'c'), (1, 'a'), (1, 'b'), (1, 'c')]   [('A', 'a', 0), ('A', 'a', 1), ('A', 'b', 0), ('A', 'b', 1), ('B', 'a ', 0), ('B', 'a', 1), ('B', 'b', 0), ('B', 'b', 1)]   [(0, 0), (0, 1), (1, 0), (1, 1)]   [('a', 'a', 'a', 'a'), ('a', 'a', 'a', 'b'), ('a', 'a', 'b', 'a'), (' a', 'a', 'b', 'b'), ('a', 'b', 'a', 'a'), ('a', 'b', 'a', 'b'), ('a', 'b', 'b', 'a'), ('a', 'b', 'b', 'b'), ('b', 'a', 'a', 'a'), ('b', 'a ', 'a', 'b'), ('b', 'a', 'b', 'a'), ('b', 'a', 'b', 'b'), ('b', 'b', 'a', 'a'), ('b', 'b', 'a', 'b'), ('b', 'b', 'b', 'a'), ('b', 'b', 'b' , 'b')]

 

?

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

from itertools import product a = (1, 2, 3) b = ('A', 'B', 'C') c = ('d', 'e', 'f') pros = product(a, b, c) count = 1 for elem in pros:     print(f'{count:02}', "---", elem)     count+=1   output: 01 --- (1, 'A', 'd') 02 --- (1, 'A', 'e') 03 --- (1, 'A', 'f') 04 --- (1, 'B', 'd') 05 --- (1, 'B', 'e') 06 --- (1, 'B', 'f') 07 --- (1, 'C', 'd') 08 --- (1, 'C', 'e') 09 --- (1, 'C', 'f') 10 --- (2, 'A', 'd') 11 --- (2, 'A', 'e') 12 --- (2, 'A', 'f') 13 --- (2, 'B', 'd') 14 --- (2, 'B', 'e') 15 --- (2, 'B', 'f') 16 --- (2, 'C', 'd') 17 --- (2, 'C', 'e') 18 --- (2, 'C', 'f') 19 --- (3, 'A', 'd') 20 --- (3, 'A', 'e') 21 --- (3, 'A', 'f') 22 --- (3, 'B', 'd') 23 --- (3, 'B', 'e') 24 --- (3, 'B', 'f') 25 --- (3, 'C', 'd') 26 --- (3, 'C', 'e') 27 --- (3, 'C', 'f')

例子2：二进制数三位数的所有可能

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

a = (0, 1) b = (0, 1) c = (0, 1) pros = product(a, b, c) count = 1 for elem in pros:     print(f'{count:02}', "---", elem)     count+=1   output: 01 --- (0, 0, 0) 02 --- (0, 0, 1) 03 --- (0, 1, 0) 04 --- (0, 1, 1) 05 --- (1, 0, 0) 06 --- (1, 0, 1) 07 --- (1, 1, 0) 08 --- (1, 1, 1)

chain 就是合并成一个 iter

?

1 2 3 4 5

from itertools import chain [e for e in chain([2, 3], {3, 4}, (3,4))]   output: [2, 3, 3, 4, 3, 4]

accumulate 可以实现将可迭代对象进行累加的效果，形成一个新的可迭代对象

?

1 2 3 4

>>> a = accumulate([1, 2, 3, 4])   >>> [i for i in a] [1, 3, 6, 10]