【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

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')]

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

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

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

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