给定一个可迭代的函数，如何在每种可能的组合中应用函数？

Question

Given the iterable [A, B, C] and the function f(x) I want to get the following:

[  A,     B,     C]  
[  A,     B,   f(C)]  
[  A,   f(B),    C]
[  A,   f(B),  f(C)]
[f(A),    B,     C]
[f(A),    B,   f(C)]
[f(A),  f(B),    C]
[f(A),  f(B),  f(C)]

Unfortunately I didn't find anything suitable in the itertools module.

Answer 1

>>> from itertools import product
>>> L = ["A", "B", "C"]
>>> def f(c): return c.lower()
... 
>>> fL = [f(x) for x in L]
>>> for i in product(*zip(L, fL)):
...     print i
... 
('A', 'B', 'C')
('A', 'B', 'c')
('A', 'b', 'C')
('A', 'b', 'c')
('a', 'B', 'C')
('a', 'B', 'c')
('a', 'b', 'C')
('a', 'b', 'c')

Explanation:

Call f for each item in L to generate fL

>>> fL
['a', 'b', 'c']

Use zip to zip the two lists into pairs

>>> zip(L, fL)
[('A', 'a'), ('B', 'b'), ('C', 'c')]

Take the cartesian product of those tuples using itertools.product

product(*zip(L, fL))

is equivalent to

product(*[('A', 'a'), ('B', 'b'), ('C', 'c')])

and that is equivalent to

product(('A', 'a'), ('B', 'b'), ('C', 'c'))

looping over that product, gives exactly the result we need.

Answer 2

You can use itertools.combinations , like this

def f(char):
    return char.lower()

iterable = ["A", "B", "C"]
indices = range(len(iterable))
from itertools import combinations
for i in range(len(iterable) + 1):
    for items in combinations(indices, i):
        print [f(iterable[j]) if j in items else iterable[j] for j in range(len(iterable))]

Output

['A', 'B', 'C']
['a', 'B', 'C']
['A', 'b', 'C']
['A', 'B', 'c']
['a', 'b', 'C']
['a', 'B', 'c']
['A', 'b', 'c']
['a', 'b', 'c']

Answer 3

import itertools
def func_combinations(f, l):
    return itertools.product(*zip(l, map(f, l)))

Demo:

>>> for combo in func_combinations(str, range(3)):
...     print combo
...
(0, 1, 2)
(0, 1, '2')
(0, '1', 2)
(0, '1', '2')
('0', 1, 2)
('0', 1, '2')
('0', '1', 2)
('0', '1', '2')

This function first computes f once for every element of the input. Then, it uses zip to turn the input and the list of f values into a list of input-output pairs. Finally, it uses itertools.product to produce each possible way to select either input or output.