Python：使用for循环创建未知数量的变量

Question

I want to create a generator that takes in any number of keyword arguments and it will return the product of the values (give as tuples as shown below).

I am having trouble avoiding hardcoding the for loop variables ( hardcoded_a , hardcoded_b ). In this scenario if I use more or less than two arguments it gives a ValueError . I don't care about hardcoding the 'okay' variables. How can I go about it so no matter how many items I pass to kwargs, I can still yield the product?

This is what I have written so far:

from itertools import product
def gen(**kwargs):
    options = {}
    [options.update({k: v}) for k, v in kwargs.iteritems()]
    for hardcoded_a, hardcoded_b in product(*(tuple(options.values()))):
       yield hardcoded_a, hardcoded_b

for okay_var1, okay_var2 in gen(dollar=(2, 20), hungry=(True, False)):
    print okay_var1, okay_var2

Answer 1

I think what you're looking for is the following:

def gen(**kwargs):
    options = {}
    [options.update({k: v}) for k, v in kwargs.iteritems()]
    for prod in product(*(tuple(options.values()))):
       yield dict(zip(options, prod))

or an even cleaner solution:

def gen(**kwargs):
    for prod in product(*kwargs.values()):
       yield dict(zip(kwargs, prod))

Answer 2

Unless I misunderstand your objective, the following should work:

from itertools import product

def gen(**kwargs):
    return product(*kwargs.values())

Example:

>>> print(list(gen(dollar=(2,20), hungry=(True,False))))
[(True, 2), (True, 20), (False, 2), (False, 20)]

It's worth noting, however, that kwargs.values() is not guaranteed to have a specific order. This will make sure they are ordered by the "natural ordering" (string comparison) of the keywords:

def gen2(**kwargs):
    return product(*map(lambda t:t[1],sorted(k.items())))

So now:

>>> print(list(gen2(dollar=(2,20), hungry=(True,False))))
[(2, True), (2, False), (20, True), (20, False)]