您不能，不仅仅因为能够在第二个循环中生成b值而保留所有生成器输出.就内存而言，这可能会付出高昂的代价.

You can't, not without ending up holding all generator output just to be able to produce b values in the second loop. That can get costly in terms of memory.

您将使用 itertools.tee() 来复制生成器:

You'd use itertools.tee() to 'duplicate' the generator:

from itertools import tee

def split_gen(gen):
    gen_a, gen_b = tee(gen, 2)
    return (a for a, b in gen_a), (b for a, b in gen_b)

gen1_split_a, gen1_split_b = split_gen(gen1)

for a in gen1_split_a:
    print a

for b in gen1_split_b:
    print b

，但是在这种情况下，发生的情况是tee对象最终将不得不存储 gen1产生的所有东西.从文档中:

but what happens in this case is that the tee object will end up having to store everything gen1 produces. From the documentation:

此itertool可能需要大量辅助存储(取决于需要存储多少临时数据).通常，如果一个迭代器在另一个迭代器启动之前使用了大部分或全部数据，则使用list()而不是tee()会更快.

根据该建议，只需将b值放入第二个循环的列表中即可:

Following that advice, just put the b values into a list for the second loop:

b_values = []
for a, b in gen1():
    print a
    b_values.append(a)

for b in b_values:
    print b

或更妙的是，只需在一个循环中同时处理a和b.

or better yet, just process both a and b in the one loop.