解决方案

from itertools import tee

def unpack_args(fn):
    return lambda t: fn(*t)

def separate(fn, lx):
    return map(
        unpack_args(
            lambda i, ly: filter(
                lambda el: bool(i) == fn(el),
                ly)),
        enumerate(tee(lx, 2)))

test

[even, odd] = separate(
    lambda x: bool(x % 2),
    [1, 2, 3, 4, 5])
print(list(even) == [2, 4])
print(list(odd) == [1, 3, 5])

不确定这是否是一个好方法，但也可以这样做

IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')
files = [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi')]
images, anims = reduce(lambda (i, a), f: (i + [f], a) if f[2] in IMAGE_TYPES else (i, a + [f]), files, ([], []))

如果列表由组和间歇分隔符组成，您可以使用:

def split(items, p):
    groups = [[]]
    for i in items:
        if p(i):
            groups.append([])
        groups[-1].append(i)
    return groups

用法:

split(range(1,11), lambda x: x % 3 == 0)
# gives [[1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

一个基于生成器的版本，如果你能忍受一个或两个原始列表的反转。

设置…

random.seed(1234)
a = list(range(10))
random.shuffle(a)
a
[2, 8, 3, 5, 6, 4, 9, 0, 1, 7]

至于分裂……

(list((a.pop(j) for j, y in [(len(a)-i-1, x) for i,x in enumerate(a[::-1])] if y%2 == 0))[::-1], a)
([2, 8, 6, 4, 0], [3, 5, 9, 1, 7])

Another list of tuples of locations and each element is built in reverse order. In a generator wrapped round that each element is tested against the predicate (here test for even) and if True then the element is poped using previously computed locations. We are working backwards along the list so poping elements out does not change positions closer to the beginning of the list. A wrapping list() evaluates the generator and a final revers [::-1] puts the elements back in the right order. The original list "a" now only contains elements that for which the predicate is False.

就我个人而言，我喜欢你引用的版本，假设你已经有了一个好的列表。如果没有，就像这样:

good = filter(lambda x: is_good(x), mylist)
bad = filter(lambda x: not is_good(x), mylist)

当然，这真的非常类似于使用列表理解，就像你最初做的，但用一个函数而不是一个查找:

good = [x for x in mylist if is_good(x)]
bad  = [x for x in mylist if not is_good(x)]

总的来说，我发现列表推导式的美学非常令人满意。当然，如果您实际上不需要保留顺序，也不需要重复，那么在集合上使用交集和差分方法也会很好。

如果你想用FP风格:

good, bad = [ sum(x, []) for x in zip(*(([y], []) if y in goodvals else ([], [y])
                                        for y in mylist)) ]

不是最易读的解决方案，但至少只遍历mylist一次。