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.

images = [f for f in files if f[2].lower() in IMAGE_TYPES]
anims  = [f for f in files if f not in images]

当条件较长时很好，例如在您的示例中。读者不需要弄清楚否定条件以及它是否适用于所有其他情况。

例如，按偶数和奇数拆分列表

arr = range(20)
even, odd = reduce(lambda res, next: res[next % 2].append(next) or res, arr, ([], []))

或者概括地说:

def split(predicate, iterable):
    return reduce(lambda res, e: res[predicate(e)].append(e) or res, iterable, ([], []))

优点:

最短路径 Predicate对每个元素只应用一次

缺点

需要函数式编程范例的知识

Good = [x for x in mylist if x in goodvals] Bad = [x for x in mylist if x not in goodvals] 我怎样才能做得更优雅呢?

代码已经非常优雅了。

使用集合可能会有轻微的性能改进，但差异是微不足道的。基于集合的方法也会丢弃重复项，并且不会保留元素的顺序。我发现列表理解也更容易阅读。

事实上，我们甚至可以更简单地使用for循环:

good, bad = [], []

for x in mylist:
    if x in goodvals:
        good.append(f)
    else:
        bad.append(f)

这种方法可以更容易地添加额外的逻辑。例如，代码很容易被修改为丢弃None值:

good, bad = [], []

for x in mylist:
    if x is None:
        continue
    if x in goodvals:
        good.append(f)
    else:
        bad.append(f)

受到@gnibbler伟大(但简洁!)回答的启发，我们可以应用该方法映射到多个分区:

from collections import defaultdict

def splitter(l, mapper):
    """Split an iterable into multiple partitions generated by a callable mapper."""

    results = defaultdict(list)

    for x in l:
        results[mapper(x)] += [x]

    return results

然后可以使用splitter，如下所示:

>>> l = [1, 2, 3, 4, 2, 3, 4, 5, 6, 4, 3, 2, 3]
>>> split = splitter(l, lambda x: x % 2 == 0)  # partition l into odds and evens
>>> split.items()
>>> [(False, [1, 3, 3, 5, 3, 3]), (True, [2, 4, 2, 4, 6, 4, 2])]

这适用于有更复杂映射的两个以上分区(也适用于迭代器):

>>> import math
>>> l = xrange(1, 23)
>>> split = splitter(l, lambda x: int(math.log10(x) * 5))
>>> split.items()
[(0, [1]),
 (1, [2]),
 (2, [3]),
 (3, [4, 5, 6]),
 (4, [7, 8, 9]),
 (5, [10, 11, 12, 13, 14, 15]),
 (6, [16, 17, 18, 19, 20, 21, 22])]

或者用字典来映射:

>>> map = {'A': 1, 'X': 2, 'B': 3, 'Y': 1, 'C': 2, 'Z': 3}
>>> l = ['A', 'B', 'C', 'C', 'X', 'Y', 'Z', 'A', 'Z']
>>> split = splitter(l, map.get)
>>> split.items()
(1, ['A', 'Y', 'A']), (2, ['C', 'C', 'X']), (3, ['B', 'Z', 'Z'])]

如果你不介意使用一个外部库，有两个我知道本机实现这个操作:

>>> files = [ ('file1.jpg', 33, '.jpg'), ('file2.avi', 999, '.avi')]
>>> IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')

iteration_utilities.partition: >>> from iteration_utilities import partition >>> notimages, images = partition(files, lambda x: x[2].lower() in IMAGE_TYPES) >>> notimages [('file2.avi', 999, '.avi')] >>> images [('file1.jpg', 33, '.jpg')] more_itertools.partition >>> from more_itertools import partition >>> notimages, images = partition(lambda x: x[2].lower() in IMAGE_TYPES, files) >>> list(notimages) # returns a generator so you need to explicitly convert to list. [('file2.avi', 999, '.avi')] >>> list(images) [('file1.jpg', 33, '.jpg')]