itertools。Groupby几乎可以满足您的要求，除了它要求对条目进行排序以确保您获得一个连续的范围之外，因此您需要首先根据键进行排序(否则您将为每种类型获得多个交错的组)。如。

def is_good(f):
    return f[2].lower() in IMAGE_TYPES

files = [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi'), ('file3.gif', 123L, '.gif')]

for key, group in itertools.groupby(sorted(files, key=is_good), key=is_good):
    print key, list(group)

给:

False [('file2.avi', 999L, '.avi')]
True [('file1.jpg', 33L, '.jpg'), ('file3.gif', 123L, '.gif')]

与其他解决方案类似，可以将键func定义为任意数量的组。

def partition(pred, seq):
  return reduce( lambda (yes, no), x: (yes+[x], no) if pred(x) else (yes, no+[x]), seq, ([], []) )

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')]

还有另一个答案，简短但“邪恶”(用于理解列表的副作用)。

digits = list(range(10))
odd = [x.pop(i) for i, x in enumerate(digits) if x % 2]

>>> odd
[1, 3, 5, 7, 9]

>>> digits
[0, 2, 4, 6, 8]

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'])]