我有这样的代码:

good = [x for x in mylist if x in goodvals]
bad = [x for x in mylist if x not in goodvals]

目标是根据mylist的内容是否满足条件,将它们拆分为另外两个列表。

我怎样才能做得更优雅呢?我能避免在mylist上做两个单独的迭代吗?我可以通过这样做来提高性能吗?


当前回答

这里已经有很多解了,但另一种方法是

anims = []
images = [f for f in files if (lambda t: True if f[2].lower() in IMAGE_TYPES else anims.append(t) and False)(f)]

只在列表上迭代一次,看起来更python化,因此对我来说是可读的。

>>> files = [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi'), ('file1.bmp', 33L, '.bmp')]
>>> IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')
>>> anims = []
>>> images = [f for f in files if (lambda t: True if f[2].lower() in IMAGE_TYPES else anims.append(t) and False)(f)]
>>> print '\n'.join([str(anims), str(images)])
[('file2.avi', 999L, '.avi')]
[('file1.jpg', 33L, '.jpg'), ('file1.bmp', 33L, '.bmp')]
>>>

其他回答

简单的生成器版本,在内存中保存尽可能少的值,并且只调用pred一次:

from collections import deque
from typing import Callable, TypeVar, Iterable
_T = TypeVar('_T')

def iter_split(pred: Callable[[_T], bool],
               iterable: Iterable[_T]) -> tuple[Iterable[_T], Iterable[_T]]:
    """Split an iterable into two iterables based on a predicate.
    
    The predicate will only be called once per element.
    
    Returns:
        A tuple of two iterables, the first containing all elements for which
        the predicate returned True, the second containing all elements for
        which the predicate returned False.
    """
    iterator = iter(iterable)
    true_values: deque[_T] = deque()
    false_values: deque[_T] = deque()
    
    def true_generator():
        while True:
            while true_values:
                yield true_values.popleft()
            
            for item in iterator:
                if pred(item):
                    yield item
                    break
                false_values.append(item)
            else:
                break
            
    def false_generator():
        while True:
            while false_values:
                yield false_values.popleft()
            
            for item in iterator:
                if not pred(item):
                    yield item
                    break
                true_values.append(item)
            else:
                break

    return true_generator(), false_generator()

第一步(pre-OP-edit):使用集合:

mylist = [1,2,3,4,5,6,7]
goodvals = [1,3,7,8,9]

myset = set(mylist)
goodset = set(goodvals)

print list(myset.intersection(goodset))  # [1, 3, 7]
print list(myset.difference(goodset))    # [2, 4, 5, 6]

这对可读性(IMHO)和性能都有好处。

第二步(post-OP-edit):

创建一个好的扩展列表:

IMAGE_TYPES = set(['.jpg','.jpeg','.gif','.bmp','.png'])

这将提高性能。否则,你现在的情况在我看来还不错。

我基本上喜欢安德斯的方法,因为它非常普遍。下面的版本将分类器放在前面(以匹配过滤器语法),并使用defaultdict(假定已导入)。

def categorize(func, seq):
    """Return mapping from categories to lists
    of categorized items.
    """
    d = defaultdict(list)
    for item in seq:
        d[func(item)].append(item)
    return d

你可以在Python中进行惰性函数编程,像这样:

partition = lambda l, c: map(
  lambda iii: (i for ii in iii for i in ii),
  zip(*(([], [e]) if c(e) else ([e], []) for e in l)))

函数式编程很优雅,但在Python中不是这样。如果你知道你的列表中没有None值,也可以参考这个例子:

partition = lambda l, c: map(
  filter(lambda x: x is not None, l),
  zip(*((None, e) if c(e) else (e, None) for e in l)))

有时候你并不需要清单上的另一半。 例如:

import sys
from itertools import ifilter

trustedPeople = sys.argv[1].split(',')
newName = sys.argv[2]

myFriends = ifilter(lambda x: x.startswith('Shi'), trustedPeople)

print '%s is %smy friend.' % (newName, newName not in myFriends 'not ' or '')