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。

手动迭代，使用条件选择每个元素将被追加到的列表:

good, bad = [], []
for x in mylist:
    (bad, good)[x in goodvals].append(x)

我认为基于N个条件来划分一个可迭代对象是很方便的

from collections import OrderedDict
def partition(iterable,*conditions):
    '''Returns a list with the elements that satisfy each of condition.
       Conditions are assumed to be exclusive'''
    d= OrderedDict((i,list())for i in range(len(conditions)))        
    for e in iterable:
        for i,condition in enumerate(conditions):
            if condition(e):
                d[i].append(e)
                break                    
    return d.values()

例如:

ints,floats,other = partition([2, 3.14, 1, 1.69, [], None],
                              lambda x: isinstance(x, int), 
                              lambda x: isinstance(x, float),
                              lambda x: True)

print " ints: {}\n floats:{}\n other:{}".format(ints,floats,other)

 ints: [2, 1]
 floats:[3.14, 1.69]
 other:[[], None]

如果元素可以满足多个条件，则删除断点。

使用布尔逻辑将数据分配给两个数组

>>> images, anims = [[i for i in files if t ^ (i[2].lower() in IMAGE_TYPES) ] for t in (0, 1)]
>>> images
[('file1.jpg', 33, '.jpg')]
>>> anims
[('file2.avi', 999, '.avi')]

bad = []
good = [x for x in mylist if x in goodvals or bad.append(x)]

append返回None，所以它可以工作。

我转向numpy来解决这个问题，以限制行数，并使其成为一个简单的函数。

我能够得到一个条件满足，将一个列表分为两个，使用np。在哪里分离出一个列表。这适用于数字，但这可以扩展使用字符串和列表，我相信。

在这儿……

from numpy import where as wh, array as arr

midz = lambda a, mid: (a[wh(a > mid)], a[wh((a =< mid))])
p_ = arr([i for i in [75, 50, 403, 453, 0, 25, 428] if i])
high,low = midz(p_, p_.mean())