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

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'])]

其他回答

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, ([], []))

下面是惰性迭代器方法:

from itertools import tee

def split_on_condition(seq, condition):
    l1, l2 = tee((condition(item), item) for item in seq)
    return (i for p, i in l1 if p), (i for p, i in l2 if not p)

它对每个项计算一次条件,并返回两个生成器,第一个生成条件为真时序列中的值,另一个生成条件为假时序列中的值。

因为它是惰性的,你可以在任何迭代器上使用它,甚至是无限迭代器:

from itertools import count, islice

def is_prime(n):
    return n > 1 and all(n % i for i in xrange(2, n))

primes, not_primes = split_on_condition(count(), is_prime)
print("First 10 primes", list(islice(primes, 10)))
print("First 10 non-primes", list(islice(not_primes, 10)))

通常情况下,非惰性列表返回方法会更好:

def split_on_condition(seq, condition):
    a, b = [], []
    for item in seq:
        (a if condition(item) else b).append(item)
    return a, b

编辑:对于您更具体的用例,将项目按某些键分割到不同的列表中,这里有一个通用函数:

DROP_VALUE = lambda _:_
def split_by_key(seq, resultmapping, keyfunc, default=DROP_VALUE):
    """Split a sequence into lists based on a key function.

        seq - input sequence
        resultmapping - a dictionary that maps from target lists to keys that go to that list
        keyfunc - function to calculate the key of an input value
        default - the target where items that don't have a corresponding key go, by default they are dropped
    """
    result_lists = dict((key, []) for key in resultmapping)
    appenders = dict((key, result_lists[target].append) for target, keys in resultmapping.items() for key in keys)

    if default is not DROP_VALUE:
        result_lists.setdefault(default, [])
        default_action = result_lists[default].append
    else:
        default_action = DROP_VALUE

    for item in seq:
        appenders.get(keyfunc(item), default_action)(item)

    return result_lists

用法:

def file_extension(f):
    return f[2].lower()

split_files = split_by_key(files, {'images': IMAGE_TYPES}, keyfunc=file_extension, default='anims')
print split_files['images']
print split_files['anims']

所有提出的解决方案的问题是,它将扫描和应用过滤功能两次。我会做一个简单的小函数,像这样:

def split_into_two_lists(lst, f):
    a = []
    b = []
    for elem in lst:
        if f(elem):
            a.append(elem)
        else:
            b.append(elem)
    return a, b

这样你就不会重复处理任何东西,也不会重复代码。

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')]

优雅快捷

受到DanSalmo评论的启发,这里有一个简洁、优雅的解决方案,同时也是最快的解决方案之一。

good_set = set(goodvals)
good, bad = [], []
for item in my_list:
    good.append(item) if item in good_set else bad.append(item)

提示:将goodvals转换为一组可以很容易地提高速度。

最快

为了获得最大速度,我们取最快的答案,并通过将good_list转换为一个集合来对其进行涡轮增压。仅这一项就为我们提供了40%以上的速度提升,我们最终得到了比最慢的解决方案快5.5倍以上的解决方案,即使它仍然可读。

good_list_set = set(good_list)  # 40%+ faster than a tuple.

good, bad = [], []
for item in my_origin_list:
    if item in good_list_set:
        good.append(item)
    else:
        bad.append(item)

稍微短一点

这是之前答案的一个更简洁的版本。

good_list_set = set(good_list)  # 40%+ faster than a tuple.

good, bad = [], []
for item in my_origin_list:
    out = good if item in good_list_set else bad
    out.append(item)

优雅可能有点主观,但一些鲁布·戈德堡风格的解决方案很可爱,很巧妙,不应该用于任何语言的产品代码中,更不用说本质上优雅的python了。


基准测试结果:

filter_BJHomer                  80/s       --   -3265%   -5312%   -5900%   -6262%   -7273%   -7363%   -8051%   -8162%   -8244%
zip_Funky                       118/s    4848%       --   -3040%   -3913%   -4450%   -5951%   -6085%   -7106%   -7271%   -7393%
two_lst_tuple_JohnLaRoy         170/s   11332%    4367%       --   -1254%   -2026%   -4182%   -4375%   -5842%   -6079%   -6254%
if_else_DBR                     195/s   14392%    6428%    1434%       --    -882%   -3348%   -3568%   -5246%   -5516%   -5717%
two_lst_compr_Parand            213/s   16750%    8016%    2540%     967%       --   -2705%   -2946%   -4786%   -5083%   -5303%
if_else_1_line_DanSalmo         292/s   26668%   14696%    7189%    5033%    3707%       --    -331%   -2853%   -3260%   -3562%
tuple_if_else                   302/s   27923%   15542%    7778%    5548%    4177%     343%       --   -2609%   -3029%   -3341%
set_1_line                      409/s   41308%   24556%   14053%   11035%    9181%    3993%    3529%       --    -569%    -991%
set_shorter                     434/s   44401%   26640%   15503%   12303%   10337%    4836%    4345%     603%       --    -448%
set_if_else                     454/s   46952%   28358%   16699%   13349%   11290%    5532%    5018%    1100%     469%       --

Python 3.7的完整基准代码(从FunkySayu修改而来):

good_list = ['.jpg','.jpeg','.gif','.bmp','.png']

import random
import string
my_origin_list = []
for i in range(10000):
    fname = ''.join(random.choice(string.ascii_lowercase) for i in range(random.randrange(10)))
    if random.getrandbits(1):
        fext = random.choice(list(good_list))
    else:
        fext = "." + ''.join(random.choice(string.ascii_lowercase) for i in range(3))

    my_origin_list.append((fname + fext, random.randrange(1000), fext))

# Parand
def two_lst_compr_Parand(*_):
    return [e for e in my_origin_list if e[2] in good_list], [e for e in my_origin_list if not e[2] in good_list]

# dbr
def if_else_DBR(*_):
    a, b = list(), list()
    for e in my_origin_list:
        if e[2] in good_list:
            a.append(e)
        else:
            b.append(e)
    return a, b

# John La Rooy
def two_lst_tuple_JohnLaRoy(*_):
    a, b = list(), list()
    for e in my_origin_list:
        (b, a)[e[2] in good_list].append(e)
    return a, b

# # Ants Aasma
# def f4():
#     l1, l2 = tee((e[2] in good_list, e) for e in my_origin_list)
#     return [i for p, i in l1 if p], [i for p, i in l2 if not p]

# My personal way to do
def zip_Funky(*_):
    a, b = zip(*[(e, None) if e[2] in good_list else (None, e) for e in my_origin_list])
    return list(filter(None, a)), list(filter(None, b))

# BJ Homer
def filter_BJHomer(*_):
    return list(filter(lambda e: e[2] in good_list, my_origin_list)), list(filter(lambda e: not e[2] in good_list,                                                                             my_origin_list))

# ChaimG's answer; as a list.
def if_else_1_line_DanSalmo(*_):
    good, bad = [], []
    for e in my_origin_list:
        _ = good.append(e) if e[2] in good_list else bad.append(e)
    return good, bad

# ChaimG's answer; as a set.
def set_1_line(*_):
    good_list_set = set(good_list)
    good, bad = [], []
    for e in my_origin_list:
        _ = good.append(e) if e[2] in good_list_set else bad.append(e)
    return good, bad

# ChaimG set and if else list.
def set_shorter(*_):
    good_list_set = set(good_list)
    good, bad = [], []
    for e in my_origin_list:
        out = good if e[2] in good_list_set else bad
        out.append(e)
    return good, bad

# ChaimG's best answer; if else as a set.
def set_if_else(*_):
    good_list_set = set(good_list)
    good, bad = [], []
    for e in my_origin_list:
        if e[2] in good_list_set:
            good.append(e)
        else:
            bad.append(e)
    return good, bad

# ChaimG's best answer; if else as a set.
def tuple_if_else(*_):
    good_list_tuple = tuple(good_list)
    good, bad = [], []
    for e in my_origin_list:
        if e[2] in good_list_tuple:
            good.append(e)
        else:
            bad.append(e)
    return good, bad

def cmpthese(n=0, functions=None):
    results = {}
    for func_name in functions:
        args = ['%s(range(256))' % func_name, 'from __main__ import %s' % func_name]
        t = Timer(*args)
        results[func_name] = 1 / (t.timeit(number=n) / n) # passes/sec

    functions_sorted = sorted(functions, key=results.__getitem__)
    for f in functions_sorted:
        diff = []
        for func in functions_sorted:
            if func == f:
                diff.append("--")
            else:
                diff.append(f"{results[f]/results[func]*100 - 100:5.0%}")
        diffs = " ".join(f'{x:>8s}' for x in diff)

        print(f"{f:27s} \t{results[f]:,.0f}/s {diffs}")


if __name__=='__main__':
    from timeit import Timer
cmpthese(1000, 'two_lst_compr_Parand if_else_DBR two_lst_tuple_JohnLaRoy zip_Funky filter_BJHomer if_else_1_line_DanSalmo set_1_line set_if_else tuple_if_else set_shorter'.split(" "))