在Python中展平浅层列表

Question

是否有一种简单的方法可以使用列表推导来展平迭代列表,或者失败,你会认为什么是平衡这样的浅层列表,平衡性能和可读性的最佳方法？

我尝试使用嵌套列表理解来压缩这样的列表,如下所示:

[image for image in menuitem for menuitem in list_of_menuitems]

但我在NameError那里遇到麻烦,因为name 'menuitem' is not defined.谷歌搜索并浏览Stack Overflow后,我得到了一个reduce声明所需的结果:

reduce(list.__add__, map(lambda x: list(x), list_of_menuitems))

但是这个方法相当难以理解,因为我需要那个list(x)调用,因为x是一个Django QuerySet对象.

结论:

感谢所有为此问题做出贡献的人.以下是我学到的内容摘要.我也将其作为社区维基,以防其他人想要添加或更正这些观察结果.

我原来的reduce语句是多余的,用这种方式编写得更好:

>>> reduce(list.__add__, (list(mi) for mi in list_of_menuitems))

这是嵌套列表理解的正确语法(Brilliant summary dF!):

>>> [image for mi in list_of_menuitems for image in mi]

但这些方法都不如使用效率高itertools.chain:

>>> from itertools import chain
>>> list(chain(*list_of_menuitems))

正如@cdleary指出的那样,通过使用chain.from_iterable如下所示来避免*操作符魔术可能是更好的风格:

>>> chain = itertools.chain.from_iterable([[1,2],[3],[5,89],[],[6]])
>>> print(list(chain))
>>> [1, 2, 3, 5, 89, 6]

Answer 1

如果您只是想要迭代数据结构的扁平化版本并且不需要可索引序列,请考虑使用itertools.chain和company.

>>> list_of_menuitems = [['image00', 'image01'], ['image10'], []]
>>> import itertools
>>> chain = itertools.chain(*list_of_menuitems)
>>> print(list(chain))
['image00', 'image01', 'image10']

它将适用于任何可迭代的东西,其中应包括Django的可迭代QuerySets,它似乎是你在问题中使用的.

编辑:无论如何,这可能与reduce一样好,因为reduce会将项目复制到正在扩展的列表中.chain如果你list(chain)在最后运行,只会产生这个(相同的)开销.

元编辑:实际上,它比问题的建议解决方案的开销更少,因为当您使用临时扩展原始文件时,您会丢弃您创建的临时列表.

编辑:正如JF Sebastian所说, itertools.chain.from_iterable避免拆包,你应该使用它来避免*魔术,但timeit应用程序显示可忽略不计的性能差异.

Answer 2

你几乎拥有它!该做的嵌套列表理解的方式是把for语句以相同的顺序,因为他们会去正规的嵌套for语句.

因此,这

for inner_list in outer_list:
    for item in inner_list:
        ...

对应于

[... for inner_list in outer_list for item in inner_list]

所以你要

[image for menuitem in list_of_menuitems for image in menuitem]

Answer 3

@ S.Lott:你激励我写一个timeit app.

我认为它也会根据分区数量(容器列表中的迭代器数量)而有所不同 - 您的评论没有提到30个项目中有多少个分区.这个图在每次运行中展平了一千个项目,分区数量不同.项目均匀分布在分区中.

代码(Python 2.6):

#!/usr/bin/env python2.6

"""Usage: %prog item_count"""

from __future__ import print_function

import collections
import itertools
import operator
from timeit import Timer
import sys

import matplotlib.pyplot as pyplot

def itertools_flatten(iter_lst):
    return list(itertools.chain(*iter_lst))

def itertools_iterable_flatten(iter_iter):
    return list(itertools.chain.from_iterable(iter_iter))

def reduce_flatten(iter_lst):
    return reduce(operator.add, map(list, iter_lst))

def reduce_lambda_flatten(iter_lst):
    return reduce(operator.add, map(lambda x: list(x), [i for i in iter_lst]))

def comprehension_flatten(iter_lst):
    return list(item for iter_ in iter_lst for item in iter_)

METHODS = ['itertools', 'itertools_iterable', 'reduce', 'reduce_lambda',
           'comprehension']

def _time_test_assert(iter_lst):
    """Make sure all methods produce an equivalent value.
    :raise AssertionError: On any non-equivalent value."""
    callables = (globals()[method + '_flatten'] for method in METHODS)
    results = [callable(iter_lst) for callable in callables]
    if not all(result == results[0] for result in results[1:]):
        raise AssertionError

def time_test(partition_count, item_count_per_partition, test_count=10000):
    """Run flatten methods on a list of :param:`partition_count` iterables.
    Normalize results over :param:`test_count` runs.
    :return: Mapping from method to (normalized) microseconds per pass.
    """
    iter_lst = [[dict()] * item_count_per_partition] * partition_count
    print('Partition count:    ', partition_count)
    print('Items per partition:', item_count_per_partition)
    _time_test_assert(iter_lst)
    test_str = 'flatten(%r)' % iter_lst
    result_by_method = {}
    for method in METHODS:
        setup_str = 'from test import %s_flatten as flatten' % method
        t = Timer(test_str, setup_str)
        per_pass = test_count * t.timeit(number=test_count) / test_count
        print('%20s: %.2f usec/pass' % (method, per_pass))
        result_by_method[method] = per_pass
    return result_by_method

if __name__ == '__main__':
    if len(sys.argv) != 2:
        raise ValueError('Need a number of items to flatten')
    item_count = int(sys.argv[1])
    partition_counts = []
    pass_times_by_method = collections.defaultdict(list)
    for partition_count in xrange(1, item_count):
        if item_count % partition_count != 0:
            continue
        items_per_partition = item_count / partition_count
        result_by_method = time_test(partition_count, items_per_partition)
        partition_counts.append(partition_count)
        for method, result in result_by_method.iteritems():
            pass_times_by_method[method].append(result)
    for method, pass_times in pass_times_by_method.iteritems():
        pyplot.plot(partition_counts, pass_times, label=method)
    pyplot.legend()
    pyplot.title('Flattening Comparison for %d Items' % item_count)
    pyplot.xlabel('Number of Partitions')
    pyplot.ylabel('Microseconds')
    pyplot.show()

编辑:决定使其成为社区维基.

注意: METHODS应该与装饰者积累,但我认为人们更容易阅读这种方式.

Answer 4

sum(list_of_lists, []) 会使它变平.

l = [['image00', 'image01'], ['image10'], []]
print sum(l,[]) # prints ['image00', 'image01', 'image10']

Answer 5

此解决方案适用于任意嵌套深度 - 不仅仅是"列表列表"深度,其他解决方案的一些(全部？)仅限于:

def flatten(x):
    result = []
    for el in x:
        if hasattr(el, "__iter__") and not isinstance(el, basestring):
            result.extend(flatten(el))
        else:
            result.append(el)
    return result

它是允许任意深度嵌套的递归 - 直到你达到最大递归深度,当然......

Answer 6

表现结果.修订.

import itertools
def itertools_flatten( aList ):
    return list( itertools.chain(*aList) )

from operator import add
def reduce_flatten1( aList ):
    return reduce(add, map(lambda x: list(x), [mi for mi in aList]))

def reduce_flatten2( aList ):
    return reduce(list.__add__, map(list, aList))

def comprehension_flatten( aList ):
    return list(y for x in aList for y in x)

我将30个项目的2级列表展平1000次

itertools_flatten     0.00554
comprehension_flatten 0.00815
reduce_flatten2       0.01103
reduce_flatten1       0.01404

减少总是一个糟糕的选择.

Answer 7

在Python 2.6中,使用chain.from_iterable():

>>> from itertools import chain
>>> list(chain.from_iterable(mi.image_set.all() for mi in h.get_image_menu()))

它避免了创建中间列表.

Answer 8

似乎有一种混乱operator.add!当您将两个列表一起添加时,正确的术语是concat,而不是添加.operator.concat是你需要使用的.

如果您正在考虑功能,它就像这样简单::

>>> list2d = ((1,2,3),(4,5,6), (7,), (8,9))
>>> reduce(operator.concat, list2d)
(1, 2, 3, 4, 5, 6, 7, 8, 9)

你看到reduce尊重序列类型,所以当你提供一个元组时,你会得到一个元组.让我们尝试一下清单::

>>> list2d = [[1,2,3],[4,5,6], [7], [8,9]]
>>> reduce(operator.concat, list2d)
[1, 2, 3, 4, 5, 6, 7, 8, 9]

啊哈,你得到一份清单.

性能怎么样::

>>> list2d = [[1,2,3],[4,5,6], [7], [8,9]]
>>> %timeit list(itertools.chain.from_iterable(list2d))
1000000 loops, best of 3: 1.36 µs per loop

from_iterable非常快!但是用concat减少它是不可比的.

>>> list2d = ((1,2,3),(4,5,6), (7,), (8,9))
>>> %timeit reduce(operator.concat, list2d)
1000000 loops, best of 3: 492 ns per loop

Answer 9

在我的头顶,你可以消除lambda:

reduce(list.__add__, map(list, [mi.image_set.all() for mi in list_of_menuitems]))

或者甚至消除地图,因为你已经有了list-comp:

reduce(list.__add__, [list(mi.image_set.all()) for mi in list_of_menuitems])

您也可以将其表达为列表总和:

sum([list(mi.image_set.all()) for mi in list_of_menuitems], [])

Answer 10

这是使用列表推导的正确解决方案(它们在问题中是向后的):

>>> join = lambda it: (y for x in it for y in x)
>>> list(join([[1,2],[3,4,5],[]]))
[1, 2, 3, 4, 5]

在你的情况下,它会

[image for menuitem in list_of_menuitems for image in menuitem.image_set.all()]

或者你可以使用join和说

join(menuitem.image_set.all() for menuitem in list_of_menuitems)

在任何一种情况下,陷阱都是for循环的嵌套.