@。洛特:你启发我写了一个timeit应用程序。

我认为它也会根据分区的数量(容器列表中的迭代器的数量)而变化——你的评论没有提到这30个项目中有多少个分区。这个图在每次运行中平摊1000个项目，使用不同数量的分区。这些物品均匀地分布在各个分区中。

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:决定让它成为社区维基。

注意:METHODS可能应该使用装饰器进行积累，但我认为这样更容易让人们阅读。

如果列表中的每个项都是字符串(并且这些字符串中的任何字符串都使用" "而不是' ')，则可以使用正则表达式(re module)

>>> flattener = re.compile("\'.*?\'")
>>> flattener
<_sre.SRE_Pattern object at 0x10d439ca8>
>>> stred = str(in_list)
>>> outed = flattener.findall(stred)

上面的代码将in_list转换为字符串，使用regex找到引号内的所有子字符串(即列表中的每一项)，并将它们作为列表输出。

Pylab提供了一个flatten: 链接到numpy flatten

这个版本是一个生成器。如果你想要一个列表，可以调整一下。

def list_or_tuple(l):
    return isinstance(l,(list,tuple))
## predicate will select the container  to be flattened
## write your own as required
## this one flattens every list/tuple


def flatten(seq,predicate=list_or_tuple):        
    ## recursive generator 
    for i in seq:
        if predicate(seq):
            for j in flatten(i):
                yield j
        else:
            yield i

如果想将满足条件的谓词平展，可以添加谓词

摘自python食谱

根据我的经验，最有效的将列表的列表扁平化的方法是:

flat_list = []
map(flat_list.extend, list_of_list)

有时与其他提出的方法进行比较:

list_of_list = [range(10)]*1000
%timeit flat_list=[]; map(flat_list.extend, list_of_list)
#10000 loops, best of 3: 119 µs per loop
%timeit flat_list=list(itertools.chain.from_iterable(list_of_list))
#1000 loops, best of 3: 210 µs per loop
%timeit flat_list=[i for sublist in list_of_list for i in sublist]
#1000 loops, best of 3: 525 µs per loop
%timeit flat_list=reduce(list.__add__,list_of_list)
#100 loops, best of 3: 18.1 ms per loop

现在，当处理更长的子列表时，效率增益会更好:

list_of_list = [range(1000)]*10
%timeit flat_list=[]; map(flat_list.extend, list_of_list)
#10000 loops, best of 3: 60.7 µs per loop
%timeit flat_list=list(itertools.chain.from_iterable(list_of_list))
#10000 loops, best of 3: 176 µs per loop

这个方法也适用于任何迭代对象:

class SquaredRange(object):
    def __init__(self, n): 
        self.range = range(n)
    def __iter__(self):
        for i in self.range: 
            yield i**2

list_of_list = [SquaredRange(5)]*3
flat_list = []
map(flat_list.extend, list_of_list)
print flat_list
#[0, 1, 4, 9, 16, 0, 1, 4, 9, 16, 0, 1, 4, 9, 16]

下面是一个使用集合处理多层列表的版本。Iterable:

import collections

def flatten(o, flatten_condition=lambda i: isinstance(i,
               collections.Iterable) and not isinstance(i, str)):
    result = []
    for i in o:
        if flatten_condition(i):
            result.extend(flatten(i, flatten_condition))
        else:
            result.append(i)
    return result