@。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。

是什么:

from operator import add
reduce(add, map(lambda x: list(x.image_set.all()), [mi for mi in list_of_menuitems]))

但是，Guido建议不要在一行代码中执行太多操作，因为这会降低可读性。在单行中执行与在多行中执行您想要的操作相比，性能收益最小(如果有的话)。

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()))

避免了中间列表的创建。

你试过扁平化吗? 从matplotlib.cbook。Flatten (seq, scalarp=) ?

l=[[1,2,3],[4,5,6], [7], [8,9]]*33

run("list(flatten(l))")
         3732 function calls (3303 primitive calls) in 0.007 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.007    0.007 <string>:1(<module>)
      429    0.001    0.000    0.001    0.000 cbook.py:475(iterable)
      429    0.002    0.000    0.003    0.000 cbook.py:484(is_string_like)
      429    0.002    0.000    0.006    0.000 cbook.py:565(is_scalar_or_string)
  727/298    0.001    0.000    0.007    0.000 cbook.py:605(flatten)
      429    0.000    0.000    0.001    0.000 core.py:5641(isMaskedArray)
      858    0.001    0.000    0.001    0.000 {isinstance}
      429    0.000    0.000    0.000    0.000 {iter}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



l=[[1,2,3],[4,5,6], [7], [8,9]]*66

run("list(flatten(l))")
         7461 function calls (6603 primitive calls) in 0.007 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.007    0.007 <string>:1(<module>)
      858    0.001    0.000    0.001    0.000 cbook.py:475(iterable)
      858    0.002    0.000    0.003    0.000 cbook.py:484(is_string_like)
      858    0.002    0.000    0.006    0.000 cbook.py:565(is_scalar_or_string)
 1453/595    0.001    0.000    0.007    0.000 cbook.py:605(flatten)
      858    0.000    0.000    0.001    0.000 core.py:5641(isMaskedArray)
     1716    0.001    0.000    0.001    0.000 {isinstance}
      858    0.000    0.000    0.000    0.000 {iter}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



l=[[1,2,3],[4,5,6], [7], [8,9]]*99

run("list(flatten(l))")
         11190 function calls (9903 primitive calls) in 0.010 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.010    0.010 <string>:1(<module>)
     1287    0.002    0.000    0.002    0.000 cbook.py:475(iterable)
     1287    0.003    0.000    0.004    0.000 cbook.py:484(is_string_like)
     1287    0.002    0.000    0.009    0.000 cbook.py:565(is_scalar_or_string)
 2179/892    0.001    0.000    0.010    0.000 cbook.py:605(flatten)
     1287    0.001    0.000    0.001    0.000 core.py:5641(isMaskedArray)
     2574    0.001    0.000    0.001    0.000 {isinstance}
     1287    0.000    0.000    0.000    0.000 {iter}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



l=[[1,2,3],[4,5,6], [7], [8,9]]*132

run("list(flatten(l))")
         14919 function calls (13203 primitive calls) in 0.013 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.013    0.013 <string>:1(<module>)
     1716    0.002    0.000    0.002    0.000 cbook.py:475(iterable)
     1716    0.004    0.000    0.006    0.000 cbook.py:484(is_string_like)
     1716    0.003    0.000    0.011    0.000 cbook.py:565(is_scalar_or_string)
2905/1189    0.002    0.000    0.013    0.000 cbook.py:605(flatten)
     1716    0.001    0.000    0.001    0.000 core.py:5641(isMaskedArray)
     3432    0.001    0.000    0.001    0.000 {isinstance}
     1716    0.001    0.000    0.001    0.000 {iter}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler'

更新 这给了我另一个想法:

l=[[1,2,3],[4,5,6], [7], [8,9]]*33

run("flattenlist(l)")
         564 function calls (432 primitive calls) in 0.000 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    133/1    0.000    0.000    0.000    0.000 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.000    0.000    0.000    0.000 <string>:1(<module>)
      429    0.000    0.000    0.000    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



l=[[1,2,3],[4,5,6], [7], [8,9]]*66

run("flattenlist(l)")
         1125 function calls (861 primitive calls) in 0.001 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    265/1    0.001    0.000    0.001    0.001 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.000    0.000    0.001    0.001 <string>:1(<module>)
      858    0.000    0.000    0.000    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



l=[[1,2,3],[4,5,6], [7], [8,9]]*99

run("flattenlist(l)")
         1686 function calls (1290 primitive calls) in 0.001 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    397/1    0.001    0.000    0.001    0.001 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.000    0.000    0.001    0.001 <string>:1(<module>)
     1287    0.000    0.000    0.000    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



l=[[1,2,3],[4,5,6], [7], [8,9]]*132

run("flattenlist(l)")
         2247 function calls (1719 primitive calls) in 0.002 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    529/1    0.001    0.000    0.002    0.002 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.000    0.000    0.002    0.002 <string>:1(<module>)
     1716    0.001    0.000    0.001    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



l=[[1,2,3],[4,5,6], [7], [8,9]]*1320

run("flattenlist(l)")
         22443 function calls (17163 primitive calls) in 0.016 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
   5281/1    0.011    0.000    0.016    0.016 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.000    0.000    0.016    0.016 <string>:1(<module>)
    17160    0.005    0.000    0.005    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

因此，为了测试当递归深入时它的有效性:深入多少?

l=[[1,2,3],[4,5,6], [7], [8,9]]*1320

new=[l]*33

run("flattenlist(new)")
         740589 function calls (566316 primitive calls) in 0.418 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
 174274/1    0.281    0.000    0.417    0.417 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.001    0.001    0.418    0.418 <string>:1(<module>)
   566313    0.136    0.000    0.136    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



new=[l]*66

run("flattenlist(new)")
         1481175 function calls (1132629 primitive calls) in 0.809 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
 348547/1    0.542    0.000    0.807    0.807 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.002    0.002    0.809    0.809 <string>:1(<module>)
  1132626    0.266    0.000    0.266    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



new=[l]*99

run("flattenlist(new)")
         2221761 function calls (1698942 primitive calls) in 1.211 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
 522820/1    0.815    0.000    1.208    1.208 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.002    0.002    1.211    1.211 <string>:1(<module>)
  1698939    0.393    0.000    0.393    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



new=[l]*132

run("flattenlist(new)")
         2962347 function calls (2265255 primitive calls) in 1.630 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
 697093/1    1.091    0.000    1.627    1.627 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.003    0.003    1.630    1.630 <string>:1(<module>)
  2265252    0.536    0.000    0.536    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}



new=[l]*1320

run("flattenlist(new)")
         29623443 function calls (22652523 primitive calls) in 16.103 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
6970921/1   10.842    0.000   16.069   16.069 <ipython-input-55-39b139bad497>:4(flattenlist)
        1    0.034    0.034   16.103   16.103 <string>:1(<module>)
 22652520    5.227    0.000    5.227    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

我打赌“扁平化”我将使用这个而不是matplobib很长很长一段时间，除非我想要一个yield生成器和快速的结果“扁平化”使用matplobib .cbook

这个，是快的。

这是代码

:

typ=(list,tuple)


def flattenlist(d):
    thelist = []
    for x in d:
        if not isinstance(x,typ):
            thelist += [x]
        else:
            thelist += flattenlist(x)
    return thelist

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

这个解决方案适用于任意的嵌套深度——而不仅仅是一些(所有?)其他解决方案所限制的“列表的列表”深度:

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

它是递归，允许任意深度嵌套-直到你达到最大递归深度，当然…