将列表拆分为值上的嵌套列表

Question

假设我有一个这样的列表：

[1, 4, None, 6, 9, None, 3, 9, 4]

我决定将其拆分为 None 上的嵌套列表，以获得此：

[[1, 4], [6, 9], [3, 9, 4]]

当然，我本可以在 (9, None) 上执行此操作，在这种情况下，我们会得到：

[[1, 4], [6], [3], [4]]

通过迭代（在 for 循环中）使用列表追加很简单

我很想知道这是否可以更快地完成——比如列表理解？

如果不是，为什么不呢？ （例如，列表推导式每次迭代不能返回多个列表元素？）

Answer 1

>>> def isplit(iterable,splitters):
    return [list(g) for k,g in itertools.groupby(iterable,lambda x:x in splitters) if not k]

>>> isplit(L,(None,))
[[1, 4], [6, 9], [3, 9, 4]]
>>> isplit(L,(None,9))
[[1, 4], [6], [3], [4]]

基准代码：

import timeit    

kabie=("isplit_kabie",
"""
import itertools
def isplit_kabie(iterable,splitters):
    return [list(g) for k,g in itertools.groupby(iterable,lambda x:x in splitters) if not k]
""" )

ssplit=("ssplit",
"""
def ssplit(seq,splitters):
    seq=list(seq)
    if splitters and seq:
        result=[]
        begin=0
        for end in range(len(seq)):
            if seq[end] in splitters:
                if end > begin:
                    result.append(seq[begin:end])
                begin=end+1
        if begin<len(seq):
            result.append(seq[begin:])
        return result
    return [seq]
""" )

ssplit2=("ssplit2",
"""
def ssplit2(seq,splitters):
    seq=list(seq)
    if splitters and seq:
        splitters=set(splitters).intersection(seq)
        if splitters:
            result=[]
            begin=0
            for end in range(len(seq)):
                if seq[end] in splitters:
                    if end > begin:
                        result.append(seq[begin:end])
                    begin=end+1
            if begin<len(seq):
                result.append(seq[begin:])
            return result
    return [seq]
""" )

emile=("magicsplit",
"""
def _itersplit(l, *splitters):
    current = []
    for item in l:
        if item in splitters:
            yield current
            current = []
        else:
            current.append(item)
    yield current

def magicsplit(l, splitters):
    return [subl for subl in _itersplit(l, *splitters) if subl]
""" )

emile_improved=("magicsplit2",
"""
def _itersplit(l, *splitters):
    current = []
    for item in l:
        if item in splitters:
            if current:
                yield current
                current = []
        else:
            current.append(item)
    if current:
        yield current

def magicsplit2(l, splitters):
    if splitters and l:
        return [i for i in _itersplit(l, *splitters)]
    return [list(l)]
""" )

karl=("ssplit_karl",
"""
def ssplit_karl(original,splitters):
    indices = [i for (i, x) in enumerate(original) if x in splitters]
    ends = indices + [len(original)]
    begins = [0] + [x + 1 for x in indices]
    return [original[begin:end] for (begin, end) in zip(begins, ends)]
""" )

ryan=("split_on",
"""
from functools import reduce
def split_on (seq, delims, remove_empty=True):
    '''Split seq into lists using delims as a delimiting elements.

    For example, split_on(delims=2, list=xrange(0,5)) yields [ [0,1], [3,4] ].

    delims can be either a single delimiting element or a list or
    tuple of multiple delimiting elements. If you wish to use a list
    or tuple as a delimiter, you must enclose it in another list or
    tuple.

    If remove_empty is False, then consecutive delimiter elements or delimiter elements at the beginning or end of the longlist'''
    delims=set(delims)
    def reduce_fun(lists, elem):
        if elem in delims:
            if remove_empty and lists[-1] == []:
                # Avoid adding multiple empty lists
                pass
            else:
                lists.append([])
        else:
            lists[-1].append(elem)
        return lists
    result_list = reduce(reduce_fun, seq, [ [], ])
    # Maybe remove trailing empty list
    if remove_empty and result_list[-1] == []:
        result_list.pop()
    return result_list
""" )

cases=(kabie, emile, emile_improved, ssplit ,ssplit2 ,ryan)

data=(
    ([1, 4, None, 6, 9, None, 3, 9, 4 ],(None,)),
    ([1, 4, None, 6, 9, None, 3, 9, 4 ]*5,{None,9,7}),
    ((),()),
    (range(1000),()),
    ("Split me",('','')),
    ("split me "*100,' '),
    ("split me,"*100,' ,'*20),
    ("split me, please!"*100,' ,!'),
    (range(100),range(100)),
    (range(100),range(101,1000)),
    (range(100),range(50,150)),
    (list(range(100))*30,(99,)),
    )

params="seq,splitters"

def benchmark(func,code,data,params='',times=10000,rounds=3,debug=''):
    assert(func.isidentifier())
    tester = timeit.Timer(stmt='{func}({params})'.format(
                                func=func,params=params),
                          setup="{code}\n".format(code=code)+
            (params and "{params}={data}\n".format(params=params,data=data)) +
            (debug and """ret=repr({func}({params}))
print({func}.__name__.rjust(16),":",ret[:30]+"..."+ret[-15:] if len(ret)>50 else ret)
                       """.format(func=func,params=params)))
    results = [tester.timeit(times) for i in range(rounds)]
    if not debug:
        print("{:>16s} takes:{:6.4f},avg:{:.2e},best:{:.4f},worst:{:.4f}".format(
            func,sum(results),sum(results)/times/rounds,min(results),max(results)))

def testAll(cases,data,params='',times=10000,rounds=3,debug=''):
    if debug:
        times,rounds = 1,1
    for dat in data:
        sdat = tuple(map(repr,dat))
        print("{}x{} times:".format(times,rounds),
              ','.join("{}".format(d[:8]+"..."+d[-5:] if len(d)>16 else d)for d in map(repr,dat)))
        for func,code in cases:
            benchmark(func,code,dat,params,times,rounds,debug)

if __name__=='__main__':
    testAll(cases,data,params,500,10)#,debug=True)

在 i3-530、Windows7、Python 3.1.2 上的输出：

500x10 times: [1, 4, N...9, 4],(None,)
    isplit_kabie takes:0.0605,avg:1.21e-05,best:0.0032,worst:0.0074
      magicsplit takes:0.0287,avg:5.74e-06,best:0.0016,worst:0.0036
     magicsplit2 takes:0.0174,avg:3.49e-06,best:0.0017,worst:0.0018
          ssplit takes:0.0149,avg:2.99e-06,best:0.0015,worst:0.0016
         ssplit2 takes:0.0198,avg:3.96e-06,best:0.0019,worst:0.0021
        split_on takes:0.0229,avg:4.59e-06,best:0.0023,worst:0.0024
500x10 times: [1, 4, N...9, 4],{9, None, 7}
    isplit_kabie takes:0.1448,avg:2.90e-05,best:0.0144,worst:0.0146
      magicsplit takes:0.0636,avg:1.27e-05,best:0.0063,worst:0.0065
     magicsplit2 takes:0.0891,avg:1.78e-05,best:0.0064,worst:0.0162
          ssplit takes:0.0593,avg:1.19e-05,best:0.0058,worst:0.0061
         ssplit2 takes:0.1004,avg:2.01e-05,best:0.0069,worst:0.0142
        split_on takes:0.0929,avg:1.86e-05,best:0.0090,worst:0.0096
500x10 times: (),()
    isplit_kabie takes:0.0041,avg:8.14e-07,best:0.0004,worst:0.0004
      magicsplit takes:0.0040,avg:8.04e-07,best:0.0004,worst:0.0004
     magicsplit2 takes:0.0022,avg:4.35e-07,best:0.0002,worst:0.0002
          ssplit takes:0.0023,avg:4.59e-07,best:0.0002,worst:0.0003
         ssplit2 takes:0.0023,avg:4.53e-07,best:0.0002,worst:0.0002
        split_on takes:0.0072,avg:1.45e-06,best:0.0007,worst:0.0009
500x10 times: range(0, 1000),()
    isplit_kabie takes:0.8892,avg:1.78e-04,best:0.0881,worst:0.0895
      magicsplit takes:0.6614,avg:1.32e-04,best:0.0654,worst:0.0673
     magicsplit2 takes:0.0958,avg:1.92e-05,best:0.0094,worst:0.0099
          ssplit takes:0.0943,avg:1.89e-05,best:0.0093,worst:0.0095
         ssplit2 takes:0.0943,avg:1.89e-05,best:0.0093,worst:0.0096
        split_on takes:1.3348,avg:2.67e-04,best:0.1328,worst:0.1340
500x10 times: 'Split me',('', '')
    isplit_kabie takes:0.0234,avg:4.68e-06,best:0.0023,worst:0.0024
      magicsplit takes:0.0126,avg:2.52e-06,best:0.0012,worst:0.0013
     magicsplit2 takes:0.0138,avg:2.76e-06,best:0.0013,worst:0.0015
          ssplit takes:0.0119,avg:2.39e-06,best:0.0012,worst:0.0012
         ssplit2 takes:0.0075,avg:1.50e-06,best:0.0007,worst:0.0008
        split_on takes:0.0191,avg:3.83e-06,best:0.0018,worst:0.0023
500x10 times: 'split m... me ',' '
    isplit_kabie takes:2.0803,avg:4.16e-04,best:0.2060,worst:0.2098
      magicsplit takes:0.9219,avg:1.84e-04,best:0.0920,worst:0.0925
     magicsplit2 takes:1.0221,avg:2.04e-04,best:0.1018,worst:0.1034
          ssplit takes:0.8294,avg:1.66e-04,best:0.0818,worst:0.0834
         ssplit2 takes:0.9911,avg:1.98e-04,best:0.0983,worst:0.1014
        split_on takes:1.5672,avg:3.13e-04,best:0.1543,worst:0.1694
500x10 times: 'split m... me,',' , , , ... , ,'
    isplit_kabie takes:2.1847,avg:4.37e-04,best:0.2164,worst:0.2275
      magicsplit takes:3.7135,avg:7.43e-04,best:0.3693,worst:0.3783
     magicsplit2 takes:3.8104,avg:7.62e-04,best:0.3795,worst:0.3884
          ssplit takes:0.9522,avg:1.90e-04,best:0.0939,worst:0.0956
         ssplit2 takes:1.0140,avg:2.03e-04,best:0.1009,worst:0.1023
        split_on takes:1.5747,avg:3.15e-04,best:0.1563,worst:0.1615
500x10 times: 'split m...ase!',' ,!'
    isplit_kabie takes:3.3443,avg:6.69e-04,best:0.3324,worst:0.3380
      magicsplit takes:2.0594,avg:4.12e-04,best:0.2054,worst:0.2076
     magicsplit2 takes:2.1850,avg:4.37e-04,best:0.2180,worst:0.2191
          ssplit takes:1.4881,avg:2.98e-04,best:0.1484,worst:0.1493
         ssplit2 takes:1.8779,avg:3.76e-04,best:0.1868,worst:0.1920
        split_on takes:2.9596,avg:5.92e-04,best:0.2946,worst:0.2980
500x10 times: range(0, 100),range(0, 100)
    isplit_kabie takes:0.9445,avg:1.89e-04,best:0.0933,worst:0.1023
      magicsplit takes:0.5878,avg:1.18e-04,best:0.0583,worst:0.0593
     magicsplit2 takes:0.5597,avg:1.12e-04,best:0.0554,worst:0.0588
          ssplit takes:0.8568,avg:1.71e-04,best:0.0852,worst:0.0874
         ssplit2 takes:0.1399,avg:2.80e-05,best:0.0121,worst:0.0242
        split_on takes:0.1462,avg:2.92e-05,best:0.0145,worst:0.0148
500x10 times: range(0, 100),range(101, 1000)
    isplit_kabie takes:19.9749,avg:3.99e-03,best:1.9789,worst:2.0330
      magicsplit takes:9.4997,avg:1.90e-03,best:0.9369,worst:0.9640
     magicsplit2 takes:9.4394,avg:1.89e-03,best:0.9267,worst:0.9665
          ssplit takes:19.2363,avg:3.85e-03,best:1.8936,worst:1.9516
         ssplit2 takes:0.2032,avg:4.06e-05,best:0.0201,worst:0.0205
        split_on takes:0.3329,avg:6.66e-05,best:0.0323,worst:0.0344
500x10 times: range(0, 100),range(50, 150)
    isplit_kabie takes:1.1394,avg:2.28e-04,best:0.1130,worst:0.1153
      magicsplit takes:0.7288,avg:1.46e-04,best:0.0721,worst:0.0760
     magicsplit2 takes:0.7220,avg:1.44e-04,best:0.0705,worst:0.0774
          ssplit takes:1.0835,avg:2.17e-04,best:0.1059,worst:0.1116
         ssplit2 takes:0.1092,avg:2.18e-05,best:0.0105,worst:0.0116
        split_on takes:0.1639,avg:3.28e-05,best:0.0162,worst:0.0168
500x10 times: [0, 1, 2..., 99],(99,)
    isplit_kabie takes:3.2579,avg:6.52e-04,best:0.3225,worst:0.3360
      magicsplit takes:2.2937,avg:4.59e-04,best:0.2274,worst:0.2344
     magicsplit2 takes:2.6054,avg:5.21e-04,best:0.2587,worst:0.2642
          ssplit takes:1.5251,avg:3.05e-04,best:0.1495,worst:0.1729
         ssplit2 takes:1.7298,avg:3.46e-04,best:0.1696,worst:0.1858
        split_on takes:4.1041,avg:8.21e-04,best:0.4033,worst:0.4291

稍微修改了 Ryan 的代码，希望你不要介意。 ssplit 是基于 Karl 的想法。添加了处理一些特殊情况的语句成为 ssplit2，这是我可能提供的最佳解决方案。

Answer 2

这是一个使用 reduce 的实现，有一些花里胡哨：

#!/usr/bin/env python

def split_on (delims, seq, remove_empty=True):
    '''Split seq into lists using delims as a delimiting elements.

    For example, split_on(delims=2, list=xrange(0,5)) yields [ [0,1], [3,4] ].

    delims can be either a single delimiting element or a list or
    tuple of multiple delimiting elements. If you wish to use a list
    or tuple as a delimiter, you must enclose it in another list or
    tuple.

    If remove_empty is False, then consecutive delimiter elements or delimiter elements at the beginning or end of the longlist'''
    if type(delims) not in (type(list()), type(tuple())):
        delims = ( delims, )
    def reduce_fun(lists, elem):
        if elem in delims:
            if remove_empty and lists[-1] == []:
                # Avoid adding multiple empty lists
                pass
            else:
                lists.append([])
        else:
            lists[-1].append(elem)
        return lists
    result_list = reduce(reduce_fun, seq, [ [], ])
    # Maybe remove trailing empty list
    if remove_empty and result_list[-1] == []:
        result_list.pop()
    return result_list

mylist = [1, 4, None, 6, 9, None, 3, 9, 4 ]

print(split_on(None, mylist))
print(split_on((9, None), mylist))

Answer 3

尝试对此使用列表推导的问题在于推导本质上是无状态的，您需要状态来执行拆分操作。特别是，您需要记住，从一个元素到下一个元素，在前一个“拆分”标记之后找到了哪些元素。

但是，您可以使用一个列表推导式来提取拆分元素的索引，然后使用另一个列表推导式来使用这些索引来拆分列表。我们需要将拆分索引转换为必要的“片段”的（开始，结束）索引。我们要做的是将拆分索引列表转换为两个单独的“开始”和“结束”列表，然后将它们压缩在一起。

整个事情看起来像：

splitters = (9, None) # for example
indices = [i for (i, x) in enumerate(original) if x in splitters]
ends = indices + [len(original)]
begins = [0] + [x + 1 for x in indices]
result = [original[begin:end] for (begin, end) in zip(begins, ends)]

Answer 4

类似这样的：

def _itersplit(l, splitters):
    current = []
    for item in l:
        if item in splitters:
            yield current
            current = []
        else:
            current.append(item)
    yield current

def magicsplit(l, *splitters):
    return [subl for subl in _itersplit(l, splitters) if subl]

得到我：

>>> l = [1, 4, None, 6, 9, None, 3, 9, 4 ]
>>> magicsplit(l, None)
[[1, 4], [6, 9], [3, 9, 4]]
>>> magicsplit(l, None, 9)
[[1, 4], [6], [3], [4]]

Answer 5

您可以找到“分隔符”元素的索引。例如，无的索引是 2 和 5（从零开始）。 您可以使用这些以及列表的长度来构造一个元组列表[ (0,1), (3,4), (6,8) ]，表示子列表的开始索引和结束索引。然后，您可以对这个元组列表使用列表推导来提取子列表。