说我有一个列表，像这样：拆分列表为嵌套的列表上的值
[1, 4, None, 6, 9, None, 3, 9, 4 ]
我决定把它们分割嵌套的列表上None，得到这个：
[ [ 1, 4 ], [ 6, 9 ], [ 3, 9, 4 ] ]
中当然，我也一直想做到这一点就在这种情况下(9, None)，我们会得到：
[ [ 1, 4 ], [ 6 ], [ 3 ], [ 4 ] ]
这为t rivial使用列表附加迭代（在一个for循环）
我很想知道这是否可以做更快的事情 - 就像列表理解？
如果不是，为什么不（例如，列表理解不能返回每次迭代多个列表元素？）
>>> 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]]
基准码：上
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)
输出i3-530处理器，Windows7中的Python 3.1.2：
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
稍加修改Ryan的代码，希望大家不要介意。 ssplit是基于卡尔的想法。添加了一些处理特殊情况的语句，成为ssplit2，这是我可能提供的最佳解决方案。
+1优雅。 – 2010-12-01 09:27:01
很好！ +1 – mshsayem 2010-12-01 09:28:02
你可以找到“分隔符”元素的索引。例如，None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表示子列表的开始索引和结束索引的元组列表[ (0,1), (3,4), (6,8) ]。然后你可以使用列表理解来覆盖这个元组列表来提取子列表。
事情是这样的：
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]
打动了我：
>>> 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]]
与试图用一个列表理解这样做的问题是，理解本质上是无状态的，你需要的状态进行分裂操作。特别是，您需要记住，从一个元素到下一个元素，在先前的“分割”标记之后找到了哪些元素。
但是，您可以使用一个列表理解来提取拆分元素的索引，然后使用这些索引来切分列表。我们需要将分割指数转化为（开始，结束）指数以确定必要的“部分”。我们要做的是将分割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。
整个事情是这样的：
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)]
下面是一个使用减少的实现，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))
:-D。没有与它真的。我只是问自己这个问题，不能快速提出一些问题。 – PoorLuzer 2010-12-01 09:35:40