将列表拆分为值上的嵌套列表
说我有一个像这样的列表：
[1, 4, None, 6, 9, None, 3, 9, 4 ] 我决定把它分成None嵌套列表，得到这个：
[ [ 1, 4 ], [ 6, 9 ], [ 3, 9, 4 ] ]
当然，我本来可以这样做(9, None)在这种情况下，我们会得到：
[ [ 1, 4 ], [ 6 ], [ 3 ], [ 4 ] ]
这是通过迭代使用列表追加（在for循环中）
我很有兴趣知道这是否能以更快的速度完成 – 就像列表理解一样？
如果没有，为什么不（例如，列表理解每次迭代都不能返回多个列表元素？）
>>> 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)
输出在i3-530，Windows7，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
略微修改Ryan的代码，希望你不介意。 斯普利特是基于卡尔的想法。 增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好的解决scheme。
像这样的东西：
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]
让我：
>>> 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]]
你可以find“分隔符”元素的索引。 例如，None的索引是2和5（从零开始）。 您可以使用这些列表以及列表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) ] 。 然后你可以使用这个列表的元组列表来提取子列表。
试图使用列表理解的问题是，理解本质上是无状态的，你需要状态来执行分裂操作。 尤其是，您需要记住，从一个元素到下一个元素，在之前的“分割”标记之后find了什么元素。
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割列表。 我们需要将分割指数翻译成（开始，结束）指数，用于必要的“件”。 我们要做的是将分割索引列表转换成两个单独的“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)]
这里有一个使用reduce的实现，有一些花里胡哨的function：
#!/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))
