una tuplas tamaño misma metodos llenar listas lista eliminar elementos comparar anidadas agregar python list-comprehension nested-lists

python - tuplas - Divida una lista en listas anidadas en un valor



tamaño de una lista python (5)

Digamos que tengo una lista como esta:

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

Decido dividir esto en listas anidadas en None , para obtener esto:

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

Por supuesto, podría haber querido hacer esto el (9, None) en cuyo caso, tendríamos:

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

Esto es trivial de hacer usando lista anexar a través de iteración (en un ciclo for)

Me interesa saber si esto se puede hacer en algo más rápido, como una lista de comprensión.

Si no, ¿por qué no (por ejemplo, una lista de comprensión no puede devolver más de un elemento de lista por iteración?)


Algo como esto:

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]

me atrapa:

>>> 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]]


Aquí hay una implementación que usa reducir, con algunas sonrisas y silbidos:

#!/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))


El problema al tratar de usar una lista de comprensión para esto es que la comprensión es inherentemente sin estado, y usted necesita un estado para realizar la operación de división. En particular, debe recordar, de un elemento al siguiente, qué elementos se encontraron después del marcador de "división" anterior.

Sin embargo, puede usar una lista de comprensión para extraer los índices de los elementos divididos, y luego otra para usar esos índices para cortar la lista. Necesitamos traducir los índices divididos en índices (inicio, fin) para las ''piezas'' necesarias. Lo que haremos será transformar la lista de índices divididos en dos listas separadas de ''comienza'' y ''finaliza'', y luego los comprimiremos juntos.

Todo se ve así:

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)]


Puede encontrar los índices de los elementos "delimitadores". Por ejemplo, los índices de Ninguno son 2 y 5 (basados ​​en cero). Puede usar estos, junto con la longitud de la lista, para construir una lista de tuplas [ (0,1), (3,4), (6,8) ] representan los índices de inicio y los índices finales de las sublistas. Luego puede usar una lista de comprensión sobre esta lista de tuplas para extraer sublistas.


>>> 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]]

código de referencia:

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)

Salida en 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

Código de Ryan ligeramente modificado, espero que no te importe. ssplit se basó en la idea de Karl. Las declaraciones agregadas que manejan algunos casos especiales se convierten en ssplit2, que es la mejor solución que puedo proporcionar.