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

tamaño de una lista python

Question

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?)

Answer 1

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]]

Answer 2

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))

Answer 3

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)]

Answer 4

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.

Answer 5

>>> 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.