如何将任意长度的列表拆分为大小相等的块?
请参阅如果数据结果将直接用于循环,并且不需要存储,则如何以块形式遍历列表。
对于字符串输入的同一问题,请参见每n个字符拆分字符串?。相同的技术通常适用,但也有一些变化。
如何将任意长度的列表拆分为大小相等的块?
请参阅如果数据结果将直接用于循环,并且不需要存储,则如何以块形式遍历列表。
对于字符串输入的同一问题,请参见每n个字符拆分字符串?。相同的技术通常适用,但也有一些变化。
当前回答
任何可迭代的通用分块器,使用户可以选择如何在结尾处处理部分分块。
在Python 3上测试。
分块.py
from enum import Enum
class PartialChunkOptions(Enum):
INCLUDE = 0
EXCLUDE = 1
PAD = 2
ERROR = 3
class PartialChunkException(Exception):
pass
def chunker(iterable, n, on_partial=PartialChunkOptions.INCLUDE, pad=None):
"""
A chunker yielding n-element lists from an iterable, with various options
about what to do about a partial chunk at the end.
on_partial=PartialChunkOptions.INCLUDE (the default):
include the partial chunk as a short (<n) element list
on_partial=PartialChunkOptions.EXCLUDE
do not include the partial chunk
on_partial=PartialChunkOptions.PAD
pad to an n-element list
(also pass pad=<pad_value>, default None)
on_partial=PartialChunkOptions.ERROR
raise a RuntimeError if a partial chunk is encountered
"""
on_partial = PartialChunkOptions(on_partial)
iterator = iter(iterable)
while True:
vals = []
for i in range(n):
try:
vals.append(next(iterator))
except StopIteration:
if vals:
if on_partial == PartialChunkOptions.INCLUDE:
yield vals
elif on_partial == PartialChunkOptions.EXCLUDE:
pass
elif on_partial == PartialChunkOptions.PAD:
yield vals + [pad] * (n - len(vals))
elif on_partial == PartialChunkOptions.ERROR:
raise PartialChunkException
return
return
yield vals
测试.py
import chunker
chunk_size = 3
for it in (range(100, 107),
range(100, 109)):
print("\nITERABLE TO CHUNK: {}".format(it))
print("CHUNK SIZE: {}".format(chunk_size))
for option in chunker.PartialChunkOptions.__members__.values():
print("\noption {} used".format(option))
try:
for chunk in chunker.chunker(it, chunk_size, on_partial=option):
print(chunk)
except chunker.PartialChunkException:
print("PartialChunkException was raised")
print("")
test.py的输出
ITERABLE TO CHUNK: range(100, 107)
CHUNK SIZE: 3
option PartialChunkOptions.INCLUDE used
[100, 101, 102]
[103, 104, 105]
[106]
option PartialChunkOptions.EXCLUDE used
[100, 101, 102]
[103, 104, 105]
option PartialChunkOptions.PAD used
[100, 101, 102]
[103, 104, 105]
[106, None, None]
option PartialChunkOptions.ERROR used
[100, 101, 102]
[103, 104, 105]
PartialChunkException was raised
ITERABLE TO CHUNK: range(100, 109)
CHUNK SIZE: 3
option PartialChunkOptions.INCLUDE used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
option PartialChunkOptions.EXCLUDE used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
option PartialChunkOptions.PAD used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
option PartialChunkOptions.ERROR used
[100, 101, 102]
[103, 104, 105]
[106, 107, 108]
其他回答
我意识到这个问题已经过时了(在谷歌上被它绊倒了),但肯定像下面这样的问题比任何复杂的建议都要简单和清晰得多,而且只使用切片:
def chunker(iterable, chunksize):
for i,c in enumerate(iterable[::chunksize]):
yield iterable[i*chunksize:(i+1)*chunksize]
>>> for chunk in chunker(range(0,100), 10):
... print list(chunk)
...
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
[20, 21, 22, 23, 24, 25, 26, 27, 28, 29]
... etc ...
非常简单的事情:
def chunks(xs, n):
n = max(1, n)
return (xs[i:i+n] for i in range(0, len(xs), n))
对于Python 2,使用xrange()代替range()。
简单而优雅
L = range(1, 1000)
print [L[x:x+10] for x in xrange(0, len(L), 10)]
或者如果您愿意:
def chunks(L, n): return [L[x: x+n] for x in xrange(0, len(L), n)]
chunks(L, 10)
我很好奇不同方法的性能,这里是:
在Python 3.5.1上测试
import time
batch_size = 7
arr_len = 298937
#---------slice-------------
print("\r\nslice")
start = time.time()
arr = [i for i in range(0, arr_len)]
while True:
if not arr:
break
tmp = arr[0:batch_size]
arr = arr[batch_size:-1]
print(time.time() - start)
#-----------index-----------
print("\r\nindex")
arr = [i for i in range(0, arr_len)]
start = time.time()
for i in range(0, round(len(arr) / batch_size + 1)):
tmp = arr[batch_size * i : batch_size * (i + 1)]
print(time.time() - start)
#----------batches 1------------
def batch(iterable, n=1):
l = len(iterable)
for ndx in range(0, l, n):
yield iterable[ndx:min(ndx + n, l)]
print("\r\nbatches 1")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
tmp = x
print(time.time() - start)
#----------batches 2------------
from itertools import islice, chain
def batch(iterable, size):
sourceiter = iter(iterable)
while True:
batchiter = islice(sourceiter, size)
yield chain([next(batchiter)], batchiter)
print("\r\nbatches 2")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in batch(arr, batch_size):
tmp = x
print(time.time() - start)
#---------chunks-------------
def chunks(l, n):
"""Yield successive n-sized chunks from l."""
for i in range(0, len(l), n):
yield l[i:i + n]
print("\r\nchunks")
arr = [i for i in range(0, arr_len)]
start = time.time()
for x in chunks(arr, batch_size):
tmp = x
print(time.time() - start)
#-----------grouper-----------
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)
def grouper(iterable, n, padvalue=None):
"grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)
arr = [i for i in range(0, arr_len)]
print("\r\ngrouper")
start = time.time()
for x in grouper(arr, batch_size):
tmp = x
print(time.time() - start)
结果:
slice
31.18285083770752
index
0.02184295654296875
batches 1
0.03503894805908203
batches 2
0.22681021690368652
chunks
0.019841909408569336
grouper
0.006506919860839844
我在这个问题的副本中看到了最棒的Python式答案:
from itertools import zip_longest
a = range(1, 16)
i = iter(a)
r = list(zip_longest(i, i, i))
>>> print(r)
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)]
您可以为任何n创建n元组。如果a=范围(1,15),那么结果将是:
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, None)]
如果列表被平均划分,那么可以用zip替换zip_langest,否则三元组(13、14、None)将丢失。上面使用了Python 3。对于Python 2,请使用izip_length。