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
参见本参考
>>> orange = range(1, 1001)
>>> otuples = list( zip(*[iter(orange)]*10))
>>> print(otuples)
[(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), ... (991, 992, 993, 994, 995, 996, 997, 998, 999, 1000)]
>>> olist = [list(i) for i in otuples]
>>> print(olist)
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], ..., [991, 992, 993, 994, 995, 996, 997, 998, 999, 1000]]
>>>
蟒蛇3
其他回答
假设列表是第一个
import math
# length of the list len(lst) is ln
# size of a chunk is size
for num in range ( math.ceil(ln/size) ):
start, end = num*size, min((num+1)*size, ln)
print(lst[start:end])
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]
使用列表综合:
l = [1,2,3,4,5,6,7,8,9,10,11,12]
k = 5 #chunk size
print [tuple(l[x:y]) for (x, y) in [(x, x+k) for x in range(0, len(l), k)]]
呵呵,单行版本
In [48]: chunk = lambda ulist, step: map(lambda i: ulist[i:i+step], xrange(0, len(ulist), step))
In [49]: chunk(range(1,100), 10)
Out[49]:
[[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, 30],
[31, 32, 33, 34, 35, 36, 37, 38, 39, 40],
[41, 42, 43, 44, 45, 46, 47, 48, 49, 50],
[51, 52, 53, 54, 55, 56, 57, 58, 59, 60],
[61, 62, 63, 64, 65, 66, 67, 68, 69, 70],
[71, 72, 73, 74, 75, 76, 77, 78, 79, 80],
[81, 82, 83, 84, 85, 86, 87, 88, 89, 90],
[91, 92, 93, 94, 95, 96, 97, 98, 99]]
您可以使用numpy的array_split函数,例如np.array_split(np.array(data),20),将其拆分为20个大小几乎相等的块。
要确保块的大小完全相等,请使用np.split。