Python 如何将列表拆分为大小相等的块？
我有一个任意长度的列表，我需要将它分成大小相等的块，并对其进行操作。有一些明显的方法可以做到这一点，比如保留一个计数器和两个列表，当第二个列表填满时，将其添加到第一个列表中，并清空第二个列表以获得下一轮数据，但这可能非常昂贵 我想知道是否有人对任何长度的列表都有很好的解决方案，例如使用生成器 我在Python 如何将列表拆分为大小相等的块？,python,list,split,chunks,Python,List,Split,Chunks,我有一个任意长度的列表，我需要将它分成大小相等的块，并对其进行操作。有一些明显的方法可以做到这一点，比如保留一个计数器和两个列表，当第二个列表填满时，将其添加到第一个列表中，并清空第二个列表以获得下一轮数据，但这可能非常昂贵 我想知道是否有人对任何长度的列表都有很好的解决方案，例如使用生成器 我在itertools中寻找有用的东西，但找不到任何明显有用的东西。但我可能错过了 相关问题：这里有一个生成器，可以生成您想要的块： def chunks(lst, n): """Yield suc
itertools相关问题：这里有一个生成器，可以生成您想要的块：
def chunks(lst, n):
"""Yield successive n-sized chunks from lst."""
for i in range(0, len(lst), n):
yield lst[i:i + n]
如果您使用的是Python 2，那么应该使用
xrange（）range（）此外，您还可以简单地使用列表理解而不是编写函数，不过最好在命名函数中封装这样的操作，以便代码更易于理解。Python 3：
[lst[i:i + n] for i in range(0, len(lst), n)]
[lst[i:i + n] for i in xrange(0, len(lst), n)]
def SplitList(mylist, chunk_size):
return [mylist[offs:offs+chunk_size] for offs in range(0, len(mylist), chunk_size)]
在后一种情况下，如果您可以确保序列始终包含给定大小的整数块（即没有不完整的最后一个块），则可以用更漂亮的方式对其进行重新表述。这里有一个生成器，可用于任意iterables：
def split_seq(iterable, size):
it = iter(iterable)
item = list(itertools.islice(it, size))
while item:
yield item
item = list(itertools.islice(it, size))
def isplitter(l, n):
l = iter(l)
chunk = list(islice(l, n))
while chunk:
yield chunk
chunk = list(islice(l, n))
>>> import pprint
>>> pprint.pprint(list(split_seq(xrange(75), 10)))
[[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],
[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]]
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]]
#from itertools import izip_longest as zip_longest # for Python 2.x
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)
def grouper(n, iterable, padvalue=None):
"grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)
[iter（iterable）]*nn次izip_longest
然后有效地执行“每个”迭代器的循环；因为这是同一个迭代器，所以每次这样的调用都会对它进行改进，从而导致每次这样的zip循环生成一个n
项的元组
def split_seq(seq, num_pieces):
start = 0
for i in xrange(num_pieces):
stop = start + len(seq[i::num_pieces])
yield seq[start:stop]
start = stop
用法：
seq = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for seq in split_seq(seq, 3):
print seq
如果你想要超简单的东西：
def chunks(l, n):
n = max(1, n)
return (l[i:i+n] for i in range(0, len(l), n))
在Python 2.x的情况下，使用xrange（）
而不是range（）
，而不调用len（），这对大型列表很有用：
def splitter(l, n):
i = 0
chunk = l[:n]
while chunk:
yield chunk
i += n
chunk = l[i:i+n]
这是为伊特拉伯雷人准备的：
def split_seq(iterable, size):
it = iter(iterable)
item = list(itertools.islice(it, size))
while item:
yield item
item = list(itertools.islice(it, size))
def isplitter(l, n):
l = iter(l)
chunk = list(islice(l, n))
while chunk:
yield chunk
chunk = list(islice(l, n))
上述产品的功能性风味：
def isplitter2(l, n):
return takewhile(bool,
(tuple(islice(start, n))
for start in repeat(iter(l))))
或：
或：
素雅
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)
例如，如果块大小为3，则可以执行以下操作：
zip(*[iterable[i::3] for i in range(3)])
资料来源：
当我的区块大小是我可以输入的固定数字时，我会使用它，例如“3”，并且永远不会改变。考虑使用片段
例如：
import matplotlib.cbook as cbook
segments = cbook.pieces(np.arange(20), 3)
for s in segments:
print s
>>> AA=range(10,21);SS=3
>>> [AA[i:i+SS] for i in range(len(AA))[::SS]]
[[10, 11, 12], [13, 14, 15], [16, 17, 18], [19, 20]]
# or [range(10, 13), range(13, 16), range(16, 19), range(19, 21)] in py3
def块（iterable，n）：
“”“假定n是大于0的整数
"""
iterable=iter（iterable）
尽管如此：
结果=[]
对于范围（n）中的i：
尝试：
a=下一个（可编辑）
除停止迭代外：
打破
其他：
结果.追加（a）
如果结果为：
产量结果
其他：
打破
g1=（范围（10）内i的i*i）
g2=块（g1,3）
打印g2
''
打印列表（g2）
'[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81]]'
请参见
蟒蛇3我知道这有点古老，但还没有人提到：
我非常喜欢tzot和J.F.Sebastian提出的Python文档版本，
但它有两个缺点：
- 这不是很明确
- 我通常不希望最后一个块中有填充值
我在代码中经常使用这个：
from itertools import islice
def chunks(n, iterable):
iterable = iter(iterable)
while True:
yield tuple(islice(iterable, n)) or iterable.next()
更新：惰性块版本：
from itertools import chain, islice
def chunks(n, iterable):
iterable = iter(iterable)
while True:
yield chain([next(iterable)], islice(iterable, n-1))
该库具有分区功能，用于：
from toolz.itertoolz.core import partition
list(partition(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]
如何将列表拆分为大小相等的块？
对我来说，“大小均匀的块”意味着它们都是相同的长度，或者排除这个选项，在长度上差异最小。例如，21个项目的5个篮子可能有以下结果：
>>> import statistics
>>> statistics.variance([5,5,5,5,1])
3.2
>>> statistics.variance([5,4,4,4,4])
0.19999999999999998
更倾向于后一种结果的一个实际原因是：如果您使用这些函数来分配工作，您已经内置了这样一种前景，即一个函数可能比其他函数完成得早，因此当其他函数继续努力工作时，它将无所事事
对其他答案的评论
当我最初写这个答案时，没有一个答案是大小相等的块——它们都在最后留下了一块矮小的块，所以它们不是很平衡，并且长度的差异比必要的要大
例如，当前最重要的答案以以下内容结尾：
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
其他如list（石斑鱼（3，范围（7）））
和chunk（范围（7，3）
都返回：[（0，1，2），（3，4，5），（6，无，无）]
。None
只是填充，在我看来相当不雅观。他们并没有均匀地把易卜拉比分块
为什么我们不能把这些分开
循环解
使用itertools.cycle
的高级平衡解决方案，这是我今天可能采用的方法。以下是设置：
来自itertools导入周期的
项目=范围（10,75）
篮子的数量=10
现在我们需要我们的列表来填充元素：
baskets = [[] for _ in range(number_of_baskets)]
最后，我们将要分配的元素压缩到一个篮子循环中，直到元素用完，从语义上讲，这正是我们想要的：
for element, basket in zip(items, cycle(baskets)):
basket.append(element)
结果如下：
>>> from pprint import pprint
>>> pprint(baskets)
[[10, 20, 30, 40, 50, 60, 70],
[11, 21, 31, 41, 51, 61, 71],
[12, 22, 32, 42, 52, 62, 72],
[13, 23, 33, 43, 53, 63, 73],
[14, 24, 34, 44, 54, 64, 74],
[15, 25, 35, 45, 55, 65],
[16, 26, 36, 46, 56, 66],
[17, 27, 37, 47, 57, 67],
[18, 28, 38, 48, 58, 68],
[19, 29, 39, 49, 59, 69]]
为了实现此解决方案，我们编写了一个函数，并提供了类型注释：
from itertools import cycle
from typing import List, Any
def cycle_baskets(items: List[Any], maxbaskets: int) -> List[List[Any]]:
baskets = [[] for _ in range(min(maxbaskets, len(items)))]
for item, basket in zip(items, cycle(baskets)):
basket.append(item)
return baskets
在上面，我们列出了我们的物品清单，以及篮子的最大数量。我们创建一个空列表列表，在其中以循环方式追加每个元素
片
另一个优雅的解决方案是使用slices，特别是slices中不太常用的步骤参数。i、 e:
start = 0
stop = None
step = number_of_baskets
first_basket = items[start:stop:step]
这尤其优雅，因为切片不在乎数据的长度——结果，我们的第一个篮子，只需要它需要的长度。我们只需要增加每个篮筐的起点
事实上，这可能是一条单行线，但是
>>> import statistics
>>> statistics.variance([5,5,5,5,1])
3.2
>>> statistics.variance([5,4,4,4,4])
0.19999999999999998
[60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]]
baskets = [[] for _ in range(number_of_baskets)]
for element, basket in zip(items, cycle(baskets)):
basket.append(element)
>>> from pprint import pprint
>>> pprint(baskets)
[[10, 20, 30, 40, 50, 60, 70],
[11, 21, 31, 41, 51, 61, 71],
[12, 22, 32, 42, 52, 62, 72],
[13, 23, 33, 43, 53, 63, 73],
[14, 24, 34, 44, 54, 64, 74],
[15, 25, 35, 45, 55, 65],
[16, 26, 36, 46, 56, 66],
[17, 27, 37, 47, 57, 67],
[18, 28, 38, 48, 58, 68],
[19, 29, 39, 49, 59, 69]]
from itertools import cycle
from typing import List, Any
def cycle_baskets(items: List[Any], maxbaskets: int) -> List[List[Any]]:
baskets = [[] for _ in range(min(maxbaskets, len(items)))]
for item, basket in zip(items, cycle(baskets)):
basket.append(item)
return baskets
start = 0
stop = None
step = number_of_baskets
first_basket = items[start:stop:step]
from typing import List, Any
def slice_baskets(items: List[Any], maxbaskets: int) -> List[List[Any]]:
n_baskets = min(maxbaskets, len(items))
return [items[i::n_baskets] for i in range(n_baskets)]
from pprint import pprint
items = list(range(10, 75))
pprint(cycle_baskets(items, 10))
pprint(slice_baskets(items, 10))
pprint([list(s) for s in yield_islice_baskets(items, 10)])
def baskets_from(items, maxbaskets=25):
baskets = [[] for _ in range(maxbaskets)]
for i, item in enumerate(items):
baskets[i % maxbaskets].append(item)
return filter(None, baskets)
def iter_baskets_from(items, maxbaskets=3):
'''generates evenly balanced baskets from indexable iterable'''
item_count = len(items)
baskets = min(item_count, maxbaskets)
for x_i in range(baskets):
yield [items[y_i] for y_i in range(x_i, item_count, baskets)]
def iter_baskets_contiguous(items, maxbaskets=3, item_count=None):
'''
generates balanced baskets from iterable, contiguous contents
provide item_count if providing a iterator that doesn't support len()
'''
item_count = item_count or len(items)
baskets = min(item_count, maxbaskets)
items = iter(items)
floor = item_count // baskets
ceiling = floor + 1
stepdown = item_count % baskets
for x_i in range(baskets):
length = ceiling if x_i < stepdown else floor
yield [items.next() for _ in range(length)]
print(baskets_from(range(6), 8))
print(list(iter_baskets_from(range(6), 8)))
print(list(iter_baskets_contiguous(range(6), 8)))
print(baskets_from(range(22), 8))
print(list(iter_baskets_from(range(22), 8)))
print(list(iter_baskets_contiguous(range(22), 8)))
print(baskets_from('ABCDEFG', 3))
print(list(iter_baskets_from('ABCDEFG', 3)))
print(list(iter_baskets_contiguous('ABCDEFG', 3)))
print(baskets_from(range(26), 5))
print(list(iter_baskets_from(range(26), 5)))
print(list(iter_baskets_contiguous(range(26), 5)))
[[0], [1], [2], [3], [4], [5]]
[[0], [1], [2], [3], [4], [5]]
[[0], [1], [2], [3], [4], [5]]
[[0, 8, 16], [1, 9, 17], [2, 10, 18], [3, 11, 19], [4, 12, 20], [5, 13, 21], [6, 14], [7, 15]]
[[0, 8, 16], [1, 9, 17], [2, 10, 18], [3, 11, 19], [4, 12, 20], [5, 13, 21], [6, 14], [7, 15]]
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13, 14], [15, 16, 17], [18, 19], [20, 21]]
[['A', 'D', 'G'], ['B', 'E'], ['C', 'F']]
[['A', 'D', 'G'], ['B', 'E'], ['C', 'F']]
[['A', 'B', 'C'], ['D', 'E'], ['F', 'G']]
[[0, 5, 10, 15, 20, 25], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18, 23], [4, 9, 14, 19, 24]]
[[0, 5, 10, 15, 20, 25], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18, 23], [4, 9, 14, 19, 24]]
[[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]]
from itertools import islice
def chunk(it, size):
it = iter(it)
return iter(lambda: tuple(islice(it, size)), ())
>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
from itertools import islice, chain, repeat
def chunk_pad(it, size, padval=None):
it = chain(iter(it), repeat(padval))
return iter(lambda: tuple(islice(it, size)), (padval,) * size)
>>> list(chunk_pad(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk_pad(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]
_no_padding = object()
def chunk(it, size, padval=_no_padding):
if padval == _no_padding:
it = iter(it)
sentinel = ()
else:
it = chain(iter(it), repeat(padval))
sentinel = (padval,) * size
return iter(lambda: tuple(islice(it, size)), sentinel)
>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
>>> list(chunk(range(14), 3, None))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk(range(14), 3, 'a'))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 'a')]
_no_padding = object()
def chunk(it, size, padval=_no_padding):
it = iter(it)
chunker = iter(lambda: tuple(islice(it, size)), ())
if padval == _no_padding:
yield from chunker
else:
for ch in chunker:
yield ch if len(ch) == size else ch + (padval,) * (size - len(ch))
>>> list(chunk([1, 2, (), (), 5], 2))
[(1, 2), ((), ()), (5,)]
>>> list(chunk([1, 2, None, None, 5], 2, None))
[(1, 2), (None, None), (5, None)]
def chunkList(initialList, chunkSize):
"""
This function chunks a list into sub lists
that have a length equals to chunkSize.
Example:
lst = [3, 4, 9, 7, 1, 1, 2, 3]
print(chunkList(lst, 3))
returns
[[3, 4, 9], [7, 1, 1], [2, 3]]
"""
finalList = []
for i in range(0, len(initialList), chunkSize):
finalList.append(initialList[i:i+chunkSize])
return finalList
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)]
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, None)]
def split_list(the_list, chunk_size):
result_list = []
while the_list:
result_list.append(the_list[:chunk_size])
the_list = the_list[chunk_size:]
return result_list
a_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print split_list(a_list, 3)
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
CHUNK = 4
[a[i*CHUNK:(i+1)*CHUNK] for i in xrange((len(a) + CHUNK - 1) / CHUNK )]
def chunks(li, n):
if li == []:
return
yield li[:n]
for e in chunks(li[n:], n):
yield e
def chunks(li, n):
if li == []:
return
yield li[:n]
yield from chunks(li[n:], n)
def dec(gen):
def new_gen(li, n):
for e in gen(li, n):
if e == []:
return
yield e
return new_gen
@dec
def chunks(li, n):
yield li[:n]
for e in chunks(li[n:], n):
yield e
[AA[i:i+SS] for i in range(len(AA))[::SS]]
>>> AA=range(10,21);SS=3
>>> [AA[i:i+SS] for i in range(len(AA))[::SS]]
[[10, 11, 12], [13, 14, 15], [16, 17, 18], [19, 20]]
# or [range(10, 13), range(13, 16), range(16, 19), range(19, 21)] in py3
from boltons import iterutils
list(iterutils.chunked_iter(list(range(50)), 11))
[[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, 30, 31, 32],
[33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43],
[44, 45, 46, 47, 48, 49]]
>>> from utilspie import iterutils
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> list(iterutils.get_chunks(a, 5))
[[1, 2, 3, 4, 5], [6, 7, 8, 9]]
sudo pip install utilspie
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)
slice
31.18285083770752
index
0.02184295654296875
batches 1
0.03503894805908203
batches 2
0.22681021690368652
chunks
0.019841909408569336
grouper
0.006506919860839844
import itertools as it
import collections as ct
import more_itertools as mit
iterable = range(11)
n = 3
list(it.zip_longest(*[iter(iterable)] * n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]
d = {}
for i, x in enumerate(iterable):
d.setdefault(i//n, []).append(x)
list(d.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]
dd = ct.defaultdict(list)
for i, x in enumerate(iterable):
dd[i//n].append(x)
list(dd.values())
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]
list(mit.chunked(iterable, n))
# [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]
list(mit.sliced(iterable, n))
# [range(0, 3), range(3, 6), range(6, 9), range(9, 11)]
list(mit.grouper(n, iterable))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]
list(mit.windowed(iterable, len(iterable)//n, step=n))
# [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, None)]