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
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 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()
而不是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)]
Python 2版本:
[lst[i:i + n] for i in xrange(0, len(lst), n)]
这是一个生成器,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()
而不是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)]
Python 2版本:
[lst[i:i + n] for i in xrange(0, len(lst), n)]
如果您知道列表大小:
def SplitList(mylist, chunk_size):
return [mylist[offs:offs+chunk_size] for offs in range(0, len(mylist), chunk_size)]
如果您没有(迭代器):
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块(即,没有不完整的最后一块),则可以用更漂亮的方式重新表述它。如果您知道列表大小:
def SplitList(mylist, chunk_size):
return [mylist[offs:offs+chunk_size] for offs in range(0, len(mylist), chunk_size)]
如果您没有(迭代器):
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))
例如:
>>> 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]]
这是一个可以处理任意iterables的生成器:
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))
例如:
>>> 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]]
嘿,一行版本
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]]
嘿,一行版本
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]]
直接来自(旧)Python文档(itertools的配方):
J.F.Sebastian建议的当前版本:
#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)
我猜圭多的时间机器工作了,会工作的,会工作的,又工作了
这些解决方案之所以有效,是因为[iter(iterable)]*n
(或早期版本中的等效项)创建了一个迭代器,在列表中重复n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其进行改进,从而导致每次这样的zip循环生成一个n
项的元组。直接来自(旧)Python文档(itertools的配方):
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
J.F.Sebastian建议的当前版本:
#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)
我猜圭多的时间机器工作了,会工作的,会工作的,又工作了
这些解决方案之所以有效,是因为[iter(iterable)]*n
(或早期版本中的等效项)创建了一个迭代器,在列表中重复n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
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()
,如果您想要超级简单的东西:
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:
不调用len(),这对大型列表很有好处:
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)
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键入的固定数字时,我会使用此选项,例如“3”,并且永远不会更改。如果区块大小为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
考虑使用碎片
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]]'
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3参见
蟒蛇3我知道这有点古老,但还没有人提到:
我知道这有点古老,但还没有人提到:
我喜欢作者提出的Python文档版本
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)]