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
根据这个答案,得票最多的答案在最后留下一个“矮子”。这是我的解决方案,可以在没有矮子的情况下,尽可能地获得大小均匀的块。它基本上试图准确选择应该拆分列表的小数点,但只需将其舍入到最接近的整数:
from __future__ import division # not needed in Python 3
def n_even_chunks(l, n):
"""Yield n as even chunks as possible from l."""
last = 0
for i in range(1, n+1):
cur = int(round(i * (len(l) / n)))
yield l[last:cur]
last = cur
演示:
>>> pprint.pprint(list(n_even_chunks(list(range(100)), 9)))
[[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, 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]]
>>> pprint.pprint(list(n_even_chunks(list(range(100)), 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, 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]]
与排名前几的答案进行比较:
>>> pprint.pprint(list(chunks(list(range(100)), 100//9)))
[[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, 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]]
>>> pprint.pprint(list(chunks(list(range(100)), 100//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, 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]]
其他回答
def main():
print(chunkify([1,2,3,4,5,6],2))
def chunkify(list, n):
chunks = []
for i in range(0, len(list), n):
chunks.append(list[i:i+n])
return chunks
main()
我认为这很简单,可以为您提供数组的一部分。
由于我必须这样做,下面是我的解决方案,给出了一个生成器和一个批量大小:
def pop_n_elems_from_generator(g, n):
elems = []
try:
for idx in xrange(0, n):
elems.append(g.next())
return elems
except StopIteration:
return elems
我想我没有看到这个选项,所以只需添加另一个:):
def chunks(iterable, chunk_size):
i = 0;
while i < len(iterable):
yield iterable[i:i+chunk_size]
i += chunk_size
这项任务可以在公认答案中使用生成器轻松完成。我正在添加实现长度方法的类实现,这可能对某些人有用。我需要知道进度(使用tqdm),所以生成器应该返回块的数量。
class ChunksIterator(object):
def __init__(self, data, n):
self._data = data
self._l = len(data)
self._n = n
def __iter__(self):
for i in range(0, self._l, self._n):
yield self._data[i:i + self._n]
def __len__(self):
rem = 1 if self._l % self._n != 0 else 0
return self._l // self._n + rem
用法:
it = ChunksIterator([1,2,3,4,5,6,7,8,9], 2)
print(len(it))
for i in it:
print(i)
这个问题让我想起Raku(以前的Perl6).comb(n)方法。它将字符串分成n个大小的块。(还有更多,但我会省略细节。)
在Python3中实现一个类似的函数作为lambda表达式非常简单:
comb = lambda s,n: (s[i:i+n] for i in range(0,len(s),n))
然后你可以这样称呼它:
some_list = list(range(0, 20)) # creates a list of 20 elements
generator = comb(some_list, 4) # creates a generator that will generate lists of 4 elements
for sublist in generator:
print(sublist) # prints a sublist of four elements, as it's generated
当然,您不必将生成器分配给变量;你可以直接这样循环:
for sublist in comb(some_list, 4):
print(sublist) # prints a sublist of four elements, as it's generated
另外,此comb()函数还对字符串进行操作:
list( comb('catdogant', 3) ) # returns ['cat', 'dog', 'ant']