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
def chunk(lst):
out = []
for x in xrange(2, len(lst) + 1):
if not len(lst) % x:
factor = len(lst) / x
break
while lst:
out.append([lst.pop(0) for x in xrange(factor)])
return out
其他回答
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
[AA[i:i+SS] for i in range(len(AA))[::SS]]
其中AA是数组,SS是块大小。例如:
>>> 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
要扩展py3中的范围,请执行以下操作
(py3) >>> [list(AA[i:i+SS]) for i in range(len(AA))[::SS]]
[[10, 11, 12], [13, 14, 15], [16, 17, 18], [19, 20]]
itertools模块中的配方提供了两种方法来实现这一点,具体取决于您希望如何处理最终的奇数大小的批次(保留它、用填充值填充它、忽略它或引发异常):
from itertools import islice, izip_longest
def batched(iterable, n):
"Batch data into lists of length n. The last batch may be shorter."
# batched('ABCDEFG', 3) --> ABC DEF G
it = iter(iterable)
while True:
batch = list(islice(it, n))
if not batch:
return
yield batch
def grouper(iterable, n, *, incomplete='fill', fillvalue=None):
"Collect data into non-overlapping fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, fillvalue='x') --> ABC DEF Gxx
# grouper('ABCDEFG', 3, incomplete='strict') --> ABC DEF ValueError
# grouper('ABCDEFG', 3, incomplete='ignore') --> ABC DEF
args = [iter(iterable)] * n
if incomplete == 'fill':
return zip_longest(*args, fillvalue=fillvalue)
if incomplete == 'strict':
return zip(*args, strict=True)
if incomplete == 'ignore':
return zip(*args)
else:
raise ValueError('Expected fill, strict, or ignore')
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']
一种老式的方法,不需要itertools,但仍然可以使用任意生成器:
def chunks(g, n):
"""divide a generator 'g' into small chunks
Yields:
a chunk that has 'n' or less items
"""
n = max(1, n)
buff = []
for item in g:
buff.append(item)
if len(buff) == n:
yield buff
buff = []
if buff:
yield buff