我有一个Python脚本，它把一个整数列表作为输入，我需要一次处理四个整数。不幸的是，我无法控制输入，否则我将它作为一个四元素元组列表传入。目前，我以这种方式迭代它:
for i in range(0, len(ints), 4):
# dummy op for example code
foo += ints[i] * ints[i + 1] + ints[i + 2] * ints[i + 3]
不过，它看起来很像“C-think”，这让我怀疑有一种更python的方式来处理这种情况。该列表在迭代后被丢弃，因此不需要保留。也许这样会更好?
while ints:
foo += ints[0] * ints[1] + ints[2] * ints[3]
ints[0:4] = []
不过，感觉还是不太对。: - /
相关问题:在Python中如何将列表分割成大小均匀的块?
当前回答
我需要一个解决方案，也将工作与集和生成器。我写不出很短很漂亮的东西，但至少可读性很好。
def chunker(seq, size):
res = []
for el in seq:
res.append(el)
if len(res) == size:
yield res
res = []
if res:
yield res
列表:
>>> list(chunker([i for i in range(10)], 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
Set:
>>> list(chunker(set([i for i in range(10)]), 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
发电机:
>>> list(chunker((i for i in range(10)), 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
其他回答
我需要一个解决方案，也将工作与集和生成器。我写不出很短很漂亮的东西，但至少可读性很好。
def chunker(seq, size):
res = []
for el in seq:
res.append(el)
if len(res) == size:
yield res
res = []
if res:
yield res
列表:
>>> list(chunker([i for i in range(10)], 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
Set:
>>> list(chunker(set([i for i in range(10)]), 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
发电机:
>>> list(chunker((i for i in range(10)), 3))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
import itertools
def chunks(iterable,size):
it = iter(iterable)
chunk = tuple(itertools.islice(it,size))
while chunk:
yield chunk
chunk = tuple(itertools.islice(it,size))
# though this will throw ValueError if the length of ints
# isn't a multiple of four:
for x1,x2,x3,x4 in chunks(ints,4):
foo += x1 + x2 + x3 + x4
for chunk in chunks(ints,4):
foo += sum(chunk)
另一种方法:
import itertools
def chunks2(iterable,size,filler=None):
it = itertools.chain(iterable,itertools.repeat(filler,size-1))
chunk = tuple(itertools.islice(it,size))
while len(chunk) == size:
yield chunk
chunk = tuple(itertools.islice(it,size))
# x2, x3 and x4 could get the value 0 if the length is not
# a multiple of 4.
for x1,x2,x3,x4 in chunks2(ints,4,0):
foo += x1 + x2 + x3 + x4
我从来不想填充我的块，所以这个要求是必要的。我发现在任何可迭代对象上工作的能力也是必需的。鉴于此，我决定扩展公认的答案，https://stackoverflow.com/a/434411/1074659。
如果由于需要比较和筛选填充值而不需要填充，则这种方法的性能会受到轻微的影响。然而，对于大块大小，这个实用程序是非常高性能的。
#!/usr/bin/env python3
from itertools import zip_longest
_UNDEFINED = object()
def chunker(iterable, chunksize, fillvalue=_UNDEFINED):
"""
Collect data into chunks and optionally pad it.
Performance worsens as `chunksize` approaches 1.
Inspired by:
https://docs.python.org/3/library/itertools.html#itertools-recipes
"""
args = [iter(iterable)] * chunksize
chunks = zip_longest(*args, fillvalue=fillvalue)
yield from (
filter(lambda val: val is not _UNDEFINED, chunk)
if chunk[-1] is _UNDEFINED
else chunk
for chunk in chunks
) if fillvalue is _UNDEFINED else chunks
这个答案拆分了一个字符串列表，f.ex。达到符合pep8线长的要求:
def split(what, target_length=79):
'''splits list of strings into sublists, each
having string length at most 79'''
out = [[]]
while what:
if len("', '".join(out[-1])) + len(what[0]) < target_length:
out[-1].append(what.pop(0))
else:
if not out[-1]: # string longer than target_length
out[-1] = [what.pop(0)]
out.append([])
return out
使用
>>> split(['deferred_income', 'long_term_incentive', 'restricted_stock_deferred', 'shared_receipt_with_poi', 'loan_advances', 'from_messages', 'other', 'director_fees', 'bonus', 'total_stock_value', 'from_poi_to_this_person', 'from_this_person_to_poi', 'restricted_stock', 'salary', 'total_payments', 'exercised_stock_options'], 75)
[['deferred_income', 'long_term_incentive', 'restricted_stock_deferred'], ['shared_receipt_with_poi', 'loan_advances', 'from_messages', 'other'], ['director_fees', 'bonus', 'total_stock_value', 'from_poi_to_this_person'], ['from_this_person_to_poi', 'restricted_stock', 'salary', 'total_payments'], ['exercised_stock_options']]
在Python 3.8中，您可以使用walrus操作符和itertools.islice。
from itertools import islice
list_ = [i for i in range(10, 100)]
def chunker(it, size):
iterator = iter(it)
while chunk := list(islice(iterator, size)):
print(chunk)
In [2]: chunker(list_, 10)
[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]
