以块为单位迭代列表的最“蟒蛇”方式是什么？

Question

我有一个Python脚本，它接受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，我不能控制输入，否则我会把它作为四元素元组的列表传递进来。目前，我是这样迭代的：

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具蟒蛇风格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。也许像这样的东西会更好？

while ints:
    foo += ints[0] * ints[1] + ints[2] * ints[3]
    ints[0:4] = []

然而，仍然不太“感觉”正确。:-/

相关问题：How do you split a list into evenly sized chunks in Python?

Answer 1

修改自Python的itertools文档的Recipes部分：

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

示例

grouper('ABCDEFG', 3, 'x')  # --> 'ABC' 'DEF' 'Gxx'

注意:Python2上的使用izip_longest而不是zip_longest。

Answer 2

def chunker(seq, size):
    return (seq[pos:pos + size] for pos in range(0, len(seq), size))
# (in python 2 use xrange() instead of range() to avoid allocating a list)

适用于任何序列：

text = "I am a very, very helpful text"

for group in chunker(text, 7):
   print(repr(group),)
# 'I am a ' 'very, v' 'ery hel' 'pful te' 'xt'

print '|'.join(chunker(text, 10))
# I am a ver|y, very he|lpful text

animals = ['cat', 'dog', 'rabbit', 'duck', 'bird', 'cow', 'gnu', 'fish']

for group in chunker(animals, 3):
    print(group)
# ['cat', 'dog', 'rabbit']
# ['duck', 'bird', 'cow']
# ['gnu', 'fish']

Answer 3

chunk_size = 4
for i in range(0, len(ints), chunk_size):
    chunk = ints[i:i+chunk_size]
    # process chunk of size <= chunk_size