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
def chunker(seq, size):
return (seq[pos:pos + size] for pos in range(0, len(seq), size))
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']
其他回答
一行程序，特别的解决方案，以4 -大小的块迭代列表x
for a, b, c, d in zip(x[0::4], x[1::4], x[2::4], x[3::4]):
... do something with a, b, c and d ...
def chunker(iterable, n):
"""Yield iterable in chunk sizes.
>>> chunks = chunker('ABCDEF', n=4)
>>> chunks.next()
['A', 'B', 'C', 'D']
>>> chunks.next()
['E', 'F']
"""
it = iter(iterable)
while True:
chunk = []
for i in range(n):
try:
chunk.append(next(it))
except StopIteration:
yield chunk
raise StopIteration
yield chunk
if __name__ == '__main__':
import doctest
doctest.testmod()
如果列表很大，执行效率最高的方法是使用生成器:
def get_chunk(iterable, chunk_size):
result = []
for item in iterable:
result.append(item)
if len(result) == chunk_size:
yield tuple(result)
result = []
if len(result) > 0:
yield tuple(result)
for x in get_chunk([1,2,3,4,5,6,7,8,9,10], 3):
print x
(1, 2, 3)
(4, 5, 6)
(7, 8, 9)
(10,)
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]
除非我遗漏了一些内容，否则没有提到以下使用生成器表达式的简单解决方案。它假设块的大小和数量都是已知的(通常情况下)，并且不需要填充:
def chunks(it, n, m):
"""Make an iterator over m first chunks of size n.
"""
it = iter(it)
# Chunks are presented as tuples.
return (tuple(next(it) for _ in range(n)) for _ in range(m))
