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

我喜欢这种方法。它感觉简单而不神奇,支持所有可迭代类型,并且不需要导入。

def chunk_iter(iterable, chunk_size):
it = iter(iterable)
while True:
    chunk = tuple(next(it) for _ in range(chunk_size))
    if not chunk:
        break
    yield chunk

其他回答

这里非常python化(也可以内联split_groups函数体)

import itertools
def split_groups(iter_in, group_size):
    return ((x for _, x in item) for _, item in itertools.groupby(enumerate(iter_in), key=lambda x: x[0] // group_size))

for x, y, z, w in split_groups(range(16), 4):
    foo += x * y + z * w
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
from itertools import izip_longest

def chunker(iterable, chunksize, filler):
    return izip_longest(*[iter(iterable)]*chunksize, fillvalue=filler)
chunk_size = 4
for i in range(0, len(ints), chunk_size):
    chunk = ints[i:i+chunk_size]
    # process chunk of size <= chunk_size

我希望通过将迭代器从列表中删除,我不是简单地复制列表的一部分。生成器可以被切片,它们将自动仍然是一个生成器,而列表将被切片成1000个条目的大块,这是较低的效率。

def iter_group(iterable, batch_size:int):
    length = len(iterable)
    start = batch_size*-1
    end = 0
    while(end < length):
        start += batch_size
        end += batch_size
        if type(iterable) == list:
            yield (iterable[i] for i in range(start,min(length-1,end)))
        else:
            yield iterable[start:end]

用法:

items = list(range(1,1251))

for item_group in iter_group(items, 1000):
    for item in item_group:
        print(item)