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

NumPy很简单:

ints = array([1, 2, 3, 4, 5, 6, 7, 8])
for int1, int2 in ints.reshape(-1, 2):
    print(int1, int2)

输出:

1 2
3 4
5 6
7 8

类似于其他提案,但不完全相同,我喜欢这样做,因为它简单易读:

it = iter([1, 2, 3, 4, 5, 6, 7, 8, 9])
for chunk in zip(it, it, it, it):
    print chunk

>>> (1, 2, 3, 4)
>>> (5, 6, 7, 8)

这样你就不会得到最后一部分。如果你想获取(9,None, None, None)作为最后一个块,只需使用itertools中的izip_longest。

你可以使用函数库中的partition或chunks函数:

from funcy import partition

for a, b, c, d in partition(4, ints):
    foo += a * b * c * d

这些函数还有迭代器版本ipartition和ichunks,在这种情况下更有效。

您还可以查看它们的实现。

使用map()而不是zip()修复填充问题在J.F.塞巴斯蒂安的回答:

>>> def chunker(iterable, chunksize):
...   return map(None,*[iter(iterable)]*chunksize)

例子:

>>> s = '1234567890'
>>> chunker(s, 3)
[('1', '2', '3'), ('4', '5', '6'), ('7', '8', '9'), ('0', None, None)]
>>> chunker(s, 4)
[('1', '2', '3', '4'), ('5', '6', '7', '8'), ('9', '0', None, None)]
>>> chunker(s, 5)
[('1', '2', '3', '4', '5'), ('6', '7', '8', '9', '0')]
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()