python - 迭代列表的所有子组

Question

假设我有一个列表 [1,2,3,4,5,6]，我想迭代 len 2 [1,2] 的所有子组[3,4] [5,6]。

简单的做法

    L = [1,2,3,4,5,6]
    N = len(L)//2
    for k in range(N):
        slice = L[k*2:(k+1)*2]
        for val in slice:
            #Do things with the slice

但是我想知道是否有更Pythonic的方法来迭代“分区”列表。我也接受使用 numpy 数组的解决方案。像这样的东西:

    L = [1,2,3,4,5,6]
    slices = f(L,2) # A nice "f" here? 
    for slice in slices:
        for val in slice:
            #Do things with the slice

非常感谢!

Answer 1

使用grouper来自 itertools 库的配方:

import itertools

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return itertools.zip_longest(*args, fillvalue=fillvalue)

L = [1,2,3,4,5,6]
for slice in grouper(L, 2):
    print(slice)