python - 如何从列表列表中选择元素的滑动窗口？

Question

假设我有以下列表:

x = [[1,2,3],[4,5,6],[7,8,9,10]]

我希望选择所有大小的“窗口”，例如n=4，交错距离，例如d=2:

    [[1,2,3],[4]]                # Starts at position `0`
        [[3],[4,5,6]]            # Starts at position `d`
              [[5,6],[7,8]]      # Starts at position `2d`
                    [[7,8,9,10]] # Starts at position `3d`

即我希望在窗口与子列表重叠的地方采用相交的“切片”。

我该怎么做？

Answer 1

如果你预先计算了一些索引，你可以用一个虚拟的单行重建任何窗口:

import itertools
import operator

def window(x, start, stop):
    first = indices[start][0]
    last = indices[stop-1][0]
    return [
        [x[i][j] for i, j in g] if k in (first, last) else x[k]
        for k, g in itertools.groupby(
            indices[start:stop],
            key=operator.itemgetter(0))
        ]

def flat_len(x):
    """Return length of flattened list."""
    return sum(len(sublist) for sublist in x)

n=4; d=2
x = [[1,2,3],[4,5,6],[7,8,9,10]]

indices = [(i, j) for i, sublist in enumerate(x) for j in range(len(sublist))]

for i in range(0,flat_len(x)-n+1,d):
    print(window(x,i,i+n,indices))

>>> [[1, 2, 3], [4]]
>>> [[3], [4, 5, 6]]
>>> [[5, 6], [7, 8]]