python - 将 Python 列表拆分为重叠 block 列表

Question

这个问题类似于 Slicing a list into a list of sub-lists ，但在我的例子中，我想将每个先前子列表的最后一个元素包含为下一个子列表中的第一个元素。而且我必须考虑到最后一个子列表必须至少有两个元素。

例如:

list_ = ['a','b','c','d','e','f','g','h']

大小为 3 的子列表的结果:

resultant_list = [['a','b','c'],['c','d','e'],['e','f','g'],['g','h']]

Answer 1

answer you linked 中的列表理解通过简单地缩短传递给范围的“步骤”参数，很容易适应支持重叠 block :

>>> list_ = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
>>> n = 3  # group size
>>> m = 1  # overlap size
>>> [list_[i:i+n] for i in range(0, len(list_), n-m)]
[['a', 'b', 'c'], ['c', 'd', 'e'], ['e', 'f', 'g'], ['g', 'h']]

这个问题的其他访问者可能没有使用输入列表(可切片、已知长度、有限)的奢侈。这是一个基于生成器的解决方案，可以处理任意可迭代对象:

from collections import deque

def chunks(iterable, chunk_size=3, overlap=0):
    # we'll use a deque to hold the values because it automatically
    # discards any extraneous elements if it grows too large
    if chunk_size < 1:
        raise Exception("chunk size too small")
    if overlap >= chunk_size:
        raise Exception("overlap too large")
    queue = deque(maxlen=chunk_size)
    it = iter(iterable)
    i = 0
    try:
        # start by filling the queue with the first group
        for i in range(chunk_size):
            queue.append(next(it))
        while True:
            yield tuple(queue)
            # after yielding a chunk, get enough elements for the next chunk
            for i in range(chunk_size - overlap):
                queue.append(next(it))
    except StopIteration:
        # if the iterator is exhausted, yield any remaining elements
        i += overlap
        if i > 0:
            yield tuple(queue)[-i:]

注意:我已经在 wimpy.util.chunks 中发布了这个实现.如果您不介意添加依赖项，您可以pip install wimpy 并使用from wimpy import chunks 而不是复制粘贴代码。