将 Python 列表拆分为重叠块的列表 [英] Splitting a Python list into a list of overlapping chunks

问题描述

这个问题类似于切片列表进入子列表列表，但在我的情况下，我想包含每个前一个子列表的最后一个元素，作为下一个子列表中的第一个元素.并且必须考虑到最后一个元素必须始终至少有两个元素.

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']]

解决方案

您链接的答案中的列表推导式是通过简单地缩短传递给范围的step"参数，很容易适应支持重叠块:

<预><代码>>>>list_ = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']>>>n = 3 # 组大小>>>m = 1 #重叠大小>>>[list_[i:i+n] for i in range(0, len(list_), n-m)][['a', 'b', 'c'], ['c', 'd', 'e'], ['e', 'f', 'g'], ['g', 'h']]

此问题的其他访问者可能无法使用输入列表(可切片、已知长度、有限).这是一个基于生成器的解决方案，可以处理任意迭代:

from collections import deque定义块(可迭代，块大小= 3，重叠= 0):# 我们将使用双端队列来保存值，因为它会自动# 如果它变得太大，则丢弃任何无关元素如果 chunk_size <1:引发异常(块大小太小")如果重叠 >= chunk_size:引发异常(重叠太大")队列 = 双端队列(最大长度 = 块大小)it = iter(可迭代)我 = 0尝试:# 首先用第一组填充队列对于我在范围内(chunk_size):queue.append(next(it))为真:产量元组(队列)# 在产生一个块后，为下一个块获取足够的元素对于 i 在范围内(chunk_size - 重叠):queue.append(next(it))除了停止迭代:# 如果迭代器耗尽，则产生任何剩余的元素我 += 重叠如果我＞0:产量元组(队列)[-i:]

注意:我已经在 wimpy.util.chunks.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.

This question is similar to Slicing a list into a list of sub-lists, but in my case I want to include the last element of the each previous sub-list, as the first element in the next sub-list. And have to take into account that the last element have always to have at least two elements.

For example:

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

The result for a size 3 sub-list:

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

解决方案

The list comprehension in the answer you linked is easily adapted to support overlapping chunks by simply shortening the "step" parameter passed to the range:

>>> 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']]

Other visitors to this question mightn't have the luxury of working with an input list (slicable, known length, finite). Here is a generator-based solution that can work with arbitrary iterables:

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:]

Note: I've since released this implementation in wimpy.util.chunks. If you don't mind adding the dependency, you can pip install wimpy and use from wimpy import chunks rather than copy-pasting the code.

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