Splitting a Python list into a list of overlapping chunks

Question

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 e

Answer 1

more_itertools has a windowing tool for overlapping iterables.

Given

import more_itertools as mit

iterable = list("abcdefgh")
iterable
# ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']

Code

windows = list(mit.windowed(iterable, n=3, step=2))
windows
# [('a', 'b', 'c'), ('c', 'd', 'e'), ('e', 'f', 'g'), ('g', 'h', None)]

If required, you can drop the None fillvalue by filtering the windows:

[list(filter(None, w)) for w in windows]
# [['a', 'b', 'c'], ['c', 'd', 'e'], ['e', 'f', 'g'], ['g', 'h']]

See also more_itertools docs for details on more_itertools.windowed

Answer 2

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.

Answer 3

Here's what I came up with:

l = [1, 2, 3, 4, 5, 6]
x = zip (l[:-1], l[1:])
for i in x:
    print (i)

(1, 2)
(2, 3)
(3, 4)
(4, 5)
(5, 6)

Zip accepts any number of iterables, there is also zip_longest

Answer 4

[list_[i:i+n] for i in xrange(0,len(list_), n-m)]