Split list into sublists of length x (with or without overlap)-CodePudding

There are many similar questions on here, but I can't find exactly what I'm looking for.

I want to split a list into sublists, each of which is exactly length x. This can include overlap, and the area of overlap doesn't matter so much. For example:

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

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

# or 

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

# or 

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

It doesn't matter how many final sublists there are, though I don't want any more than necessary.

It also doesn't matter where the overlap happens, I just need to capture all the individual items in the original list and have each sublist result in the same number of items.

Thanks!

CodePudding user response：

You can adjust the accepted answer by NedBatchelder in this thread to work for the described scenario.

This is a generator function which I think is a pretty neat solution.

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    # n must not be 0
    for i in range(0, len(lst), n):
        if i   n >= len(lst):
            yield lst[-n:]
        else:
            yield lst[i:i   n]


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

for i in range(1, 11):
    print(list(chunks(l, i)))

Expected output:

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

CodePudding user response：

The trick as I see it is to iterate an index in steps of x but "clip" the last one to be no less than x from the end:

>>> a = list(range(1, 11))
>>> x = 3
>>> [a[i:i x] for i in (min(i, len(a) - x) for i in range(0, len(a), x))]
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [8, 9, 10]]

CodePudding user response：

Taken straight from itertools' recipes:

from itertools import zip_longest
def grouper(iterable, n, *, incomplete='fill', fillvalue=None):
    "Collect data into non-overlapping fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, fillvalue='x') --> ABC DEF Gxx
    # grouper('ABCDEFG', 3, incomplete='strict') --> ABC DEF ValueError
    # grouper('ABCDEFG', 3, incomplete='ignore') --> ABC DEF
    args = [iter(iterable)] * n
    if incomplete == 'fill':
        return zip_longest(*args, fillvalue=fillvalue)
    if incomplete == 'strict':
        return zip(*args, strict=True)
    if incomplete == 'ignore':
        return zip(*args)
    else:
        raise ValueError('Expected fill, strict, or ignore')

You may then use the "fill" option, and add the last item continuously if you wish to guarantee the size of the sublist:

>>> list_to_split = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> max_len = 3
>>> list(grouper(list_to_split, max_len, fillvalue=list_to_split[-1]))
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 10, 10)]

CodePudding user response：

Iterate over the list, slicing max_len elements at a time. We start the slice at min(idx, len(list_to_split) - max_len)) in case we're too close to the end of the list:

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

result = []
for idx in range(0, len(list_to_split), max_len):
    start = min(idx, len(list_to_split) - max_len)
    result.append(list_to_split[start:start   max_len])

print(result)

You can turn this into a list comprehension, but it's admittedly not very readable:

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

result = [
    list_to_split[
    min(idx, len(list_to_split) - max_len):
    min(idx, len(list_to_split) - max_len)   max_len]
    for idx in range(0, len(list_to_split), max_len)
]

print(result)

Both of these output:

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

CodePudding user response：

Following code should do without any libraries, though there are many libraries too that you can use

def main():
    '''The Main'''

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


    print([l[i:i x] for i in range(0, len(l), x)])


if __name__ == '__main__':
    main()

Output

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