Sliding window of width n over the given iterable

I have a sequence, window size and step:

``````seq = [0,1,2,3,4]
n=4
step=2

from more_itertools import windowed
list(windowed([0,1,2,3,4], n, fillvalue=0, step=step))
``````

result:

``````[(0, 1, 2, 3), (2, 3, 4, 0)]
``````

but I need:

``````[(0, 1, 2, 3), (2, 3, 4, 0), (4, 0, 0, 0)]
``````

On Best Solutions

``````seq = [0,1,2,3,4]
n=4
step=2

list(windowed(padded(seq, 0, n=n, next_multiple=True), n, step=step))
``````
On

Just write your own `windowed` function:

``````def windowed(iterable, size, fillvalue=None, step=1):
for i in range(0, len(iterable), step):
window = iterable[i:i+size]
window += [fillvalue] * (size - len(window))
yield window
``````
``````>>> list(windowed([0,1,2,3,4], 4, fillvalue=0, step=2))
[[0, 1, 2, 3], [2, 3, 4, 0], [4, 0, 0, 0]]
``````
On

this should also work with iterables and not just sequences:

``````from itertools import islice

def sliding_window(seq, n, step, fillvalue=None):
it = iter(seq)
values = tuple(islice(it, n))
while values:
yield values + (n-len(values)) * (fillvalue, )
values = values[step:] + tuple(islice(it, step))
``````

the function outputs:

``````print(list(sliding_window(seq, n, step, fillvalue=0)))
# [(0, 1, 2, 3), (2, 3, 4, 0), (4, 0, 0, 0)]
``````

most of it is borrowed from the original itertools recipe for a sliding window.

On

Consider `more_itertools.stagger`:

Given

``````import itertools as it

import more_itertools as mit

iterable = [0, 1, 2, 3, 5]
``````

Code

Get all results from sliding windows:

``````windows = list(mit.stagger(iterable, offsets=(0, 1, 2, 3), longest=True, fillvalue=0))
windows
# [(0, 1, 2, 3), (1, 2, 3, 5), (2, 3, 5, 0), (3, 5, 0, 0), (5, 0, 0, 0)]
``````

Next, filter out the desired results:

``````[w for i, w in enumerate(windows) if not (i % 2)]
# [(0, 1, 2, 3), (2, 3, 5, 0), (5, 0, 0, 0)]
``````

or slice the iterable:

``````list(it.islice(windows, 0, None, 2))
# [(0, 1, 2, 3), (2, 3, 5, 0), (5, 0, 0, 0)]
``````