iterating a span of a list

[Skaperen]

i'm looking for something that can do like this span() function can:

a = ['foo','bar','boo','far','baz'] for b in span(a,3): print(*b) a = [1,2,3,4,5,6,7,8,9] for b in span(a,4): print(*b)

Output:
foo bar boo bar boo far boo far baz 1 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8 6 7 8 9

this shouldn't be hard to code, but if zip() exists, maybe something like this span() exists. if it would return an iterator, that would be fine.

[Gribouillis]

It looks like more_itertools.windowed()

[Skaperen]

implementing it was next to trivial. a slice in a comprehension.

def span(s,n): return iter([s[x:x+n]for x in range(len(s)-n+1)])

[DeaD_EyE]

import itertools from functools import partial def window(sequence, window_size, uncomplete=True, fillvalue=None): if uncomplete: izip = partial(itertools.zip_longest, fillvalue=fillvalue) else: izip = zip yield from izip(*[iter(sequence)] * window_size)

l = list(range(10)) for chunk in window(l, 3): print(chunk)

Output:
(0, 1, 2) (3, 4, 5) (6, 7, 8) (9, None, None)

If you use uncomplete=False, the zip function will be used.
The zip function stops iterating, if there are not enough values, to fill the window.

l = list(range(10)) for chunk in window(l, 3, uncomplete=False): print(chunk)

Output:
(0, 1, 2) (3, 4, 5) (6, 7, 8)

[Gribouillis]

The more_itertools implementation uses a deque, but it works with any iterable, not only iterables with random access.

@DeaD_EyE I think your function does something different.

[Skaperen]

yes, DeaD_EyE's result is different.