Rolling or sliding window iterator?

Question

I need a rolling window (aka sliding window) iterable over a sequence/iterator/generator. (Default Python iteration could be considered a special case, where the window length is 1.) I'm currently using the following code. How can I do it more elegantly and/or efficiently?

def rolling_window(seq, window_size):
    it = iter(seq)
    win = [it.next() for cnt in xrange(window_size)] # First window
    yield win
    for e in it: # Subsequent windows
        win[:-1] = win[1:]
        win[-1] = e
        yield win

if __name__=="__main__":
    for w in rolling_window(xrange(6), 3):
        print w

"""Example output:   
   [0, 1, 2]
   [1, 2, 3]
   [2, 3, 4]
   [3, 4, 5]
"""

---

_{For the specific case of window_size == 2 (i.e., iterating over adjacent, overlapping pairs in a sequence), see also How can I iterate over overlapping (current, next) pairs of values from a list?.}

Answer 1

There's one in an old version of the Python docs with itertools examples:

from itertools import islice

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result
    for elem in it:
        result = result[1:] + (elem,)
        yield result

The one from the docs is a little more succinct and uses itertools to greater effect I imagine.

---

If your iterator is a simple list/tuple a simple way to slide through it with a specified window size would be:

seq = [0, 1, 2, 3, 4, 5]
window_size = 3

for i in range(len(seq) - window_size + 1):
    print(seq[i: i + window_size])

Output:

[0, 1, 2]
[1, 2, 3]
[2, 3, 4]
[3, 4, 5]