["python","algorithm"]
Rolling or sliding window iterator?

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?.
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]