Rolling or sliding window iterator?

Viewed 155639

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 Iterate a list as pair (current, next) in Python.

28 Answers

There is a library which does exactly what you need:

import more_itertools
list(more_itertools.windowed([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],n=3, step=3))

Out: [(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)]

I use the following code as a simple sliding window that uses generators to drastically increase readability. Its speed has so far been sufficient for use in bioinformatics sequence analysis in my experience.

I include it here because I didn't see this method used yet. Again, I make no claims about its compared performance.

def slidingWindow(sequence,winSize,step=1):
"""Returns a generator that will iterate through
the defined chunks of input sequence. Input sequence
must be sliceable."""

    # Verify the inputs
    if not ((type(winSize) == type(0)) and (type(step) == type(0))):
        raise Exception("**ERROR** type(winSize) and type(step) must be int.")
    if step > winSize:
        raise Exception("**ERROR** step must not be larger than winSize.")
    if winSize > len(sequence):
        raise Exception("**ERROR** winSize must not be larger than sequence length.")

    # Pre-compute number of chunks to emit
    numOfChunks = ((len(sequence)-winSize)/step)+1

    # Do the work
    for i in range(0,numOfChunks*step,step):
        yield sequence[i:i+winSize]

Let's make it lazy!

from itertools import islice, tee

def window(iterable, size): 
    iterators = tee(iterable, size) 
    iterators = [islice(iterator, i, None) for i, iterator in enumerate(iterators)]  
    yield from zip(*iterators)

list(window(range(5), 3))
# [(0, 1, 2), (1, 2, 3), (2, 3, 4)]

I tested a few solutions along with the one I came up with. I found the one I came up with to be the fastest so I thought I would share my python3 implementation.

import itertools
import sys

def windowed(l, stride):
    return zip(*[itertools.islice(l, i, sys.maxsize) for i in range(stride)])
#Importing the numpy library
import numpy as np
arr = np.arange(6) #Sequence
window_size = 3
np.lib.stride_tricks.as_strided(arr, shape= (len(arr) - window_size +1, window_size), 
strides = arr.strides*2)

"""Example output:

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

"""

The toolz/cytoolz package has a sliding_window function.

>>> from cytoolz import sliding_window
>>> list(sliding_window(3, range(6))) # returns [(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5)]

Modified DiPaolo's answer to allow arbitrary fill and variable step size

import itertools
def window(seq, n=2,step=1,fill=None,keep=0):
    "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(itertools.islice(it, n))    
    if len(result) == n:
        yield result
    while True:        
#         for elem in it:        
        elem = tuple( next(it, fill) for _ in range(step))
        result = result[step:] + elem        
        if elem[-1] is fill:
            if keep:
                yield result
            break
        yield result

here is a one liner. I timed it and it's comprable to the performance of the top answer and gets progressively better with larger seq from 20% slower with len(seq) = 20 and 7% slower with len(seq) = 10000

zip(*[seq[i:(len(seq) - n - 1 + i)] for i in range(n)])

Trying my part, simple, one liner, pythonic way using islice. But, may not be optimally efficient.

from itertools import islice
array = range(0, 10)
window_size = 4
map(lambda i: list(islice(array, i, i + window_size)), range(0, len(array) - window_size + 1))
# output = [[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6], [4, 5, 6, 7], [5, 6, 7, 8], [6, 7, 8, 9]]

Explanation: Create window by using islice of window_size and iterate this operation using map over all array.

Optimized Function for sliding window data in Deep learning

def SlidingWindow(X, window_length, stride):
    indexer = np.arange(window_length)[None, :] + stride*np.arange(int(len(X)/stride)-window_length+4)[:, None]
    return X.take(indexer)

to apply on multidimensional array

import numpy as np
def SlidingWindow(X, window_length, stride1):
    stride=  X.shape[1]*stride1
    window_length = window_length*X.shape[1]
    indexer = np.arange(window_length)[None, :] + stride1*np.arange(int(len(X)/stride1)-window_length-1)[:, None]
    return X.take(indexer)

my two versions of window implementation

from typing import Sized, Iterable

def window(seq: Sized, n: int, strid: int = 1, drop_last: bool = False):
    for i in range(0, len(seq), strid):
        res = seq[i:i + n]
        if drop_last and len(res) < n:
            break
        yield res


def window2(seq: Iterable, n: int, strid: int = 1, drop_last: bool = False):
    it = iter(seq)
    result = []
    step = 0
    for i, ele in enumerate(it):
        result.append(ele)
        result = result[-n:]
        if len(result) == n:
            if step % strid == 0:
                yield result
            step += 1
    if not drop_last:
        yield result

Another simple way to generate window of fixed length from a list

from collections import deque

def window(ls,window_size=3):
    window = deque(maxlen=window_size)

    for element in ls:
        
        if len(window)==window_size:
            yield list(window)
        window.append(element)

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

for w in window(ls):
    print(w)

My (keep it simple) solution that I ended up using:

def sliding_window(items, size):
    return [items[start:end] for start, end
            in zip(range(0, len(items) - size + 1), range(size, len(items) + 1))]

Needless to say, the items sequence needs to be sliceable. Working with indices is not ideal, but it seems to be the least bad option given the alternatives... This can also easily be changed to a generator: just replace [...] with (...).

In Python 3.10, we have the itertools.pairwise(iterable) function to slide a window with two elements:

Here's the doc :

Return successive overlapping pairs taken from the input iterable.

The number of 2-tuples in the output iterator will be one fewer than the number of inputs. It will be empty if the input iterable has fewer than two values.

Roughly equivalent to:

def pairwise(iterable):
    # pairwise('ABCDEFG') --> AB BC CD DE EF FG
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)
Related