# Average on overlapping windows in Python

## Question:

I’m trying to compute a moving average but with a set step size between each average. For example, if I was computing the average of a 4 element window every 2 elements:

``````data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
``````

This should produce the average of [1, 2, 3, 4], [3, 4, 5, 6], [5, 6, 7, 8], [7, 8, 9, 10].

``````window_avg = [2.5, 4.5, 6.5, 8.5]
``````

My data is such that the ending will be truncated before processing so there is no problem with the length with respect to window size.

I’ve read a bit about how to do moving averages in Python and there seems to be a lot of usage of itertools; however, the iterators go one element at a time and I can’t figure out how to have a step size between each calculation of the average.
(How to calculate moving average in Python 3?)

I have also been able to do this before in MATLAB by creating a matrix of indices which are overlapping and then indexing the data vector and performing a column wise mean (Create matrix by repeatedly overlapping a vector). However, since this vector is rather large (~70 000 elements, window of 450 samples, average every 30 samples), the computation would probably require too much memory.

Any help would be greatly appreciated. I am using Python 2.7.

One way to compute the average of a sliding window across a list in Python is to use a list comprehension. You can use

``````>>> range(0, len(data), 2)
[0, 2, 4, 6, 8]
``````

to get the starting indices of each window, and then `numpy`‘s `mean` function to take the average of each window. See the demo below:

``````>>> import numpy as np
>>> window_size = 4
>>> stride = 2
>>> window_avg = [ np.mean(data[i:i+window_size]) for i in range(0, len(data), stride)
if i+window_size <= len(data) ]
>>> window_avg
[2.5, 4.5, 6.5, 8.5]
``````

Note that the list comprehension does have a condition to ensure that it only computes the average of “full windows”, or sublists with exactly `window_size` elements.

When run on a dataset of the size discussed in the OP, this method computes on my MBA in a little over 200 ms:

``````In [5]: window_size = 450
In [6]: data = range(70000)
In [7]: stride = 30
In [8]: timeit [ np.mean(data[i:i+window_size]) for i in range(0, len(data), stride)
if i+window_size <= len(data) ]
1 loops, best of 3: 220 ms per loop
``````

It is about twice as fast on my machine to the `itertools` approach presented by @Abhijit:

``````In [9]: timeit map(np.mean, izip(*(islice(it, i, None, stride) for i, it in enumerate(tee(data, window_size)))))
1 loops, best of 3: 436 ms per loop
``````

The following approach uses itertools at its fullest to create moving average window of size `4`. As then entire expression is a generator which is evaluated when calculating the average, it has a complexity of O(n).

``````>>> import numpy as np
>>> from itertools import count, tee, izip, islice
>>> map(np.mean, izip(*(islice(it,i,None,2)
for i, it in enumerate(tee(data, 4)))))
[2.5, 4.5, 6.5, 8.5]
``````

Its interesting to note, how individual itertools function works in accord.

1. `itertools.tee` n-plicates an iterator, in this case 4 times
2. enumerate creates an enumerator object which yield a tuple of index and element (which is the iterator)
3. slice the iterator with stride 2, starting from the index position.

You can use rolling function of Pandas DataFrame,

``````data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
df = pd.DataFrame(data)
>>>
0
0   1
1   2
2   3
3   4
4   5
5   6
6   7
7   8
8   9
9  10
``````

Using Pandas DataFrame’s rolling function,

``````df.rolling(4).mean().dropna()[::2]
>>>
0
3  2.5
5  4.5
7  6.5
9  8.5
``````

4 is the window size and 2 in [::2] can be assumed to be step size.
Actually, `df.rolling(4).mean().dropna()` shift the window 1-by-1 and by applying index [::2], we pick one after taking two steps.

Alternatively,
If you have Pandas version > 1.5, you can give step size. Note that, center argument must be ‘True’. The solution:

``````df.rolling(4, step=2, center=True).mean().dropna()

>>> df
0
2  2.5
4  4.5
6  6.5
8  8.5
``````
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