Creating a tumbling windows in python

Just wondering if there is a way to construct a tumbling window in python. So for example if I have list/ndarray , listA = [3,2,5,9,4,6,3,8,7,9]. Then how could I find the maximum of the first 3 items (3,2,5) -> 5, and then the next 3 items (9,4,6) -> 9 and so on… Sort of like breaking it up to sections and finding the max. So the final result would be list [5,9,8,9]

Answer

Approach #1: One-liner for windowed-max using np.maximum.reduceat –

In [118]: np.maximum.reduceat(listA,np.arange(0,len(listA),3))
Out[118]: array([5, 9, 8, 9])

Becomes more compact with np.r_ –

np.maximum.reduceat(listA,np.r_[:len(listA):3])

Approach #2: Generic ufunc way

Here’s a function for generic ufuncs and that window length as a parameter –

def windowed_ufunc(a, ufunc, W):
    a = np.asarray(a)
    n = len(a)
    L = W*(n//W)
    out = ufunc(a[:L].reshape(-1,W),axis=1)
    if n>L:
        out = np.hstack((out, ufunc(a[L:])))
    return out

Sample run –

In [81]: a = [3,2,5,9,4,6,3,8,7,9]

In [82]: windowed_ufunc(a, ufunc=np.max, W=3)
Out[82]: array([5, 9, 8, 9])

On other ufuncs –

In [83]: windowed_ufunc(a, ufunc=np.min, W=3)
Out[83]: array([2, 4, 3, 9])

In [84]: windowed_ufunc(a, ufunc=np.sum, W=3)
Out[84]: array([10, 19, 18,  9])

In [85]: windowed_ufunc(a, ufunc=np.mean, W=3)
Out[85]: array([3.33333333, 6.33333333, 6.        , 9.        ])

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Benchmarking

Timings on NumPy solutions on array data with sample data scaled up by 10000x –

In [159]: a = [3,2,5,9,4,6,3,8,7,9]

In [160]: a = np.tile(a, 10000)

# @yatu's soln
In [162]: %timeit moving_maxima(a, w=3)
435 µs ± 8.54 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

# From this post - app#1
In [167]: %timeit np.maximum.reduceat(a,np.arange(0,len(a),3))
353 µs ± 2.55 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

# From this post - app#2
In [165]: %timeit windowed_ufunc(a, ufunc=np.max, W=3)
379 µs ± 6.44 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)