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 f
If you want a one-liner, you can use list comprehension:
listA = [3,2,5,9,4,6,3,8,7,9]
listB=[max(listA[i:i+3]) for i in range(0,len(listA),3)]
print (listB)
it returns:
[5, 9, 8, 9]
Of course the codes can be written more dynamically: if you want a different window size, just change 3
to any integer.
Using numpy, you can extend the list with zeroes so its length is divisible by the window size, and reshape and compute the max
along the second axis:
def moving_maxima(a, w):
mod = len(a)%w
d = w if mod else mod
x = np.r_[a, [0]*(d-mod)]
return x.reshape(-1,w).max(1)
Some examples:
moving_maxima(listA,2)
# array([3., 9., 6., 8., 9.])
moving_maxima(listA,3)
#array([5, 9, 8, 9])
moving_maxima(listA,4)
#array([9, 8, 9])
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. ])
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)