This computes the "rolling max" of A
(similar to rolling average) over a sliding window of length K
:
import numpy as np
A = np.random.rand(100000)
K = 10
rollingmax = np.array([max(A[j:j+K]) for j in range(len(A)-K)])
but I think it is far from optimal in terms of performance.
I know that the pandas
library has rolling_max
, but in my project, I don't want to use this new dependance.
Question: is there a simple way to compute the rolling maximum with numpy only?
I guess this little trick using strides
and as_strided
will do the job:
def max_rolling1(a, window,axis =1):
shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
strides = a.strides + (a.strides[-1],)
rolling = np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
return np.max(rolling,axis=axis)
and for comaprison purpose I defined another function based on your algorithm :
def max_rolling2(A,K):
rollingmax = np.array([max(A[j:j+K]) for j in range(len(A)-K)])
return rollingmax
and the comparison by timeit
on my laptop is :
with :
A = np.random.rand(100000)
K = 10
%timeit X = max_rolling2(A,K)
170 ms ± 19.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit X = max_rolling1(A,K)
> 3.75 ms ± 479 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
The solution is totally similar to Divakar's answer here (full credit to him) but the final cropping of the array has different indices in this context:
maximum_filter1d(A, size=K)[K//2:-((K+1)//2)]
Example:
import numpy as np
from scipy.ndimage.filters import maximum_filter1d
A = np.random.randint(0, 10, (50))
K = 5
rollingmax = np.array([max(A[j-K:j]) for j in range(K,len(A))])
rollingmax2 = np.array([max(A[j:j+K]) for j in range(len(A)-K)])
rollingmax3 = maximum_filter1d(A,size=K)[K//2:-((K+1)//2)]
print A, rollingmax, rollingmax2, rollingmax3
[6 7 7 9 4 5 4 7 2 0 3 3 5 9 4 6 6 1 5 2 7 5 7 7 5 6 0 9 0 5 9 3 7 1 9 5 3 7 5 1 6 9 6 0 5 1 5 5 4 9]
[9 9 9 9 7 7 7 7 5 9 9 9 9 9 6 6 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 7 7 9 9 9 9 9 6 5 5]
[9 9 9 9 7 7 7 7 5 9 9 9 9 9 6 6 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 7 7 9 9 9 9 9 6 5 5]
[9 9 9 9 7 7 7 7 5 9 9 9 9 9 6 6 7 7 7 7 7 7 7 9 9 9 9 9 9 9 9 9 9 9 9 7 7 9 9 9 9 9 6 5 5]
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With