CODEWITHSUNDEEP
pythonnumpyscipyvectorization
Varlor
Varlor

Reputation: 1471

find_peaks does not identify a peak at the start of the array

I am trying to find a vectorized approach of finding the first position in an array where the values did not get higher than the maximum of n previous numbers. I thought about using the find_peaks method of scipy.signal to find a local maximum. I think it does exactly that if you define the distance to let's say 10 n is 10. But unfortunately, the condition for the distance has to be fulfilled in both directions - previous and upcoming numbers. Is there any other method or approach to finding such a thing?

Example:

arr1 = np.array([1.        , 0.73381293, 0.75649351, 0.77693474, 0.77884614,
       0.81055903, 0.81402439, 0.78798586, 0.78839588, 0.82967961,
       0.8448    , 0.83276451, 0.82539684, 0.81762916, 0.82722515,
       0.82101804, 0.82871127, 0.82825041, 0.82086957, 0.8347826 ,
       0.82666665, 0.82352942, 0.81270903, 0.81191224, 0.83180428,
       0.84975767, 0.84044236, 0.85057473, 0.8394649 , 0.80000001,
       0.83870965, 0.83962262, 0.85039371, 0.83359748, 0.84019768,
       0.83281732, 0.83660132])

from scipy.signal import find_peaks
peaks, _ = find_peaks(arr1, distance=10)

In this case, it finds positions 10 and 27. But also position 0 has 10 following elements which are not higher. How can I find those?

Upvotes: 9

Views: 4352

Answers (3)

Divakar
Divakar

Reputation: 221684

We can use 1D sliding-windowed max filter from SciPy. Also, it seems you are comparing against n previous elements. Since, the first element won't have any previous element, we need to make it ignore.

Hence, we would have the implementation, like so -

from scipy.ndimage.filters import maximum_filter1d

def peaks_previousN(a, n):
    W = (n-1)//2
    return np.flatnonzero(a[1:]>=maximum_filter1d(a, n, origin=W)[:-1])+1

Sample run with given sample array -

In [107]: peaks_previousN(arr1, n=10)
Out[107]: array([25, 27])

Upvotes: 1

CDJB
CDJB

Reputation: 14546

Unfortunately, find_peaks() works by comparing neighboring values - so will not identify peaks that occur at the beginning or end of the array. One workaround is to use np.concatenate() to insert the minimum value of the array at the beginning and end, and then subtract 1 from the peaks variable:

>>> import numpy as np
>>> peaks, _ = find_peaks(np.concatenate(([min(arr1)],arr1,[min(arr1)])), distance=10)
>>> peaks-1
array([ 0, 10, 27], dtype=int64)

Upvotes: 7

Frederik Bode
Frederik Bode

Reputation: 2744

def rolling_window(a, window):
    shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
    strides = a.strides + (a.strides[-1],)
    return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)

def get_peaks(arr, window):
    maxss = np.argmax(rolling_window(arr1, window), axis=1)
    return np.where(maxss == 0)[0]

>>> arr1 = np.array([1.        , 0.73381293, 0.75649351, 0.77693474, 0.77884614,
       0.81055903, 0.81402439, 0.78798586, 0.78839588, 0.82967961,
       0.8448    , 0.83276451, 0.82539684, 0.81762916, 0.82722515,
       0.82101804, 0.82871127, 0.82825041, 0.82086957, 0.8347826 ,
       0.82666665, 0.82352942, 0.81270903, 0.81191224, 0.83180428,
       0.84975767, 0.84044236, 0.85057473, 0.8394649 , 0.80000001,
       0.83870965, 0.83962262, 0.85039371, 0.83359748, 0.84019768,
       0.83281732, 0.83660132])
>>> get_peaks(arr1, 10)
array([ 0, 10, 27])

Credit for rolling window function : Rolling window for 1D arrays in Numpy?

Upvotes: 2

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