calculate histogram peaks in python

Question

In Python, how do I calcuate the peaks of a histogram?

I tried this:

import numpy as np
from scipy.signal import argrelextrema

data = [0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 1, 2, 3, 4,

        5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9,

        12,

        15, 16, 17, 18, 19, 15, 16, 17, 18, 

        19, 20, 21, 22, 23, 24,]

h = np.histogram(data, bins=[0, 5, 10, 15, 20, 25])
hData = h[0]
peaks = argrelextrema(hData, np.greater)

But the result was:

(array([3]),)

I'd expect it to find the peaks in bin 0 and bin 3.

Note that the peaks span more than 1 bin. I don't want it to consider the peaks that span more than 1 column as additional peak.

I'm open to another way to get the peaks.

Note:

>>> h[0]
array([19, 15,  1, 10,  5])
>>>

argrelextrema returns local maxima and minima - index three is the only local maxima in the array. How are you defining peak? You could write your own solution once you've defined the requirement and worked out the logic. — wwii
– wwii, Commented Aug 10, 2015 at 3:15
The accepted answer to Finding local maxima/minima with Numpy in a 1D numpy array should get you on your way. — wwii
– wwii, Commented Aug 10, 2015 at 3:30
@wwii I'm not sure how to describe in words but I want where the derivatives would be zero if this was a continuous function and the slope approaching this is positive. — brian
– brian, Commented Aug 11, 2015 at 1:02
I want the endpoints to be inclusive. Ultimately I’d want it to find all max values if this was a continuous function. — brian
– brian, Commented Aug 11, 2015 at 12:22

S. Huber · Accepted Answer · 2017-10-11 17:36:55Z

12

In computational topology, the formalism of persistent homology provides a definition of "peak" that seems to address your need. In the 1-dimensional case the peaks are illustrated by the blue bars in the following figure:

Most persistent peaks

A description of the algorithm is given in this Stack Overflow answer of a peak detection question.

The nice thing is that this method not only identifies the peaks but it quantifies the "significance" in a natural way.

A simple and efficient implementation (as fast as sorting numbers) and the source material to the above answer given in this blog article: https://www.sthu.org/blog/13-perstopology-peakdetection/index.html

answered Oct 11, 2017 at 17:36

S. Huber

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1 Comment

maxschlepzig Over a year ago

Hm, perhaps you can add a usage note/example for someone who isn't already familiar with computational topology? Say I have some list/array from which I plot my histogram - how do I use your example implementation, concretely. I noticed that get_persistent_homology() returns different results when the input is sorted or not. Also, what is returned exactly by that function? A list of Peak elements where each 'Peak' has left/right/born/died fields ... How do I get the 'real' peak from that?

erdogant · Accepted Answer · 2020-07-31 20:45:09Z

Try the findpeaks library.

pip install findpeaks

# Your input data:
data = [0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 12, 15, 16, 17, 18, 19, 15, 16, 17, 18,  19, 20, 21, 22, 23, 24,]

# import library
from findpeaks import findpeaks

# Find some peaks using the smoothing parameter.
fp = findpeaks(lookahead=1, interpolate=10)
# fit
results = fp.fit(data)
# Make plot
fp.plot()

# Results with respect to original input data.
results['df']

# Results based on interpolated smoothed data.
results['df_interp']

squaregoldfish · Accepted Answer · 2023-08-23 18:20:03Z

I wrote an easy function:

def find_peaks(a):
  x = np.array(a)
  max = np.max(x)
  length = len(a)
  ret = []
  for i in range(length):
      ispeak = True
      if i-1 > 0:
          ispeak &= (x[i] > 1.8 * x[i-1])
      if i+1 < length:
          ispeak &= (x[i] > 1.8 * x[i+1])
    
      ispeak &= (x[i] > 0.05 * max)
      if ispeak:
          ret.append(i)
  return ret

I defined a peak as a value bigger than 180% that of the neighbors and bigger than 5% of the max value. Of course you can adapt the values as you prefer in order to find the best set up for your problem.

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