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])
>>> 

Answer 1

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:

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

Answer 2

I wrote an easy function:

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