Finding local maxima/minima with Numpy in a 1D num

Question

Can you suggest a module function from numpy/scipy that can find local maxima/minima in a 1D numpy array? Obviously the simplest approach ever is to have a look at the nearest neighbours, but I would like to have an accepted solution that is part of the numpy distro.

Answer 1

As of SciPy version 1.1, you can also use find_peaks. Below are two examples taken from the documentation itself.

Using the height argument, one can select all maxima above a certain threshold (in this example, all non-negative maxima; this can be very useful if one has to deal with a noisy baseline):

import matplotlib.pyplot as plt
from scipy.misc import electrocardiogram
from scipy.signal import find_peaks
import numpy as np

x = electrocardiogram()[2000:4000]
peaks, _ = find_peaks(x, height=0)
plt.plot(x)
plt.plot(peaks, x[peaks], "x")
plt.plot(np.zeros_like(x), "--", color="gray")
plt.show()

Another extremely helpful argument is distance, which defines the minimum distance between two peaks:

peaks, _ = find_peaks(x, distance=150)
# difference between peaks is >= 150
print(np.diff(peaks))
# prints [186 180 177 171 177 169 167 164 158 162 172]

plt.plot(x)
plt.plot(peaks, x[peaks], "x")
plt.show()

Answer 2

Update: I wasn't happy with gradient so I found it more reliable to use numpy.diff. Please let me know if it does what you want.

Regarding the issue of noise, the mathematical problem is to locate maxima/minima if we want to look at noise we can use something like convolve which was mentioned earlier.

import numpy as np
from matplotlib import pyplot

a=np.array([10.3,2,0.9,4,5,6,7,34,2,5,25,3,-26,-20,-29],dtype=np.float)

gradients=np.diff(a)
print gradients


maxima_num=0
minima_num=0
max_locations=[]
min_locations=[]
count=0
for i in gradients[:-1]:
        count+=1

    if ((cmp(i,0)>0) & (cmp(gradients[count],0)<0) & (i != gradients[count])):
        maxima_num+=1
        max_locations.append(count)     

    if ((cmp(i,0)<0) & (cmp(gradients[count],0)>0) & (i != gradients[count])):
        minima_num+=1
        min_locations.append(count)


turning_points = {'maxima_number':maxima_num,'minima_number':minima_num,'maxima_locations':max_locations,'minima_locations':min_locations}  

print turning_points

pyplot.plot(a)
pyplot.show()

Answer 3

For curves with not too much noise, I recommend the following small code snippet:

from numpy import *

# example data with some peaks:
x = linspace(0,4,1e3)
data = .2*sin(10*x)+ exp(-abs(2-x)**2)

# that's the line, you need:
a = diff(sign(diff(data))).nonzero()[0] + 1 # local min+max
b = (diff(sign(diff(data))) > 0).nonzero()[0] + 1 # local min
c = (diff(sign(diff(data))) < 0).nonzero()[0] + 1 # local max


# graphical output...
from pylab import *
plot(x,data)
plot(x[b], data[b], "o", label="min")
plot(x[c], data[c], "o", label="max")
legend()
show()

The +1 is important, because diff reduces the original index number.

Answer 4

While this question is really old. I believe there is a much simpler approach in numpy (a one liner).

import numpy as np

list = [1,3,9,5,2,5,6,9,7]

np.diff(np.sign(np.diff(list))) #the one liner

#output
array([ 0, -2,  0,  2,  0,  0, -2])

To find a local max or min we essentially want to find when the difference between the values in the list (3-1, 9-3...) changes from positive to negative (max) or negative to positive (min). Therefore, first we find the difference. Then we find the sign, and then we find the changes in sign by taking the difference again. (Sort of like a first and second derivative in calculus, only we have discrete data and don't have a continuous function.)

The output in my example does not contain the extrema (the first and last values in the list). Also, just like calculus, if the second derivative is negative, you have max, and if it is positive you have a min.

Thus we have the following matchup:

[1,  3,  9,  5,  2,  5,  6,  9,  7]
    [0, -2,  0,  2,  0,  0, -2]
        Max     Min         Max

Answer 5

import numpy as np
x=np.array([6,3,5,2,1,4,9,7,8])
y=np.array([2,1,3,5,3,9,8,10,7])
sortId=np.argsort(x)
x=x[sortId]
y=y[sortId]
minm = np.array([])
maxm = np.array([])
i = 0
while i < length-1:
    if i < length - 1:
        while i < length-1 and y[i+1] >= y[i]:
            i+=1

        if i != 0 and i < length-1:
            maxm = np.append(maxm,i)

        i+=1

    if i < length - 1:
        while i < length-1 and y[i+1] <= y[i]:
            i+=1

        if i < length-1:
            minm = np.append(minm,i)
        i+=1


print minm
print maxm

minm and maxm contain indices of minima and maxima, respectively. For a huge data set, it will give lots of maximas/minimas so in that case smooth the curve first and then apply this algorithm.

Answer 6

None of these solutions worked for me since I wanted to find peaks in the center of repeating values as well. for example, in

ar = np.array([0,1,2,2,2,1,3,3,3,2,5,0])

the answer should be

array([ 3,  7, 10], dtype=int64)

I did this using a loop. I know it's not super clean, but it gets the job done.

def findLocalMaxima(ar):
# find local maxima of array, including centers of repeating elements    
maxInd = np.zeros_like(ar)
peakVar = -np.inf
i = -1
while i < len(ar)-1:
#for i in range(len(ar)):
    i += 1
    if peakVar < ar[i]:
        peakVar = ar[i]
        for j in range(i,len(ar)):
            if peakVar < ar[j]:
                break
            elif peakVar == ar[j]:
                continue
            elif peakVar > ar[j]:
                peakInd = i + np.floor(abs(i-j)/2)
                maxInd[peakInd.astype(int)] = 1
                i = j
                break
    peakVar = ar[i]
maxInd = np.where(maxInd)[0]
return maxInd