Finding local maxima minima with Numpy in a 1D numpy array

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

User · Accepted Answer

If you are looking for all entries in the 1d array a smaller than their neighbors, you can try

numpy.r_[True, a[1:] < a[:-1]] & numpy.r_[a[:-1] < a[1:], True]

You could also smooth your array before this step using numpy.convolve().

I don't think there is a dedicated function for this.

User · Answer

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  if you want to find minima  just multiply you input by -1    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  gt   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

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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  gt 0   amp   cmp gradients count  0  lt 0   amp   i    gradients count             maxima num  1         max locations append count            if   cmp i 0  lt 0   amp   cmp gradients count  0  gt 0   amp   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

User · Answer

Another one     def local maxima mask vec               Get a mask of all points in vec which are local maxima      param vec  A real-valued vector      return  A boolean mask of the same size where True elements correspond to maxima               mask   np zeros vec shape  dtype np bool      greater than the last   np diff vec  gt 0    N-1     mask 1     greater than the last     mask  -1   amp    greater than the last     return mask

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In SciPy    0 11  import numpy as np from scipy signal import argrelextrema  x   np random random 12     for local maxima argrelextrema x  np greater     for local minima argrelextrema x  np less    Produces    gt  gt  gt  x array   0 56660112   0 76309473   0 69597908   0 38260156   0 24346445      0 56021785   0 24109326   0 41884061   0 35461957   0 54398472      0 59572658   0 92377974    gt  gt  gt  argrelextrema x  np greater   array  1  5  7      gt  gt  gt  argrelextrema x  np less   array  4  6  8       Note  these are the indices of x that are local max min  To get the values  try    gt  gt  gt  x argrelextrema x  np greater  0     scipy signal also provides argrelmax and argrelmin for finding maxima and minima respectively

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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  lt  len ar -1   for i in range len ar        i    1     if peakVar  lt  ar i           peakVar   ar i          for j in range i len ar                if peakVar  lt  ar j                   break             elif peakVar    ar j                   continue             elif peakVar  gt  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

User · Answer

Why not use Scipy built-in function signal find peaks cwt to do the job    from scipy import signal import numpy as np   generate junk data  numpy 1D arr  xs   np arange 0  np pi  0 05  data   np sin xs     maxima   use builtin function to find  max  peaks max peakind   signal find peaks cwt data  np arange 1 10      inverse   in order to find minima  inv data   1 data   minima   use builtin function fo find  min  peaks  use inversed data  min peakind   signal find peaks cwt inv data  np arange 1 10     show results print  maxima    data max peakind  print  minima    data min peakind    results   maxima   0 9995736  minima   0 09146464    Regards

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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

User · Answer

Another approach  more words  less code  that may help   The locations of local maxima and minima are also the locations of the zero crossings of the first derivative   It is generally much easier to find zero crossings than it is to directly find local maxima and minima   Unfortunately  the first derivative tends to  amplify  noise  so when significant noise is present in the original data  the first derivative is best used only after the original data has had some degree of smoothing applied   Since smoothing is  in the simplest sense  a low pass filter  the smoothing is often best  well  most easily  done by using a convolution kernel  and  shaping  that kernel can provide a surprising amount of feature-preserving enhancing capability   The process of finding an optimal kernel can be automated using a variety of means  but the best may be simple brute force  plenty fast for finding small kernels    A good kernel will  as intended  massively distort the original data  but it will NOT affect the location of the peaks valleys of interest   Fortunately  quite often a suitable kernel can be created via a simple SWAG   educated guess    The width of the smoothing kernel should be a little wider than the widest expected  interesting  peak in the original data  and its shape will resemble that peak  a single-scaled wavelet    For mean-preserving kernels  what any good smoothing filter should be  the sum of the kernel elements should be precisely equal to 1 00  and the kernel should be symmetric about its center  meaning it will have an odd number of elements   Given an optimal smoothing kernel  or a small number of kernels optimized for different data content   the degree of smoothing becomes a scaling factor for  the  gain  of  the convolution kernel   Determining the  correct   optimal  degree of smoothing  convolution kernel gain  can even be automated  Compare the standard deviation of the first derivative data with the standard deviation of the smoothed data   How the ratio of the two standard deviations changes with changes in the degree of smoothing cam be used to predict effective smoothing values   A few manual data runs  that are truly representative  should be all that s needed   All the prior solutions posted above compute the first derivative  but they don t treat it as a statistical measure  nor do the above solutions attempt to performing feature preserving enhancing smoothing  to help subtle peaks  leap above  the noise    Finally  the bad news  Finding  real  peaks becomes a royal pain when the noise also has features that look like real peaks  overlapping bandwidth    The next more-complex solution is generally to use a longer convolution kernel  a  wider kernel aperture   that takes into account the relationship between adjacent  real  peaks  such as minimum or maximum rates for peak occurrence   or to use multiple convolution passes using kernels having different widths  but only if it is faster  it is a fundamental mathematical truth that linear convolutions performed in sequence can always be convolved together into a single convolution    But it is often far easier to first find a sequence of useful kernels  of varying widths  and convolve them together than it is to directly find the final kernel in a single step   Hopefully this provides enough info to let Google  and perhaps a good stats text  fill in the gaps   I really wish I had the time to provide a worked example  or a link to one   If anyone comes across one online  please post it here

User · Answer

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  lt  length-1      if i  lt  length - 1          while i  lt  length-1 and y i 1   gt   y i               i  1          if i    0 and i  lt  length-1              maxm   np append maxm i           i  1      if i  lt  length - 1          while i  lt  length-1 and y i 1   lt   y i               i  1          if i  lt  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

User · Answer

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     gt  0  nonzero   0    1   local min c    diff sign diff data     lt  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

User · Answer

Another solution using essentially a dilate operator   import numpy as np from scipy ndimage import rank filter  def find local maxima x      x dilate   rank filter x  -1  size 3     return x dilate    x    and for the minima   def find local minima x      x erode   rank filter x  -0  size 3     return x erode    x    Also  from scipy ndimage you can replace rank filter x  -1  size 3  with grey dilation and rank filter x  0  size 3  with grey erosion  This won t require a local sort  so it is slightly faster

[python] Finding local maxima/minima with Numpy in a 1D numpy array

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