Develop Reference

Is there a way I can find the range of local maxima of histogram?

I'm wondering if there's a way I can find the range of local maxima of a histogram. For instance, suppose I have the following histogram (just ignore the orange curve):
The histogram is actually obtained from a dictionary. I'm hoping to find the range of local maxima of this histogram (on the horizontal axis), which are, say, 1.3-1.6, and 2.1-2.4 in this case. I have no idea which tools would be helpful or which techniques I may want to use. I know there's a tool to find local maxima of a 1-D array:
from scipy.signal import argrelextrema
x = np.random.random(12)
argrelextrema(x, np.greater)
but I don't think it would work here since I'm looking for a range, and there're some 'wiggles' on the histogram. Can anyone give me some suggestions/examples about how I can obtain the range I'm looking for? Thanks a lot for the help
PS: I trying to not just search for the ranges of x whose y values are above a certain limit:)

I don't know if I correctly understand what you want to do, but you can treat the histogram as a Probability Density Function (PDF) of a bimodal distribution, then find the modes and the Highest Density Intervals (HDIs) around the two modes.
So, I create some sample data
import numpy as np
import pandas as pd
import scipy.stats as sps
from scipy.signal import find_peaks, argrelextrema
import matplotlib.pyplot as plt
d1 = sps.norm(loc=1.3, scale=.2)
d2 = sps.norm(loc=2.2, scale=.3)
r1 = d1.rvs(size=5000, random_state=1)
r2 = d2.rvs(size=5000, random_state=1)
r = np.concatenate((r1, r2))
h = plt.hist(r, bins=100, density=True);
We have only h, the result of the hist function that will contains the density (100) and the ranges of the bins (101).
print(h[0].size)
100
print(h[1].size)
101
So we first need to choose the mean of each bin
density = h[0]
values = h[1][:-1] + np.diff(h[1])[0] / 2
plt.hist(r, bins=100, density=True, alpha=.25)
plt.plot(values, density);
Now we can normalize the PDF (to sum to 1) and smooth the data with moving average that we'll use only to get the peaks (maxima) and minima
norm_density = density / density.sum()
norm_density_ma = pd.Series(norm_density).rolling(7, center=True).mean().values
plt.plot(values, norm_density_ma)
plt.plot(values, norm_density);
Now we can obtain indexes of maxima
peaks = find_peaks(norm_density_ma)[0]
peaks
array([24, 57])
and minima
minima = argrelextrema(norm_density_ma, np.less)[0]
minima
array([40])
and check they're correct
plt.plot(values, norm_density_ma)
plt.plot(values, norm_density)
for peak in peaks:
plt.axvline(values[peak], color='r')
plt.axvline(values[minima], color='k', ls='--');
Finally, we have to find out the HDIs around the two modes (peaks) from the normalized h histogram data. We can use a simple function to get the HDI of grid (see HDI_of_grid for details and Doing Bayesian Data Analysis by John K. Kruschke)
def HDI_of_grid(probMassVec, credMass=0.95):
sortedProbMass = np.sort(probMassVec, axis=None)[::-1]
HDIheightIdx = np.min(np.where(np.cumsum(sortedProbMass) >= credMass))
HDIheight = sortedProbMass[HDIheightIdx]
HDImass = np.sum(probMassVec[probMassVec >= HDIheight])
idx = np.where(probMassVec >= HDIheight)[0]
return {'indexes':idx, 'mass':HDImass, 'height':HDIheight}
Let's say we want the HDIs to contain a mass of 0.3
# HDI around the 1st mode
hdi1 = HDI_of_grid(norm_density, credMass=.3)
plt.plot(values, norm_density_ma)
plt.plot(values, norm_density)
plt.fill_between(
values[hdi1['indexes']],
0, norm_density[hdi1['indexes']],
alpha=.25
)
for peak in peaks:
plt.axvline(values[peak], color='r')
for the 2nd mode, we'll get HDI from minima to avoid the 1st mode
# HDI around the 2nd mode
hdi2 = HDI_of_grid(norm_density[minima[0]:], credMass=.3)
plt.plot(values, norm_density_ma)
plt.plot(values, norm_density)
plt.fill_between(
values[hdi1['indexes']],
0, norm_density[hdi1['indexes']],
alpha=.25
)
plt.fill_between(
values[hdi2['indexes']+minima],
0, norm_density[hdi2['indexes']+minima],
alpha=.25
)
for peak in peaks:
plt.axvline(values[peak], color='r')
And we have the values of the two HDIs
# 1st mode
values[peaks[0]]
1.320249129265321
# 0.3 HDI
values[hdi1['indexes']].take([0, -1])
array([1.12857599, 1.45715851])
# 2nd mode
values[peaks[1]]
2.2238510564735363
# 0.3 HDI
values[hdi2['indexes']+minima].take([0, -1])
array([1.95003229, 2.47028795])

2D FFT from MATLAB to Python

Most of my programming experience is in MATLAB and I recently started get familiar with Python.
I came across some great MATLAB code here that pertains to some things I'd like to work with, so I've tried to recreate it in Python:
import numpy as np
import math
import matplotlib.pyplot as plt
x = np.linspace(-2, 2, 100) # seconds
y = np.linspace(-3, 3, 200) # seconds
xFreq = 2; # Hz
yFreq = -3; # Hz
a = np.matrix(np.matrix(np.exp(2j * np.pi * y * yFreq)))
b = np.matrix(np.exp(2j * np.pi * np.matrix(x).T * xFreq))
c = np.dot(b,a).T
plt.imshow(c.real, cmap='gray', extent = [min(x), max(x), min(y), max(y)], aspect=2/3);
plt.colorbar()
plt.xlabel('x (Sec)')
plt.ylabel('y (Sec)')
plt.show()
nfftx = len(x);
fs = 1/np.diff(x)[0];
fx = np.linspace(-1,1,nfftx) * fs/2;
nffty = len(y);
fs = 1/np.diff(y)[0];
fy = np.linspace(-1,1,nffty) * fs/2;
imF = np.fft.fftshift(np.fft.fft2(c))/np.size(c)
plt.figure()
plt.title("FFT (real)")
plt.imshow(np.real(imF), cmap='gray', extent = [min(fx), max(fx), min(fy), max(fy)], aspect=2/3)
plt.xlabel('fx (Hz)')
plt.ylabel('fy (Hz)')
Any idea why the y frequency is shown at 3 Hz vice -3 Hz
I couldn't understand what the original commentator was doing in MATLAB with these two lines:
Nfft = 4 * 2 .^ nextpow2(size(im));
imF = fftshift(fft2(im, Nfft(1), Nfft(2))) / numel(im);
which is likely why my FFT signal is so intense relative to background. Thoughts on how I could adjust my FFT in Python?

I only have a partial answer.
If you look closely, the colors on the sinusoidal image generated with your Python code and the one generated with Matlab code you linked have inverted color (check the colors of the stripes closer to edges, and the colors on the color bar).
That explains why you have inverted colors on the FFT plot, and may be why you got 3 Hz, instead of -3 Hz. Unfortunately, I do not have access to a computer with Python right now and won't be able to verify this. I guess this may be a good thing to start troubleshooting with.
EDIT:
Yes, you are right. I completely missed the flipud in the Matlab script. I do no think your c calculation is wrong. The easiest way to check that is to save the Matlab data and import it to Python.
In Matlab:
save('data.mat', 'im')
Then import that to Python using scipy:
im_matlab = scipy.io.loadmat('data.mat')['im']
np.all(np.isclose(im_matlab, im))
If the last line gives you True, then that means the calculations are correct.
About the plot, imshow assumes that the origin (0-th index of the numpy array) is the top left corner, which is the norm for images. You can change this by using origin='lower' keyword in plt.imshow.
About fftshift, I think this answer to a different StackOverflow question is what you need.

Analyzing seasonality of Google trend time series using FFT

I am trying to evaluate the amplitude spectrum of the Google trends time series using a fast Fourier transformation. If you look at the data for 'diet' in the data provided here it shows a very strong seasonal pattern:
I thought I could analyze this pattern using a FFT, which presumably should have a strong peak for a period of 1 year.
However when I apply a FFT like this (a_gtrend_ham being the time series multiplied with a Hamming window):
import matplotlib.pyplot as plt
import numpy as np
from numpy.fft import fft, fftshift
import pandas as pd
gtrend = pd.read_csv('multiTimeline.csv',index_col=0)
gtrend.index = pd.to_datetime(gtrend.index, format='%Y-%m')
# Sampling rate
fs = 12 #Points per year
a_gtrend_orig = gtrend['diet: (Worldwide)']
N_gtrend_orig = len(a_gtrend_orig)
length_gtrend_orig = N_gtrend_orig / fs
t_gtrend_orig = np.linspace(0, length_gtrend_orig, num = N_gtrend_orig, endpoint = False)
a_gtrend_sel = a_gtrend_orig.loc['2005-01-01 00:00:00':'2017-12-01 00:00:00']
N_gtrend = len(a_gtrend_sel)
length_gtrend = N_gtrend / fs
t_gtrend = np.linspace(0, length_gtrend, num = N_gtrend, endpoint = False)
a_gtrend_zero_mean = a_gtrend_sel - np.mean(a_gtrend_sel)
ham = np.hamming(len(a_gtrend_zero_mean))
a_gtrend_ham = a_gtrend_zero_mean * ham
N_gtrend = len(a_gtrend_ham)
ampl_gtrend = 1/N_gtrend * abs(fft(a_gtrend_ham))
mag_gtrend = fftshift(ampl_gtrend)
freq_gtrend = np.linspace(-0.5, 0.5, len(ampl_gtrend))
response_gtrend = 20 * np.log10(mag_gtrend)
response_gtrend = np.clip(response_gtrend, -100, 100)
My resulting amplitude spectrum does not show any dominant peak:
Where is my misunderstanding of how to use the FFT to get the spectrum of the data series?

Here is a clean implementation of what I think you are trying to accomplish. I include graphical output and a brief discussion of what it likely means.
First, we use the rfft() because the data is real valued. This saves time and effort (and reduces the bug rate) that otherwise follows from generating the redundant negative frequencies. And we use rfftfreq() to generate the frequency list (again, it is unnecessary to hand code it, and using the api reduces the bug rate).
For your data, the Tukey window is more appropriate than the Hamming and similar cos or sin based window functions. Notice also that we subtract the median before multiplying by the window function. The median() is a fairly robust estimate of the baseline, certainly more so than the mean().
In the graph you can see that the data falls quickly from its intitial value and then ends low. The Hamming and similar windows, sample the middle too narrowly for this and needlessly attenuate a lot of useful data.
For the FT graphs, we skip the zero frequency bin (the first point) since this only contains the baseline and omitting it provides a more convenient scaling for the y-axes.
You will notice some high frequency components in the graph of the FT output.
I include a sample code below that illustrates a possible origin of those high frequency components.
Okay here is the code:
import matplotlib.pyplot as plt
import numpy as np
from numpy.fft import rfft, rfftfreq
from scipy.signal import tukey
from numpy.fft import fft, fftshift
import pandas as pd
gtrend = pd.read_csv('multiTimeline.csv',index_col=0,skiprows=2)
#print(gtrend)
gtrend.index = pd.to_datetime(gtrend.index, format='%Y-%m')
#print(gtrend.index)
a_gtrend_orig = gtrend['diet: (Worldwide)']
t_gtrend_orig = np.linspace( 0, len(a_gtrend_orig)/12, len(a_gtrend_orig), endpoint=False )
a_gtrend_windowed = (a_gtrend_orig-np.median( a_gtrend_orig ))*tukey( len(a_gtrend_orig) )
plt.subplot( 2, 1, 1 )
plt.plot( t_gtrend_orig, a_gtrend_orig, label='raw data' )
plt.plot( t_gtrend_orig, a_gtrend_windowed, label='windowed data' )
plt.xlabel( 'years' )
plt.legend()
a_gtrend_psd = abs(rfft( a_gtrend_orig ))
a_gtrend_psdtukey = abs(rfft( a_gtrend_windowed ) )
# Notice that we assert the delta-time here,
# It would be better to get it from the data.
a_gtrend_freqs = rfftfreq( len(a_gtrend_orig), d = 1./12. )
# For the PSD graph, we skip the first two points, this brings us more into a useful scale
# those points represent the baseline (or mean), and are usually not relevant to the analysis
plt.subplot( 2, 1, 2 )
plt.plot( a_gtrend_freqs[1:], a_gtrend_psd[1:], label='psd raw data' )
plt.plot( a_gtrend_freqs[1:], a_gtrend_psdtukey[1:], label='windowed psd' )
plt.xlabel( 'frequency ($yr^{-1}$)' )
plt.legend()
plt.tight_layout()
plt.show()
And here is the output displayed graphically. There are strong signals at 1/year and at 0.14 (which happens to be 1/2 of 1/14 yrs), and there is a set of higher frequency signals that at first perusal might seem quite mysterious.
We see that the windowing function is actually quite effective in bringing the data to baseline and you see that the relative signal strengths in the FT are not altered very much by applying the window function.
If you look at the data closely, there seems to be some repeated variations within the year. If those occur with some regularity, they can be expected to appear as signals in the FT, and indeed the presence or absence of signals in the FT is often used to distinguish between signal and noise. But as will be shown, there is a better explanation for the high frequency signals.
Okay, now here is a sample code that illustrates one way those high frequency components can be produced. In this code, we create a single tone, and then we create a set of spikes at the same frequency as the tone. Then we Fourier transform the two signals and finally, graph the raw and FT data.
import matplotlib.pyplot as plt
import numpy as np
from numpy.fft import rfft, rfftfreq
t = np.linspace( 0, 1, 1000. )
y = np.cos( 50*3.14*t )
y2 = [ 1. if 1.-v < 0.01 else 0. for v in y ]
plt.subplot( 2, 1, 1 )
plt.plot( t, y, label='tone' )
plt.plot( t, y2, label='spikes' )
plt.xlabel('time')
plt.subplot( 2, 1, 2 )
plt.plot( rfftfreq(len(y),d=1/100.), abs( rfft(y) ), label='tone' )
plt.plot( rfftfreq(len(y2),d=1/100.), abs( rfft(y2) ), label='spikes' )
plt.xlabel('frequency')
plt.legend()
plt.tight_layout()
plt.show()
Okay, here are the graphs of the tone, and the spikes, and then their Fourier transforms. Notice that the spikes produce high frequency components that are very similar to those in our data.
In other words, the origin of the high frequency components is very likely in the short time scales associated with the spikey character of signals in the raw data.

Autocorrelation code in Python produces errors (guitar pitch detection)

This link provides code for an autocorrelation-based pitch detection algorithm. I am using it to detect pitches in simple guitar melodies.
In general, it produces very good results. For example, for the melody C4, C#4, D4, D#4, E4 it outputs:
262.743653536
272.144441273
290.826273006
310.431336809
327.094621169
Which correlates to the correct notes.
However, in some cases like this audio file (E4, F4, F#4, G4, G#4, A4, A#4, B4) it produces errors:
325.861452246
13381.6439242
367.518651703
391.479384923
414.604661221
218.345286173
466.503751322
244.994090035
More specifically, there are three errors here: 13381Hz is wrongly detected instead of F4 (~350Hz) (weird error), and also 218Hz instead of A4 (440Hz) and 244Hz instead of B4 (~493Hz), which are octave errors.
I assume the two errors are caused by something different? Here is the code:
slices = segment_signal(y, sr)
for segment in slices:
pitch = freq_from_autocorr(segment, sr)
print pitch
def segment_signal(y, sr, onset_frames=None, offset=0.1):
if (onset_frames == None):
onset_frames = remove_dense_onsets(librosa.onset.onset_detect(y=y, sr=sr))
offset_samples = int(librosa.time_to_samples(offset, sr))
print onset_frames
slices = np.array([y[i : i + offset_samples] for i
in librosa.frames_to_samples(onset_frames)])
return slices
You can see the freq_from_autocorr function in the first link above.
The only think that I have changed is this line:
corr = corr[len(corr)/2:]
Which I have replaced with:
corr = corr[int(len(corr)/2):]
UPDATE:
I noticed the smallest the offset I use (the smallest the signal segment I use to detect each pitch), the more high-frequency (10000+ Hz) errors I get.
Specifically, I noticed that the part that goes differently in those cases (10000+ Hz) is the calculation of the i_peak value. When in cases with no error it is in the range of 50-150, in the case of the error it is 3-5.

The autocorrelation function in the code snippet that you linked is not particularly robust. In order to get the correct result, it needs to locate the first peak on the left hand side of the autocorrelation curve. The method that the other developer used (calling the numpy.argmax() function) does not always find the correct value.
I've implemented a slightly more robust version, using the peakutils package. I don't promise that it's perfectly robust either, but in any case it achieves a better result than the version of the freq_from_autocorr() function that you were previously using.
My example solution is listed below:
import librosa
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import fftconvolve
from pprint import pprint
import peakutils
def freq_from_autocorr(signal, fs):
# Calculate autocorrelation (same thing as convolution, but with one input
# reversed in time), and throw away the negative lags
signal -= np.mean(signal) # Remove DC offset
corr = fftconvolve(signal, signal[::-1], mode='full')
corr = corr[len(corr)//2:]
# Find the first peak on the left
i_peak = peakutils.indexes(corr, thres=0.8, min_dist=5)[0]
i_interp = parabolic(corr, i_peak)[0]
return fs / i_interp, corr, i_interp
def parabolic(f, x):
"""
Quadratic interpolation for estimating the true position of an
inter-sample maximum when nearby samples are known.
f is a vector and x is an index for that vector.
Returns (vx, vy), the coordinates of the vertex of a parabola that goes
through point x and its two neighbors.
Example:
Defining a vector f with a local maximum at index 3 (= 6), find local
maximum if points 2, 3, and 4 actually defined a parabola.
In [3]: f = [2, 3, 1, 6, 4, 2, 3, 1]
In [4]: parabolic(f, argmax(f))
Out[4]: (3.2142857142857144, 6.1607142857142856)
"""
xv = 1/2. * (f[x-1] - f[x+1]) / (f[x-1] - 2 * f[x] + f[x+1]) + x
yv = f[x] - 1/4. * (f[x-1] - f[x+1]) * (xv - x)
return (xv, yv)
# Time window after initial onset (in units of seconds)
window = 0.1
# Open the file and obtain the sampling rate
y, sr = librosa.core.load("./Vocaroo_s1A26VqpKgT0.mp3")
idx = np.arange(len(y))
# Set the window size in terms of number of samples
winsamp = int(window * sr)
# Calcualte the onset frames in the usual way
onset_frames = librosa.onset.onset_detect(y=y, sr=sr)
onstm = librosa.frames_to_time(onset_frames, sr=sr)
fqlist = [] # List of estimated frequencies, one per note
crlist = [] # List of autocorrelation arrays, one array per note
iplist = [] # List of peak interpolated peak indices, one per note
for tm in onstm:
startidx = int(tm * sr)
freq, corr, ip = freq_from_autocorr(y[startidx:startidx+winsamp], sr)
fqlist.append(freq)
crlist.append(corr)
iplist.append(ip)
pprint(fqlist)
# Choose which notes to plot (it's set to show all 8 notes in this case)
plidx = [0, 1, 2, 3, 4, 5, 6, 7]
# Plot amplitude curves of all notes in the plidx list
fgwin = plt.figure(figsize=[8, 10])
fgwin.subplots_adjust(bottom=0.0, top=0.98, hspace=0.3)
axwin = []
ii = 1
for tm in onstm[plidx]:
axwin.append(fgwin.add_subplot(len(plidx)+1, 1, ii))
startidx = int(tm * sr)
axwin[-1].plot(np.arange(startidx, startidx+winsamp), y[startidx:startidx+winsamp])
ii += 1
axwin[-1].set_xlabel('Sample ID Number', fontsize=18)
fgwin.show()
# Plot autocorrelation function of all notes in the plidx list
fgcorr = plt.figure(figsize=[8,10])
fgcorr.subplots_adjust(bottom=0.0, top=0.98, hspace=0.3)
axcorr = []
ii = 1
for cr, ip in zip([crlist[ii] for ii in plidx], [iplist[ij] for ij in plidx]):
if ii == 1:
shax = None
else:
shax = axcorr[0]
axcorr.append(fgcorr.add_subplot(len(plidx)+1, 1, ii, sharex=shax))
axcorr[-1].plot(np.arange(500), cr[0:500])
# Plot the location of the leftmost peak
axcorr[-1].axvline(ip, color='r')
ii += 1
axcorr[-1].set_xlabel('Time Lag Index (Zoomed)', fontsize=18)
fgcorr.show()
The printed output looks like:
In [1]: %run autocorr.py
[325.81996740236065,
346.43374761017725,
367.12435233192753,
390.17291696559079,
412.9358117076161,
436.04054933498134,
465.38986619237039,
490.34120132405866]
The first figure produced by my code sample depicts the amplitude curves for the next 0.1 seconds following each detected onset time:
The second figure produced by the code shows the autocorrelation curves, as computed inside of the freq_from_autocorr() function. The vertical red lines depict the location of the first peak on the left for each curve, as estimated by the peakutils package. The method used by the other developer was getting incorrect results for some of these red lines; that's why his version of that function was occasionally returning the wrong frequencies.
My suggestion would be to test the revised version of the freq_from_autocorr() function on other recordings, see if you can find more challenging examples where even the improved version still gives incorrect results, and then get creative and try to develop an even more robust peak finding algorithm that never, ever mis-fires.

The autocorrelation method is not always right. You may want to implement a more sophisticated method like YIN:
http://audition.ens.fr/adc/pdf/2002_JASA_YIN.pdf
or MPM:
http://www.cs.otago.ac.nz/tartini/papers/A_Smarter_Way_to_Find_Pitch.pdf
Both of the above papers are good reads.

find peaks location in a spectrum numpy

I have a TOF spectrum and I would like to implement an algorithm using python (numpy) that finds all the maxima of the spectrum and returns the corresponding x values.
I have looked up online and I found the algorithm reported below.
The assumption here is that near the maximum the difference between the value before and the value at the maximum is bigger than a number DELTA. The problem is that my spectrum is composed of points equally distributed, even near the maximum, so that DELTA is never exceeded and the function peakdet returns an empty array.
Do you have any idea how to overcome this problem? I would really appreciate comments to understand better the code since I am quite new in python.
Thanks!
import sys
from numpy import NaN, Inf, arange, isscalar, asarray, array
def peakdet(v, delta, x = None):
maxtab = []
mintab = []
if x is None:
x = arange(len(v))
v = asarray(v)
if len(v) != len(x):
sys.exit('Input vectors v and x must have same length')
if not isscalar(delta):
sys.exit('Input argument delta must be a scalar')
if delta <= 0:
sys.exit('Input argument delta must be positive')
mn, mx = Inf, -Inf
mnpos, mxpos = NaN, NaN
lookformax = True
for i in arange(len(v)):
this = v[i]
if this > mx:
mx = this
mxpos = x[i]
if this < mn:
mn = this
mnpos = x[i]
if lookformax:
if this < mx-delta:
maxtab.append((mxpos, mx))
mn = this
mnpos = x[i]
lookformax = False
else:
if this > mn+delta:
mintab.append((mnpos, mn))
mx = this
mxpos = x[i]
lookformax = True
return array(maxtab), array(mintab)
Below is shown part of the spectrum. I actually have more peaks than those shown here.

This, I think could work as a starting point. I'm not a signal-processing expert, but I tried this on a generated signal Y that looks quite like yours and one with much more noise:
from scipy.signal import convolve
import numpy as np
from matplotlib import pyplot as plt
#Obtaining derivative
kernel = [1, 0, -1]
dY = convolve(Y, kernel, 'valid')
#Checking for sign-flipping
S = np.sign(dY)
ddS = convolve(S, kernel, 'valid')
#These candidates are basically all negative slope positions
#Add one since using 'valid' shrinks the arrays
candidates = np.where(dY < 0)[0] + (len(kernel) - 1)
#Here they are filtered on actually being the final such position in a run of
#negative slopes
peaks = sorted(set(candidates).intersection(np.where(ddS == 2)[0] + 1))
plt.plot(Y)
#If you need a simple filter on peak size you could use:
alpha = -0.0025
peaks = np.array(peaks)[Y[peaks] < alpha]
plt.scatter(peaks, Y[peaks], marker='x', color='g', s=40)
The sample outcomes:
For the noisy one, I filtered peaks with alpha:
If the alpha needs more sophistication you could try dynamically setting alpha from the peaks discovered using e.g. assumptions about them being a mixed gaussian (my favourite being the Otsu threshold, exists in cv and skimage) or some sort of clustering (k-means could work).
And for reference, this I used to generate the signal:
Y = np.zeros(1000)
def peaker(Y, alpha=0.01, df=2, loc=-0.005, size=-.0015, threshold=0.001, decay=0.5):
peaking = False
for i, v in enumerate(Y):
if not peaking:
peaking = np.random.random() < alpha
if peaking:
Y[i] = loc + size * np.random.chisquare(df=2)
continue
elif Y[i - 1] < threshold:
peaking = False
if i > 0:
Y[i] = Y[i - 1] * decay
peaker(Y)
EDIT: Support for degrading base-line
I simulated a slanting base-line by doing this:
Z = np.log2(np.arange(Y.size) + 100) * 0.001
Y = Y + Z[::-1] - Z[-1]
Then to detect with a fixed alpha (note that I changed sign on alpha):
from scipy.signal import medfilt
alpha = 0.0025
Ybase = medfilt(Y, 51) # 51 should be large in comparison to your peak X-axis lengths and an odd number.
peaks = np.array(peaks)[Ybase[peaks] - Y[peaks] > alpha]
Resulting in the following outcome (the base-line is plotted as dashed black line):
EDIT 2: Simplification and a comment
I simplified the code to use one kernel for both convolves as #skymandr commented. This also removed the magic number in adjusting the shrinkage so that any size of the kernel should do.
For the choice of "valid" as option to convolve. It would probably have worked just as well with "same", but I choose "valid" so I didn't have to think about the edge-conditions and if the algorithm could detect spurios peaks there.

As of SciPy version 1.1, you can also use find_peaks:
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import find_peaks
np.random.seed(0)
Y = np.zeros(1000)
# insert #deinonychusaur's peaker function here
peaker(Y)
# make data noisy
Y = Y + 10e-4 * np.random.randn(len(Y))
# find_peaks gets the maxima, so we multiply our signal by -1
Y *= -1
# get the actual peaks
peaks, _ = find_peaks(Y, height=0.002)
# multiply back for plotting purposes
Y *= -1
plt.plot(Y)
plt.plot(peaks, Y[peaks], "x")
plt.show()
This will plot (note that we use height=0.002 which will only find peaks higher than 0.002):
In addition to height, we can also set the minimal distance between two peaks. If you use distance=100, the plot then looks as follows:
You can use
peaks, _ = find_peaks(Y, height=0.002, distance=100)
in the code above.

After looking at the answers and suggestions I decided to offer a solution I often use because it is straightforward and easier to tweak.
It uses a sliding window and counts how many times a local peak appears as a maximum as window shifts along the x-axis. As #DrV suggested, no universal definition of "local maximum" exists, meaning that some tuning parameters are unavoidable. This function uses "window size" and "frequency" to fine tune the outcome. Window size is measured in number of data points of independent variable (x) and frequency counts how sensitive should peak detection be (also expressed as a number of data points; lower values of frequency produce more peaks and vice versa). The main function is here:
def peak_finder(x0, y0, window_size, peak_threshold):
# extend x, y using window size
y = numpy.concatenate([y0, numpy.repeat(y0[-1], window_size)])
x = numpy.concatenate([x0, numpy.arange(x0[-1], x0[-1]+window_size)])
local_max = numpy.zeros(len(x0))
for ii in range(len(x0)):
local_max[ii] = x[y[ii:(ii + window_size)].argmax() + ii]
u, c = numpy.unique(local_max, return_counts=True)
i_return = numpy.where(c>=peak_threshold)[0]
return(list(zip(u[i_return], c[i_return])))
along with a snippet used to produce the figure shown below:
import numpy
from matplotlib import pyplot
def plot_case(axx, w_f):
p = peak_finder(numpy.arange(0, len(Y)), -Y, w_f[0], w_f[1])
r = .9*min(Y)/10
axx.plot(Y)
for ip in p:
axx.text(ip[0], r + Y[int(ip[0])], int(ip[0]),
rotation=90, horizontalalignment='center')
yL = pyplot.gca().get_ylim()
axx.set_ylim([1.15*min(Y), yL[1]])
axx.set_xlim([-50, 1100])
axx.set_title(f'window: {w_f[0]}, count: {w_f[1]}', loc='left', fontsize=10)
return(None)
window_frequency = {1:(15, 15), 2:(100, 100), 3:(100, 5)}
f, ax = pyplot.subplots(1, 3, sharey='row', figsize=(9, 4),
gridspec_kw = {'hspace':0, 'wspace':0, 'left':.08,
'right':.99, 'top':.93, 'bottom':.06})
for k, v in window_frequency.items():
plot_case(ax[k-1], v)
pyplot.show()
Three cases show parameter values that render (from left to right panel):
(1) too many, (2) too few, and (3) an intermediate amount of peaks.
To generate Y data, I used the function #deinonychusaur gave above, and added some noise to it from #Cleb's answer.
I hope some might find this useful, but it's efficiency primarily depends on actual peak shapes and distances.

Finding a minimum or a maximum is not that simple, because there is no universal definition for "local maximum".
Your code seems to look for a miximum and then accept it as a maximum if the signal falls after the maximum below the maximum minus some delta value. After that it starts to look for a minimum with similar criteria. It does not really matter if your data falls or rises slowly, as the maximum is recorded when it is reached and appended to the list of maxima once the level fallse below the hysteresis threshold.
This is a possible way to find local minima and maxima, but it has several shortcomings. One of them is that the method is not symmetric, i.e. if the same data is run backwards, the results are not necessarily the same.
Unfortunately, I cannot help much more, because the correct method really depends on the data you are looking at, its shape and its noisiness. If you have some samples, then we might be able to come up with some suggestions.