How to fit a double Gaussian distribution in Python? - python

I am trying to obtain a double Gaussian distribution for data (link) using Python. The raw data is of the form:
For the given data, I would like to obtain two Gaussian profiles for the peaks seen in figure. I tried it with the following code (source):
from sklearn import mixture
import matplotlib.pyplot
import matplotlib.mlab
import numpy as np
from pylab import *
data = np.genfromtxt('gaussian_fit.dat', skiprows = 1)
x = data[:, 0]
y = data[:, 1]
clf = mixture.GMM(n_components=2, covariance_type='full')
clf.fit((y, x))
m1, m2 = clf.means_
w1, w2 = clf.weights_
c1, c2 = clf.covars_
fig = plt.figure(figsize = (5, 5))
plt.subplot(111)
plotgauss1 = lambda x: plot(x,w1*matplotlib.mlab.normpdf(x,m1,np.sqrt(c1))[0], linewidth=3)
plotgauss2 = lambda x: plot(x,w2*matplotlib.mlab.normpdf(x,m2,np.sqrt(c2))[0], linewidth=3)
fig.savefig('gaussian_fit.pdf')
But I am not able to get the desired output. So, how can a double Gaussian distribution be obtained in Python?
Update
I was able to fit a single Gaussian distribution with the following code:
import pylab as plb
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy import asarray as ar,exp
import numpy as np
data = np.genfromtxt('gaussian_fit.dat', skiprows = 1)
x = data[:, 0]
y = data[:, 1]
n = len(x)
mean = sum(x*y)/n
sigma = sum(y*(x-mean)**2)/n
def gaus(x,a,x0,sigma):
return a*exp(-(x-x0)**2/(2*sigma**2))
popt,pcov = curve_fit(gaus, x, y ,p0 = [1, mean, sigma])
fig = plt.figure(figsize = (5, 5))
plt.subplot(111)
plt.plot(x, y, label='Raw')
plt.plot(x, gaus(x, *popt), 'o', markersize = 4, label='Gaussian fit')
plt.xlabel('X')
plt.ylabel('Y')
plt.legend()
fig.savefig('gaussian_fit.pdf')

You can't use scikit-learn for this, because the you are not dealing with a set of samples whose distribution you want to estimate. You could of course transform your curve to a PDF, sample it and then try to fit it using a Gaussian mixture model, but that seems to be a bit of an overkill to me.
Here's a solution using simple least square curve fitting. To get it to work I had to remove the background, i.e. ignore all data points with y < 5, and also provide a good starting vector for leastsq, which can be estimated form a plot of the data.
Finding the Starting Vector
The parameter vector that that is found by the least squares method is the vector
params = [c1, mu1, sigma1, c2, mu2, sigma2]
Here, c1 and c2 are scaling factors for the two Gaussians, i.e. their height, mu1and mu2 are the means, i.e. the horizontal positions of the peaks and sigma1 and sigma2 the standard deviations that determine the width of the Gaussians. To find a starting vector I just looked at a plot of the data and estimated the height of the two peaks ( = c1, c2, respectively) and their horizontal position (= mu1, mu1, respectively). sigma1 and sigma2 were simply set to 1.0.
Code
from sklearn import mixture
import matplotlib.pyplot
import matplotlib.mlab
import numpy as np
from pylab import *
from scipy.optimize import leastsq
data = np.genfromtxt('gaussian_fit.dat', skiprows = 1)
x = data[:, 0]
y = data[:, 1]
def double_gaussian( x, params ):
(c1, mu1, sigma1, c2, mu2, sigma2) = params
res = c1 * np.exp( - (x - mu1)**2.0 / (2.0 * sigma1**2.0) ) \
+ c2 * np.exp( - (x - mu2)**2.0 / (2.0 * sigma2**2.0) )
return res
def double_gaussian_fit( params ):
fit = double_gaussian( x, params )
return (fit - y_proc)
# Remove background.
y_proc = np.copy(y)
y_proc[y_proc < 5] = 0.0
# Least squares fit. Starting values found by inspection.
fit = leastsq( double_gaussian_fit, [13.0,-13.0,1.0,60.0,3.0,1.0] )
plot( x, y, c='b' )
plot( x, double_gaussian( x, fit[0] ), c='r' )

Related

Gaussian fitted curve showing a tail that does not go back to base-level

Based upon existing topics on Stackoverflow, I have managed to fit a Gaussian curve to my dataset. However, the fitted Gaussian shows one tail that does not go back to base-level (i.e., in the example below, the right tail suddenly stops at a higher y-value compared to the left tail). This surprises me, as per definition a Gaussian should show a perfectly symmetrical bell-shaped curve. How can I generate a Gaussian curve of which both tails are equally long (i.e., the tails stop at the same width measured from the plume center-line) and end at the same base-level (i.e., the same y-value)? The reason I would like to have this, is because in my data sometimes a second peak starts to arise while the first peak did not go back to base-level yet. I would like to separate these peaks by fitting a Gaussian that goes back to base-level, as theoretically each peak should go back to its base-level. Thanks a lot in advance!
import numpy as np
from lmfit import Model
import matplotlib.pyplot as plt
from scipy.signal import find_peaks
x = np.array([-20.0,-17.0,-14.0,-11.0,-8.0,-5.0,-2.0,1.0,4.0,7.0,10.0,13.0,16.0,19.0,22.0,25.0,28.0,31.0,34.0,37.0,40.0,43.0,46.0,49.0,52.0,55.0,58.0,61.0,64.0,67.0,70.0,73.0,76.0,79.0,82.0])
y = np.array([1.90269,1.93535,2.62402,3.08949,2.82409,3.07588,3.22015,3.18884,5.14053,10.5111,18.6118,28.6343,37.7625,46.3641,53.9163,60.7622,66.5765,71.0596,74.4948,77.7177,80.373,82.5833,83.9021,83.4652,79.0229,71.4679,61.93,52.113,43.8517,36.211,29.3815,23.8966,19.31,15.5209,12.4532])
def gaussian(x, amp, cen, wid):
return (amp / (np.sqrt(2*np.pi) * wid)) * np.exp(-(x-cen)**2 / (2*wid**2))
def line(x, slope, intercept):
return slope*x + intercept
peak_index = find_peaks(y,height=27.6)[0][0]
mean = sum(x*y)/np.sum(y) #weighted arithmetic mean
mod = Model(gaussian) + Model(line)
pars = mod.make_params(amp=max(y), cen=x[peak_index],
wid=np.sqrt(sum((x-mean)**2 * y)/sum(y)), slope=0, intercept=1)
result = mod.fit(y, pars, x=x)
comps = result.eval_components()
plt.plot(x, y, 'bo')
plt.plot(x, comps['gaussian'], 'k--')
Edit: The following example hopefully illustrates why I am interested in this. I have a long data-set in which the signal of different sources are being measured. The data-set is processed such that it generates the arrays x_measured and y_measured that contain the measured values belonging to one source. My program automatically detects the plume that occurs within the measured values, and stores the values of this plume in arrays called x and y. To these x and y arrays, I perform a Gaussian fit.
However, sometimes the measured values show that 2 plumes are overlapping, hence there is no measured plume from and back to base-level. An example is given in the code below. My program for these measured values now gives a Gaussian fit whereby the right tail goes to around y=0, but the left tail of the Gaussian fit stops around y=4.5. I would like the left tail to also go back to around y=0. This is, because theoretically I know that each plume should start and go back to the same base-level, and I want to compute the plume-width of such a Gaussian plume. For the example below, the left tail does not go back to around y=0, hence I cannot determine the width of the plume. I would like to have a Gaussian-fit of which both tails go back to the same base-level of y=0, such that I can determine the width of the plume.
x_measured = np.arange(-20,245,3)
y_measured = np.array([38.7586,38.2323,37.2958,35.9924,34.4196,32.7123,31.0257,29.5169,28.3244,27.5502,27.2458,27.4078,27.9815,28.8728,29.9643,31.1313,32.2545,33.2276,33.9594,34.373,34.4041,34.0009,33.1267,31.7649,29.9247,27.6458,24.9992,22.0845,19.0215,15.9397,12.966,10.2127,7.76834,5.69046,4.00296,2.69719,1.73733,1.06907,0.629744,0.358021,0.201123,0.11878,0.0839719,0.0813392,0.104295,0.151634,0.224209,0.321912,0.441478,0.575581,0.713504,0.843351,0.954777,1.04109,1.09974,1.13118,1.13683,1.11758,1.07369,1.0059,0.917066,0.81321,0.703288,0.597775,0.506678,0.437843,0.396256,0.384633,0.405147,0.461496,0.560387,0.71144,0.925262,1.21022,1.56925,1.99788,2.48458,3.01314,3.56626,4.12898,4.69031,5.24283,5.78014,6.29365,6.77004,7.19071,7.53399,7.78019,7.91889])
x = np.arange(10,104,3)
y = np.array([22.4548,23.4302,25.3389,27.9929,30.486,32.0528,33.5527,35.1304,35.9941,36.8606,37.1889,37.723,36.4069,35.9751,33.8824,31.0909,27.4247,23.3213,18.8772,14.3363,11.1075,7.68792,4.54899,2.2057,0,0,0,0,0,0,0.179834,0])
def gaussian(x, amp, cen, wid):
return (amp / (np.sqrt(2*np.pi) * wid)) * np.exp(-(x-cen)**2 / (2*wid**2))
def line(x, slope, intercept):
return slope*x + intercept
peak_index = find_peaks(y,height=27.6)[0][0]
mean = sum(x*y)/np.sum(y) #weighted arithmetic mean
mod = Model(gaussian) + Model(line)
pars = mod.make_params(amp=max(y), cen=x[peak_index],
wid=np.sqrt(sum((x-mean)**2 * y)/sum(y)), slope=0, intercept=1)
result = mod.fit(y, pars, x=x)
comps = result.eval_components()
plt.plot(x, y, 'bo')
plt.plot(x, comps['gaussian'], 'k--')
plt.plot(x_measured,y_measured)
It is unclear why you expect a bimodal fit with the model you defined. Use two different Gaussian functions for your fit, then evaluate the fitted functions for a longer interval x_fit to see the curves returning to baseline:
import numpy as np
from lmfit import Model
import matplotlib.pyplot as plt
from scipy.signal import find_peaks
x = np.array([-20.0,-17.0,-14.0,-11.0,-8.0,-5.0,-2.0,1.0,4.0,7.0,10.0,13.0,16.0,19.0,22.0,25.0,28.0,31.0,34.0,37.0,40.0,43.0,46.0,49.0,52.0,55.0,58.0,61.0,64.0,67.0,70.0,73.0,76.0,79.0,82.0])
y = np.array([1.90269,1.93535,2.62402,3.08949,2.82409,3.07588,3.22015,3.18884,5.14053,10.5111,18.6118,28.6343,37.7625,46.3641,53.9163,60.7622,66.5765,71.0596,74.4948,77.7177,80.373,82.5833,83.9021,83.4652,79.0229,71.4679,61.93,52.113,43.8517,36.211,29.3815,23.8966,19.31,15.5209,12.4532])
def gaussian1(x, amp1, cen1, wid1):
return (amp1 / (np.sqrt(2*np.pi) * wid1)) * np.exp(-(x-cen1)**2 / (2*wid1**2))
def gaussian2(x, amp2, cen2, wid2):
return (amp2 / (np.sqrt(2*np.pi) * wid2)) * np.exp(-(x-cen2)**2 / (2*wid2**2))
#peak_index = find_peaks(y,height=27.6)[0][0]
#mean = sum(x*y)/np.sum(y) #weighted arithmetic mean
mod = Model(gaussian1) + Model(gaussian2)
#I just filled in some start values, the details of educated guesses can be filled in later by you
pars = mod.make_params(amp1=30, amp2=40, cen1=20, cen2=40, wid1=2, wid2=2)
result = mod.fit(y, pars, x=x)
print(result.params)
x_fit=np.linspace(-30, 120, 500)
comps_elem = result.eval_components(x=x_fit)
comps_comb = result.eval(x=x_fit)
plt.plot(x, y, 'bo')
plt.plot(x_fit, comps_comb, 'k')
plt.plot(x_fit, comps_elem['gaussian1'], 'k-.')
plt.plot(x_fit, comps_elem['gaussian2'], 'k--')
plt.show()
Sample output:
The corresponding scipy.curve_fit function would look like this:
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from scipy.optimize import curve_fit
x = [-20.0,-17.0,-14.0,-11.0,-8.0,-5.0,-2.0,1.0,4.0,7.0,10.0,13.0,16.0,19.0,22.0,25.0,28.0,31.0,34.0,37.0,40.0,43.0,46.0,49.0,52.0,55.0,58.0,61.0,64.0,67.0,70.0,73.0,76.0,79.0,82.0]
y = [1.90269,1.93535,2.62402,3.08949,2.82409,3.07588,3.22015,3.18884,5.14053,10.5111,18.6118,28.6343,37.7625,46.3641,53.9163,60.7622,66.5765,71.0596,74.4948,77.7177,80.373,82.5833,83.9021,83.4652,79.0229,71.4679,61.93,52.113,43.8517,36.211,29.3815,23.8966,19.31,15.5209,12.4532]
def gauss(x, mu, sigma, A):
return A*np.exp(-(x-mu)**2/2/sigma**2)
def bimodal(x, mu1, sigma1, A1, mu2, sigma2, A2):
return gauss(x, mu1, sigma1, A1) + gauss(x, mu2, sigma2, A2)
expected = (20, 2, 30, 40, 2, 40)
params, cov = curve_fit(bimodal, x, y, expected)
sigma=np.sqrt(np.diag(cov))
x_fit = np.linspace(-20, 120, 500)
plt.plot(x_fit, bimodal(x_fit, *params), color='red', lw=3, label='model')
plt.plot(x_fit, gauss(x_fit, *params[:3]), color='red', lw=1, ls="--", label='distribution 1')
plt.plot(x_fit, gauss(x_fit, *params[3:]), color='red', lw=1, ls=":", label='distribution 2')
plt.scatter(x, y, marker="X", color="black", label="original data")
plt.legend()
print(pd.DataFrame(data={'params': params, 'sigma': sigma}, index=bimodal.__code__.co_varnames[1:]))
plt.show()

Log-normal distribution, log x-axis, scaling pdf

I want to make lognormal distribution. I'm using a logarithmic x-axis. However, I can't scale Probability density function correctly. I found one post on the forum for a full complete answer I did not find there(Scaling and fitting to a log-normal distribution using a logarithmic axis in python). Along with the change in the parameter "s" the graph has a different shape. Can the unambiguously correct distribution shape be obtained?Thank you for help.
import numpy as np
import scipy.stats as stats
import matplotlib.pyplot as plt
import pandas as pd
from math import log
# Import data
data = pd.read_excel('data2018.xlsx')
# Create bins (log)
classes = 16
s=10
bins_log10 = np.logspace(np.log10(data['diff'].min()), np.log10(data['diff'].max()), classes + 1)
bins_log10_s = np.logspace(np.log10(data['diff'].min()), np.log10(data['diff'].max()), (classes + 1) * s)
# Plot histogram
plt.style.use('ggplot')
counts, bins, _ = plt.hist(data['diff'], bins=bins_log10, edgecolor='black', linewidth=1,
label="Histogram")
# Calculation of bin centers and multiplied them
restored = [[d] * int(counts[n]) for n, d in enumerate((bins[1:] + bins[:-1]) / 2)]
# Flatten the result
restored = [item for sublist in restored for item in sublist]
# Calculate of fitting parameters. shape = sigma, log(scale) = mu
shape, loc, scale = stats.lognorm.fit(restored, floc=0)
# Calculate centers and length log_bins
cen_log_bins = (bins_log10[1:] + bins_log10[:-1]) / 2
len_log_bins = (bins_log10[1:] - bins_log10[:-1])
samples_fit_log_cntr = stats.lognorm.pdf(cen_log_bins, shape, loc=loc, scale=scale)
plt.plot(cen_log_bins,samples_fit_log_cntr * len_log_bins * counts.sum(), ls='dashed',label='PDF with centers', linewidth=2)
# Smooth pdf
bins_log10_cntr_s = (bins_log10_s[1:] + bins_log10_s[:-1]) / 2
samples_fit_log_cntr = stats.lognorm.pdf(bins_log10_cntr_s, shape, loc=loc, scale=scale)
bins_log_cntr = bins_log10_s[1:] - bins_log10_s[:-1]
plt.plot(bins_log10_cntr_s, samples_fit_log_cntr * bins_log_cntr * counts.sum() * s,color='blue',label='Smooth PDF with centers', linewidth=2)
plt.title("Fit results: $\mu = %.2f, \sigma$ = %.2f" % (log(scale), shape))
plt.xscale('log')
plt.legend()
plt.tight_layout()
plt.show()

How to do a simple Gaussian mixture sampling and PDF plotting with NumPy/SciPy?

I add three normal distributions to obtain a new distribution as shown below, how can I do sampling according to this distribution in python?
import matplotlib.pyplot as plt
import scipy.stats as ss
import numpy as np
x = np.linspace(0, 10, 1000)
y1 = [ss.norm.pdf(v, loc=5, scale=1) for v in x]
y2 = [ss.norm.pdf(v, loc=1, scale=1.3) for v in x]
y3 = [ss.norm.pdf(v, loc=9, scale=1.3) for v in x]
y = np.sum([y1, y2, y3], axis=0)/3
plt.plot(x, y, '-')
plt.xlabel('$x$')
plt.ylabel('$P(x)$')
BTW, is there a better way to plot such a probability distribution?
It seems that you're asking two questions: how do I sample from a distribution and how do I plot the PDF?
Assuming you're trying to sample from a mixture distribution of 3 normal ones shown in your code, the following code snipped performs this kind of sampling in the naïve, straightforward way as a proof-of-concept.
Basically, the idea is to
Choose an index i among the index of components, i.e. 0, 1, 2 ..., according to their probability weights.
Having chosen i, select the corresponding distribution and obtain a sample point from it.
Continue from 1 until enough sample points are collected.
However, to plot the PDF, you don't really need a sample in this case, because the theoretical solution is quite easy. In the more general case, the PDF can be approximated by a histogram from the sample.
The code below performs both sampling and PDF-plotting using the theoretical PDF.
import numpy as np
import numpy.random
import scipy.stats as ss
import matplotlib.pyplot as plt
# Set-up.
n = 10000
numpy.random.seed(0x5eed)
# Parameters of the mixture components
norm_params = np.array([[5, 1],
[1, 1.3],
[9, 1.3]])
n_components = norm_params.shape[0]
# Weight of each component, in this case all of them are 1/3
weights = np.ones(n_components, dtype=np.float64) / 3.0
# A stream of indices from which to choose the component
mixture_idx = numpy.random.choice(len(weights), size=n, replace=True, p=weights)
# y is the mixture sample
y = numpy.fromiter((ss.norm.rvs(*(norm_params[i])) for i in mixture_idx),
dtype=np.float64)
# Theoretical PDF plotting -- generate the x and y plotting positions
xs = np.linspace(y.min(), y.max(), 200)
ys = np.zeros_like(xs)
for (l, s), w in zip(norm_params, weights):
ys += ss.norm.pdf(xs, loc=l, scale=s) * w
plt.plot(xs, ys)
plt.hist(y, normed=True, bins="fd")
plt.xlabel("x")
plt.ylabel("f(x)")
plt.show()
In order to make the answer of Cong Ma work more general, I slightly modified his code. The weights work now for any number of mixture components.
import numpy as np
import numpy.random
import scipy.stats as ss
import matplotlib.pyplot as plt
# Set-up.
n = 10000
numpy.random.seed(0x5eed)
# Parameters of the mixture components
norm_params = np.array([[5, 1],
[1, 1.3],
[9, 1.3]])
n_components = norm_params.shape[0]
# Weight of each component, in this case all of them are 1/3
weights = np.ones(n_components, dtype=np.float64) / float(n_components)
# A stream of indices from which to choose the component
mixture_idx = numpy.random.choice(n_components, size=n, replace=True, p=weights)
# y is the mixture sample
y = numpy.fromiter((ss.norm.rvs(*(norm_params[i])) for i in mixture_idx),
dtype=np.float64)
# Theoretical PDF plotting -- generate the x and y plotting positions
xs = np.linspace(y.min(), y.max(), 200)
ys = np.zeros_like(xs)
for (l, s), w in zip(norm_params, weights):
ys += ss.norm.pdf(xs, loc=l, scale=s) * w
plt.plot(xs, ys)
plt.hist(y, normed=True, bins="fd")
plt.xlabel("x")
plt.ylabel("f(x)")
plt.show()

Plotting a decision boundary separating 2 classes using Matplotlib's pyplot

I could really use a tip to help me plotting a decision boundary to separate to classes of data. I created some sample data (from a Gaussian distribution) via Python NumPy. In this case, every data point is a 2D coordinate, i.e., a 1 column vector consisting of 2 rows. E.g.,
[ 1
2 ]
Let's assume I have 2 classes, class1 and class2, and I created 100 data points for class1 and 100 data points for class2 via the code below (assigned to the variables x1_samples and x2_samples).
mu_vec1 = np.array([0,0])
cov_mat1 = np.array([[2,0],[0,2]])
x1_samples = np.random.multivariate_normal(mu_vec1, cov_mat1, 100)
mu_vec1 = mu_vec1.reshape(1,2).T # to 1-col vector
mu_vec2 = np.array([1,2])
cov_mat2 = np.array([[1,0],[0,1]])
x2_samples = np.random.multivariate_normal(mu_vec2, cov_mat2, 100)
mu_vec2 = mu_vec2.reshape(1,2).T
When I plot the data points for each class, it would look like this:
Now, I came up with an equation for an decision boundary to separate both classes and would like to add it to the plot. However, I am not really sure how I can plot this function:
def decision_boundary(x_vec, mu_vec1, mu_vec2):
g1 = (x_vec-mu_vec1).T.dot((x_vec-mu_vec1))
g2 = 2*( (x_vec-mu_vec2).T.dot((x_vec-mu_vec2)) )
return g1 - g2
I would really appreciate any help!
EDIT:
Intuitively (If I did my math right) I would expect the decision boundary to look somewhat like this red line when I plot the function...
Your question is more complicated than a simple plot : you need to draw the contour which will maximize the inter-class distance. Fortunately it's a well-studied field, particularly for SVM machine learning.
The easiest method is to download the scikit-learn module, which provides a lot of cool methods to draw boundaries: scikit-learn: Support Vector Machines
Code :
# -*- coding: utf-8 -*-
import numpy as np
import matplotlib
from matplotlib import pyplot as plt
import scipy
from sklearn import svm
mu_vec1 = np.array([0,0])
cov_mat1 = np.array([[2,0],[0,2]])
x1_samples = np.random.multivariate_normal(mu_vec1, cov_mat1, 100)
mu_vec1 = mu_vec1.reshape(1,2).T # to 1-col vector
mu_vec2 = np.array([1,2])
cov_mat2 = np.array([[1,0],[0,1]])
x2_samples = np.random.multivariate_normal(mu_vec2, cov_mat2, 100)
mu_vec2 = mu_vec2.reshape(1,2).T
fig = plt.figure()
plt.scatter(x1_samples[:,0],x1_samples[:,1], marker='+')
plt.scatter(x2_samples[:,0],x2_samples[:,1], c= 'green', marker='o')
X = np.concatenate((x1_samples,x2_samples), axis = 0)
Y = np.array([0]*100 + [1]*100)
C = 1.0 # SVM regularization parameter
clf = svm.SVC(kernel = 'linear', gamma=0.7, C=C )
clf.fit(X, Y)
Linear Plot
w = clf.coef_[0]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (clf.intercept_[0]) / w[1]
plt.plot(xx, yy, 'k-')
MultiLinear Plot
C = 1.0 # SVM regularization parameter
clf = svm.SVC(kernel = 'rbf', gamma=0.7, C=C )
clf.fit(X, Y)
h = .02 # step size in the mesh
# create a mesh to plot in
x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1
y_min, y_max = X[:, 1].min() - 1, X[:, 1].max() + 1
xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
np.arange(y_min, y_max, h))
# Plot the decision boundary. For that, we will assign a color to each
# point in the mesh [x_min, m_max]x[y_min, y_max].
Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])
# Put the result into a color plot
Z = Z.reshape(xx.shape)
plt.contour(xx, yy, Z, cmap=plt.cm.Paired)
Implementation
If you want to implement it yourself, you need to solve the following quadratic equation:
The Wikipedia article
Unfortunately, for non-linear boundaries like the one you draw, it's a difficult problem relying on a kernel trick but there isn't a clear cut solution.
Based on the way you've written decision_boundary you'll want to use the contour function, as Joe noted above. If you just want the boundary line, you can draw a single contour at the 0 level:
f, ax = plt.subplots(figsize=(7, 7))
c1, c2 = "#3366AA", "#AA3333"
ax.scatter(*x1_samples.T, c=c1, s=40)
ax.scatter(*x2_samples.T, c=c2, marker="D", s=40)
x_vec = np.linspace(*ax.get_xlim())
ax.contour(x_vec, x_vec,
decision_boundary(x_vec, mu_vec1, mu_vec2),
levels=[0], cmap="Greys_r")
Which makes:
Those were some great suggestions, thanks a lot for your help! I ended up solving the equation analytically and this is the solution I ended up with (I just want to post it for future reference:
# 2-category classification with random 2D-sample data
# from a multivariate normal distribution
import numpy as np
from matplotlib import pyplot as plt
def decision_boundary(x_1):
""" Calculates the x_2 value for plotting the decision boundary."""
return 4 - np.sqrt(-x_1**2 + 4*x_1 + 6 + np.log(16))
# Generating a Gaussion dataset:
# creating random vectors from the multivariate normal distribution
# given mean and covariance
mu_vec1 = np.array([0,0])
cov_mat1 = np.array([[2,0],[0,2]])
x1_samples = np.random.multivariate_normal(mu_vec1, cov_mat1, 100)
mu_vec1 = mu_vec1.reshape(1,2).T # to 1-col vector
mu_vec2 = np.array([1,2])
cov_mat2 = np.array([[1,0],[0,1]])
x2_samples = np.random.multivariate_normal(mu_vec2, cov_mat2, 100)
mu_vec2 = mu_vec2.reshape(1,2).T # to 1-col vector
# Main scatter plot and plot annotation
f, ax = plt.subplots(figsize=(7, 7))
ax.scatter(x1_samples[:,0], x1_samples[:,1], marker='o', color='green', s=40, alpha=0.5)
ax.scatter(x2_samples[:,0], x2_samples[:,1], marker='^', color='blue', s=40, alpha=0.5)
plt.legend(['Class1 (w1)', 'Class2 (w2)'], loc='upper right')
plt.title('Densities of 2 classes with 25 bivariate random patterns each')
plt.ylabel('x2')
plt.xlabel('x1')
ftext = 'p(x|w1) ~ N(mu1=(0,0)^t, cov1=I)\np(x|w2) ~ N(mu2=(1,1)^t, cov2=I)'
plt.figtext(.15,.8, ftext, fontsize=11, ha='left')
# Adding decision boundary to plot
x_1 = np.arange(-5, 5, 0.1)
bound = decision_boundary(x_1)
plt.plot(x_1, bound, 'r--', lw=3)
x_vec = np.linspace(*ax.get_xlim())
x_1 = np.arange(0, 100, 0.05)
plt.show()
And the code can be found here
EDIT:
I also have a convenience function for plotting decision regions for classifiers that implement a fit and predict method, e.g., the classifiers in scikit-learn, which is useful if the solution cannot be found analytically. A more detailed description how it works can be found here.
You can create your own equation for the boundary:
where you have to find the positions x0 and y0, as well as the constants ai and bi for the radius equation. So, you have 2*(n+1)+2 variables. Using scipy.optimize.leastsq is straightforward for this type of problem.
The code attached below builds the residual for the leastsq penalizing the points outsize the boundary. The result for your problem, obtained with:
x, y = find_boundary(x2_samples[:,0], x2_samples[:,1], n)
ax.plot(x, y, '-k', lw=2.)
x, y = find_boundary(x1_samples[:,0], x1_samples[:,1], n)
ax.plot(x, y, '--k', lw=2.)
using n=1:
using n=2:
usng n=5:
using n=7:
import numpy as np
from numpy import sin, cos, pi
from scipy.optimize import leastsq
def find_boundary(x, y, n, plot_pts=1000):
def sines(theta):
ans = np.array([sin(i*theta) for i in range(n+1)])
return ans
def cosines(theta):
ans = np.array([cos(i*theta) for i in range(n+1)])
return ans
def residual(params, x, y):
x0 = params[0]
y0 = params[1]
c = params[2:]
r_pts = ((x-x0)**2 + (y-y0)**2)**0.5
thetas = np.arctan2((y-y0), (x-x0))
m = np.vstack((sines(thetas), cosines(thetas))).T
r_bound = m.dot(c)
delta = r_pts - r_bound
delta[delta>0] *= 10
return delta
# initial guess for x0 and y0
x0 = x.mean()
y0 = y.mean()
params = np.zeros(2 + 2*(n+1))
params[0] = x0
params[1] = y0
params[2:] += 1000
popt, pcov = leastsq(residual, x0=params, args=(x, y),
ftol=1.e-12, xtol=1.e-12)
thetas = np.linspace(0, 2*pi, plot_pts)
m = np.vstack((sines(thetas), cosines(thetas))).T
c = np.array(popt[2:])
r_bound = m.dot(c)
x_bound = popt[0] + r_bound*cos(thetas)
y_bound = popt[1] + r_bound*sin(thetas)
return x_bound, y_bound
I like the mglearn library to draw decision boundaries. Here is one example from the book "Introduction to Machine Learning with Python" by A. Mueller:
fig, axes = plt.subplots(1, 3, figsize=(10, 3))
for n_neighbors, ax in zip([1, 3, 9], axes):
clf = KNeighborsClassifier(n_neighbors=n_neighbors).fit(X, y)
mglearn.plots.plot_2d_separator(clf, X, fill=True, eps=0.5, ax=ax, alpha=.4)
mglearn.discrete_scatter(X[:, 0], X[:, 1], y, ax=ax)
ax.set_title("{} neighbor(s)".format(n_neighbors))
ax.set_xlabel("feature 0")
ax.set_ylabel("feature 1")
axes[0].legend(loc=3)
If you want to use scikit learn, you can write your code like this:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.linear_model import LogisticRegression
# read data
data = pd.read_csv('ex2data1.txt', header=None)
X = data[[0,1]].values
y = data[2]
# use LogisticRegression
log_reg = LogisticRegression()
log_reg.fit(X, y)
# Coefficient of the features in the decision function. (from theta 1 to theta n)
parameters = log_reg.coef_[0]
# Intercept (a.k.a. bias) added to the decision function. (theta 0)
parameter0 = log_reg.intercept_
# Plotting the decision boundary
fig = plt.figure(figsize=(10,7))
x_values = [np.min(X[:, 1] -5 ), np.max(X[:, 1] +5 )]
# calcul y values
y_values = np.dot((-1./parameters[1]), (np.dot(parameters[0],x_values) + parameter0))
colors=['red' if l==0 else 'blue' for l in y]
plt.scatter(X[:, 0], X[:, 1], label='Logistics regression', color=colors)
plt.plot(x_values, y_values, label='Decision Boundary')
plt.show()
see: Building-a-Logistic-Regression-with-Scikit-learn
Just solved a very similar problem with a different approach (root finding) and wanted to post this alternative as answer here for future reference:
def discr_func(x, y, cov_mat, mu_vec):
"""
Calculates the value of the discriminant function for a dx1 dimensional
sample given covariance matrix and mean vector.
Keyword arguments:
x_vec: A dx1 dimensional numpy array representing the sample.
cov_mat: numpy array of the covariance matrix.
mu_vec: dx1 dimensional numpy array of the sample mean.
Returns a float value as result of the discriminant function.
"""
x_vec = np.array([[x],[y]])
W_i = (-1/2) * np.linalg.inv(cov_mat)
assert(W_i.shape[0] > 1 and W_i.shape[1] > 1), 'W_i must be a matrix'
w_i = np.linalg.inv(cov_mat).dot(mu_vec)
assert(w_i.shape[0] > 1 and w_i.shape[1] == 1), 'w_i must be a column vector'
omega_i_p1 = (((-1/2) * (mu_vec).T).dot(np.linalg.inv(cov_mat))).dot(mu_vec)
omega_i_p2 = (-1/2) * np.log(np.linalg.det(cov_mat))
omega_i = omega_i_p1 - omega_i_p2
assert(omega_i.shape == (1, 1)), 'omega_i must be a scalar'
g = ((x_vec.T).dot(W_i)).dot(x_vec) + (w_i.T).dot(x_vec) + omega_i
return float(g)
#g1 = discr_func(x, y, cov_mat=cov_mat1, mu_vec=mu_vec_1)
#g2 = discr_func(x, y, cov_mat=cov_mat2, mu_vec=mu_vec_2)
x_est50 = list(np.arange(-6, 6, 0.1))
y_est50 = []
for i in x_est50:
y_est50.append(scipy.optimize.bisect(lambda y: discr_func(i, y, cov_mat=cov_est_1, mu_vec=mu_est_1) - \
discr_func(i, y, cov_mat=cov_est_2, mu_vec=mu_est_2), -10,10))
y_est50 = [float(i) for i in y_est50]
Here is the result:
(blue the quadratic case, red the linear case (equal variances)
I know this question has been answered in a very thorough way analytically. I just wanted to share a possible 'hack' to the problem. It is unwieldy but gets the job done.
Start by building a mesh grid of the 2d area and then based on the classifier just build a class map of the entire space. Subsequently detect changes in the decision made row-wise and store the edges points in a list and scatter plot the points.
def disc(x): # returns the class of the point based on location x = [x,y]
temp = 0.5 + 0.5*np.sign(disc0(x)-disc1(x))
# disc0() and disc1() are the discriminant functions of the respective classes
return 0*temp + 1*(1-temp)
num = 200
a = np.linspace(-4,4,num)
b = np.linspace(-6,6,num)
X,Y = np.meshgrid(a,b)
def decColor(x,y):
temp = np.zeros((num,num))
print x.shape, np.size(x,axis=0)
for l in range(num):
for m in range(num):
p = np.array([x[l,m],y[l,m]])
#print p
temp[l,m] = disc(p)
return temp
boundColorMap = decColor(X,Y)
group = 0
boundary = []
for x in range(num):
group = boundColorMap[x,0]
for y in range(num):
if boundColorMap[x,y]!=group:
boundary.append([X[x,y],Y[x,y]])
group = boundColorMap[x,y]
boundary = np.array(boundary)
Sample Decision Boundary for a simple bivariate gaussian classifier
Given two bi-variate normal distributions, you can use Gaussian Discriminant Analysis (GDA) to come up with a decision boundary as the difference between the log of the 2 pdf's.
Here's a way to do it using scipy multivariate_normal (the code is not optimized):
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import multivariate_normal
from numpy.linalg import norm
from numpy.linalg import inv
from scipy.spatial.distance import mahalanobis
def normal_scatter(mean, cov, p):
size = 100
sigma_x = cov[0,0]
sigma_y = cov[1,1]
mu_x = mean[0]
mu_y = mean[1]
x_ps, y_ps = np.random.multivariate_normal(mean, cov, size).T
x,y = np.mgrid[mu_x-3*sigma_x:mu_x+3*sigma_x:1/size, mu_y-3*sigma_y:mu_y+3*sigma_y:1/size]
grid = np.empty(x.shape + (2,))
grid[:, :, 0] = x; grid[:, :, 1] = y
z = p*multivariate_normal.pdf(grid, mean, cov)
return x_ps, y_ps, x,y,z
# Dist 1
mu_1 = np.array([1, 1])
cov_1 = .5*np.array([[1, 0], [0, 1]])
p_1 = .5
x_ps, y_ps, x,y,z = normal_scatter(mu_1, cov_1, p_1)
plt.plot(x_ps,y_ps,'x')
plt.contour(x, y, z, cmap='Blues', levels=3)
# Dist 2
mu_2 = np.array([2, 1])
#cov_2 = np.array([[2, -1], [-1, 1]])
cov_2 = cov_1
p_2 = .5
x_ps, y_ps, x,y,z = normal_scatter(mu_2, cov_2, p_2)
plt.plot(x_ps,y_ps,'.')
plt.contour(x, y, z, cmap='Oranges', levels=3)
# Decision Boundary
X = np.empty(x.shape + (2,))
X[:, :, 0] = x; X[:, :, 1] = y
g = np.log(p_1*multivariate_normal.pdf(X, mu_1, cov_1)) - np.log(p_2*multivariate_normal.pdf(X, mu_2, cov_2))
plt.contour(x, y, g, [0])
plt.grid()
plt.axhline(y=0, color='k')
plt.axvline(x=0, color='k')
plt.plot([mu_1[0], mu_2[0]], [mu_1[1], mu_2[1]], 'k')
plt.show()
If p_1 != p_2, then you get non-linear boundary. The decision boundary is given by g above.
Then to plot the decision hyper-plane (line in 2D), you need to evaluate g for a 2D mesh, then get the contour which will give a separating line.
You can also assume to have equal co-variance matrices for both distributions, which will give a linear decision boundary. In this case, you can replace the calculation of g in the above code with the following:
W = inv(cov_1).dot(mu_1-mu_2)
x_0 = 1/2*(mu_1+mu_2) - cov_1.dot(np.log(p_1/p_2)).dot((mu_1-mu_2)/mahalanobis(mu_1, mu_2, cov_1))
X = np.empty(x.shape + (2,))
X[:, :, 0] = x; X[:, :, 1] = y
g = (X-x_0).dot(W)
i use this method from this book python-machine-learning-2nd.pdf URL
from matplotlib.colors import ListedColormap
import matplotlib.pyplot as plt
def plot_decision_regions(X, y, classifier, test_idx=None, resolution=0.02):
# setup marker generator and color map
markers = ('s', 'x', 'o', '^', 'v')
colors = ('red', 'blue', 'lightgreen', 'gray', 'cyan')
cmap = ListedColormap(colors[:len(np.unique(y))])
# plot the decision surface
x1_min, x1_max = X[:, 0].min() - 1, X[:, 0].max() + 1
x2_min, x2_max = X[:, 1].min() - 1, X[:, 1].max() + 1
xx1, xx2 = np.meshgrid(np.arange(x1_min, x1_max, resolution),
np.arange(x2_min, x2_max, resolution))
Z = classifier.predict(np.array([xx1.ravel(), xx2.ravel()]).T)
Z = Z.reshape(xx1.shape)
plt.contourf(xx1, xx2, Z, alpha=0.3, cmap=cmap)
plt.xlim(xx1.min(), xx1.max())
plt.ylim(xx2.min(), xx2.max())
for idx, cl in enumerate(np.unique(y)):
plt.scatter(x=X[y == cl, 0],
y=X[y == cl, 1],
alpha=0.8,
c=colors[idx],
marker=markers[idx],
label=cl,
edgecolor='black')
# highlight test samples
if test_idx:
# plot all samples
X_test, y_test = X[test_idx, :], y[test_idx]
plt.scatter(X_test[:, 0],
X_test[:, 1],
c='',
edgecolor='black',
alpha=1.0,
linewidth=1,
marker='o',
s=100,
label='test set')
Since version 1.1, sklearn has a function for this:
https://scikit-learn.org/stable/modules/generated/sklearn.inspection.DecisionBoundaryDisplay.html#sklearn.inspection.DecisionBoundaryDisplay

Make dots in matplotlib plots selectable by mouse

scikit-learn has a very nice demo that creates an outlier analysis tool. Here is the
import numpy as np
import pylab as pl
import matplotlib.font_manager
from scipy import stats
from sklearn import svm
from sklearn.covariance import EllipticEnvelope
# Example settings
n_samples = 200
outliers_fraction = 0.25
clusters_separation = [0, 1, 2]
# define two outlier detection tools to be compared
classifiers = {
"One-Class SVM": svm.OneClassSVM(nu=0.95 * outliers_fraction + 0.05,
kernel="rbf", gamma=0.1),
"robust covariance estimator": EllipticEnvelope(contamination=.1)}
# Compare given classifiers under given settings
xx, yy = np.meshgrid(np.linspace(-7, 7, 500), np.linspace(-7, 7, 500))
n_inliers = int((1. - outliers_fraction) * n_samples)
n_outliers = int(outliers_fraction * n_samples)
ground_truth = np.ones(n_samples, dtype=int)
ground_truth[-n_outliers:] = 0
# Fit the problem with varying cluster separation
for i, offset in enumerate(clusters_separation):
np.random.seed(42)
# Data generation
X1 = 0.3 * np.random.randn(0.5 * n_inliers, 2) - offset
X2 = 0.3 * np.random.randn(0.5 * n_inliers, 2) + offset
X = np.r_[X1, X2]
# Add outliers
X = np.r_[X, np.random.uniform(low=-6, high=6, size=(n_outliers, 2))]
# Fit the model with the One-Class SVM
pl.figure(figsize=(10, 5))
for i, (clf_name, clf) in enumerate(classifiers.items()):
# fit the data and tag outliers
clf.fit(X)
y_pred = clf.decision_function(X).ravel()
threshold = stats.scoreatpercentile(y_pred,
100 * outliers_fraction)
y_pred = y_pred > threshold
n_errors = (y_pred != ground_truth).sum()
# plot the levels lines and the points
Z = clf.decision_function(np.c_[xx.ravel(), yy.ravel()])
Z = Z.reshape(xx.shape)
subplot = pl.subplot(1, 2, i + 1)
subplot.set_title("Outlier detection")
subplot.contourf(xx, yy, Z, levels=np.linspace(Z.min(), threshold, 7),
cmap=pl.cm.Blues_r)
a = subplot.contour(xx, yy, Z, levels=[threshold],
linewidths=2, colors='red')
subplot.contourf(xx, yy, Z, levels=[threshold, Z.max()],
colors='orange')
b = subplot.scatter(X[:-n_outliers, 0], X[:-n_outliers, 1], c='white')
c = subplot.scatter(X[-n_outliers:, 0], X[-n_outliers:, 1], c='black')
subplot.axis('tight')
subplot.legend(
[a.collections[0], b, c],
['learned decision function', 'true inliers', 'true outliers'],
prop=matplotlib.font_manager.FontProperties(size=11))
subplot.set_xlabel("%d. %s (errors: %d)" % (i + 1, clf_name, n_errors))
subplot.set_xlim((-7, 7))
subplot.set_ylim((-7, 7))
pl.subplots_adjust(0.04, 0.1, 0.96, 0.94, 0.1, 0.26)
pl.show()
And here is what it looks like:
Is that cool or what?
However, I want the plot to be mouse-sensitive. That is, I want to be able to click on dots and find out what they are, with either a tool-tip or with a pop-up window, or something in a scroller. And I'd also like to be able to click-to-zoom, rather than zoom with a bounding box.
Is there any way to do this?
Not to plug my own project to much, but have a look at mpldatacursor. If you'd prefer, it's also quite easy to implement from scratch.
As a quick example:
import matplotlib.pyplot as plt
import numpy as np
from mpldatacursor import datacursor
x1, y1 = np.random.random((2, 5))
x2, y2 = np.random.random((2, 5))
fig, ax = plt.subplots()
ax.plot(x1, y1, 'ro', markersize=12, label='Series A')
ax.plot(x2, y2, 'bo', markersize=12, label='Series B')
ax.legend()
datacursor()
plt.show()
For this to work with the example code you posted, you'd need to change things slightly. As it is, the artist labels are set in the call to legend, instead of when the artist is created. This means that there's no way to retrieve what's displayed in the legend for a particular artist. All you'd need to do is just pass in the labels as a kwarg to scatter instead of as the second argument to legend, and things should work as you were wanting.

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