Below I used the nympy generated draws to compare the empirical moments of Pareto distribution with analytical ones (I used the following link for the formulas of mean, variance, skewness and excess kurtosis, https://en.wikipedia.org/wiki/Pareto_distribution). Obtained results for mean and variance are similar, however the results for skewness and kurtosis are very different. What am I doing wrong?
Thank you beforehand
import numpy as np
import pandas as pd
from scipy.stats import skew
from scipy.stats import kurtosis
from prettytable import PrettyTable
x_m = [1, 2, 3, 4]
alpha = [5, 6, 7, 8]
#drawing samples from distribution
for a, x in zip(alpha, x_m):
print (a, x)
data = (np.random.default_rng().pareto(a, 10000000)+1) * x
mean = np.mean(data)
var = np.var(data)
skew = skew(data)
kurt = kurtosis(data)
#Analytical estimation
for a, x in zip(alpha, x_m):
a_mean = (a*x)/(a-1)
a_var = (a*x**2)/((a-1)**2*(a-2))
a_skew = (2*(1+a)/(a-3))*(np.sqrt(a-2/a))
a_kurt = (6*(a**3+a**2-6*a-2))/(a*(a-3)*(a-4))
#Table
header = ['Moments', 'Simulated', 'Analytical']
Moments = ['Mean', 'Variance', 'Skewness', 'Excess Kurtosis']
Simulated = [round(mean,4), round(var,4), round(skew,4), round(kurt,4)]
Analytical = [round(a_mean,4), round(a_var,4), round(a_skew,4), round(a_kurt,4)]
table = PrettyTable()
table.add_column(header[0], Moments)
table.add_column(header[1], Simulated)
table.add_column(header[2], Analytical)
print(table)
its a small typo in the analytical skewness line:
a_skew = (2*(1+a)/(a-3))*(np.sqrt(a-2/a))
should be
a_skew = (2*(1+a)/(a-3))*(np.sqrt((a-2)/a))
The Kurtosis is correct with a sufficiently large sample.
Related
I am trying to fit a quadratic function to some data, and I'm trying to do this without using numpy's polyfit function.
Mathematically I tried to follow this website https://neutrium.net/mathematics/least-squares-fitting-of-a-polynomial/ but somehow I don't think that I'm doing it right. If anyone could assist me that would be great, or If you could suggest another way to do it that would also be awesome.
What I've tried so far:
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
ones = np.ones(3)
A = np.array( ((0,1),(1,1),(2,1)))
xfeature = A.T[0]
squaredfeature = A.T[0] ** 2
b = np.array( (1,2,0), ndmin=2 ).T
b = b.reshape(3)
features = np.concatenate((np.vstack(ones), np.vstack(xfeature), np.vstack(squaredfeature)), axis = 1)
featuresc = features.copy()
print(features)
m_det = np.linalg.det(features)
print(m_det)
determinants = []
for i in range(3):
featuresc.T[i] = b
print(featuresc)
det = np.linalg.det(featuresc)
determinants.append(det)
print(det)
featuresc = features.copy()
determinants = determinants / m_det
print(determinants)
plt.scatter(A.T[0],b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
p2 = np.polyfit(A.T[0],b,2)
plt.plot(u, np.polyval(p2,u), 'b--')
plt.show()
As you can see my curve doesn't compare well to nnumpy's polyfit curve.
Update:
I went through my code and removed all the stupid mistakes and now it works, when I try to fit it over 3 points, but I have no idea how to fit over more than three points.
This is the new code:
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
ones = np.ones(3)
A = np.array( ((0,1),(1,1),(2,1)))
xfeature = A.T[0]
squaredfeature = A.T[0] ** 2
b = np.array( (1,2,0), ndmin=2 ).T
b = b.reshape(3)
features = np.concatenate((np.vstack(ones), np.vstack(xfeature), np.vstack(squaredfeature)), axis = 1)
featuresc = features.copy()
print(features)
m_det = np.linalg.det(features)
print(m_det)
determinants = []
for i in range(3):
featuresc.T[i] = b
print(featuresc)
det = np.linalg.det(featuresc)
determinants.append(det)
print(det)
featuresc = features.copy()
determinants = determinants / m_det
print(determinants)
plt.scatter(A.T[0],b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
p2 = np.polyfit(A.T[0],b,2)
plt.plot(u, np.polyval(p2,u), 'r--')
plt.show()
Instead using Cramer's Rule, actually solve the system using least squares. Remember that Cramer's Rule will only work if the total number of points you have equals the desired order of polynomial plus 1.
If you don't have this, then Cramer's Rule will not work as you're trying to find an exact solution to the problem. If you have more points, the method is unsuitable as we will create an overdetermined system of equations.
To adapt this to more points, numpy.linalg.lstsq would be a better fit as it solves the solution to the Ax = b by computing the vector x that minimizes the Euclidean norm using the matrix A. Therefore, remove the y values from the last column of the features matrix and solve for the coefficients and use numpy.linalg.lstsq to solve for the coefficients:
import numpy as np
import matplotlib.pyplot as plt
ones = np.ones(4)
xfeature = np.asarray([0,1,2,3])
squaredfeature = xfeature ** 2
b = np.asarray([1,2,0,3])
features = np.concatenate((np.vstack(ones),np.vstack(xfeature),np.vstack(squaredfeature)), axis = 1) # Change - remove the y values
determinants = np.linalg.lstsq(features, b)[0] # Change - use least squares
plt.scatter(xfeature,b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
plt.show()
I get this plot now, which matches what the dashed curve is in your graph, also matching what numpy.polyfit gives you:
I have clustered my data (12000, 3) using sklearn Gaussian mixture model algorithm (GMM). I have 3 clusters. Each point of my data represents a molecular structure. I would like to know how could I sampled each cluster. I have tried with the function:
gmm = GMM(n_components=3).fit(Data)
gmm.sample(n_samples=20)
but it does preform a sampling of the whole distribution, but I need a sample of each one of the components.
Well this is not that easy since you need to calculate the eigenvectors of all covariance matrices. Here is some example code for a problem I studied
import numpy as np
from scipy.stats import multivariate_normal
import random
from operator import truediv
import itertools
from scipy import linalg
import matplotlib.pyplot as plt
import matplotlib as mpl
from sklearn import mixture
#import some data which can be used for gmm
mix = np.loadtxt("mixture.txt", usecols=(0,1), unpack=True)
#print(mix.shape)
color_iter = itertools.cycle(['navy', 'c', 'cornflowerblue', 'gold',
'darkorange'])
def plot_results(X, Y_, means, covariances, index, title):
#function for plotting the gaussians
splot = plt.subplot(2, 1, 1 + index)
for i, (mean, covar, color) in enumerate(zip(
means, covariances, color_iter)):
v, w = linalg.eigh(covar)
v = 2. * np.sqrt(2.) * np.sqrt(v)
u = w[0] / linalg.norm(w[0])
# as the DP will not use every component it has access to
# unless it needs it, we shouldn't plot the redundant
# components.
if not np.any(Y_ == i):
continue
plt.scatter(X[Y_ == i, 0], X[Y_ == i, 1], .8, color=color)
# Plot an ellipse to show the Gaussian component
angle = np.arctan(u[1] / u[0])
angle = 180. * angle / np.pi # convert to degrees
ell = mpl.patches.Ellipse(mean, v[0], v[1], 180. + angle, color=color)
ell.set_clip_box(splot.bbox)
ell.set_alpha(0.5)
splot.add_artist(ell)
plt.xlim(-4., 3.)
plt.ylim(-4., 2.)
gmm = mixture.GaussianMixture(n_components=3, covariance_type='full').fit(mix.T)
print(gmm.predict(mix.T))
plot_results(mix.T, gmm.predict(mix.T), gmm.means_, gmm.covariances_, 0,
'Gaussian Mixture')
So for my problem the resulting plot looked like this:
Edit: here the answer to your comment. I would use pandas to do this. Assume X is your feature matrix and y are your labels, then
import pandas as pd
y_pred = gmm.predict(X)
df_all_info = pd.concat([X,y,y_pred], axis=1)
In the resulting dataframe you can check all the information you want, you can even just exclude the samples the algorithm misclassified with:
df_wrong = df_all_info[df_all_info['name of y-column'] != df_all_info['name of y_pred column']]
I use sklearn.linear_model.ElasticNetCV and I would like to get a similar figure as Matlab provides with lassoPlot with plottype=CV or R's plot(cv.glmnet(x,y)), i.e., a plot of the cross validations errors over various alphas (note, in Matlab and R this parameter is called lambda). Here is an example:
import sklearn.linear_model as lm
import matplotlib.pyplot as plt
import numpy as np
import scipy as sp
import scipy.stats as stats
# toy example
# generate 200 samples of five-dimensional artificial data X from a
# exponential distributions with various means:
X = np.zeros( (200,5 ) )
for col in range(5):
X[ :, col ] = stats.expon.rvs( scale=1.0/(col+1) )
# generate response data Y = X*r + eps where r has just two nonzero
# components, and the noise eps is normal with standard deviation 0.1:
r = np.array( [ 0, 2, 0, -3, 0 ] )
Y = np.dot(X,r) + sp.randn( 200 )*0.1
enet = lm.ElasticNetCV()
alphas,coefs, _ = enet.path( X, Y )
# plot regulization paths
plt.plot( -np.log10(alphas), coefs.T, linestyle='-' )
plt.show()
I would like to plot also in a separate figure the cross validation error for each alpha. But It seems that ElasticNetCV.path() does not return a mse vector. Is there a simliar functionality in sklearn to Matlab.lassoPlot with plottype='CV' see: http://de.mathworks.com/help/stats/lasso-and-elastic-net.html or R's cv.glmnet(x,y) https://web.stanford.edu/~hastie/glmnet/glmnet_alpha.html. Alternatively, I would implement it using sklearn.cross_validation. Do you have any suggestions?
In scipy there is no support for fitting a negative binomial distribution using data
(maybe due to the fact that the negative binomial in scipy is only discrete).
For a normal distribution I would just do:
from scipy.stats import norm
param = norm.fit(samp)
Is there something similar 'ready to use' function in any other library?
Statsmodels has discrete.discrete_model.NegativeBinomial.fit(), see here:
https://www.statsmodels.org/dev/generated/statsmodels.discrete.discrete_model.NegativeBinomial.fit.html#statsmodels.discrete.discrete_model.NegativeBinomial.fit
Not only because it is discrete, also because maximum likelihood fit to negative binomial can be quite involving, especially with an additional location parameter. That would be the reason why .fit() method is not provided for it (and other discrete distributions in Scipy), here is an example:
In [163]:
import scipy.stats as ss
import scipy.optimize as so
In [164]:
#define a likelihood function
def likelihood_f(P, x, neg=1):
n=np.round(P[0]) #by definition, it should be an integer
p=P[1]
loc=np.round(P[2])
return neg*(np.log(ss.nbinom.pmf(x, n, p, loc))).sum()
In [165]:
#generate a random variable
X=ss.nbinom.rvs(n=100, p=0.4, loc=0, size=1000)
In [166]:
#The likelihood
likelihood_f([100,0.4,0], X)
Out[166]:
-4400.3696690513316
In [167]:
#A simple fit, the fit is not good and the parameter estimate is way off
result=so.fmin(likelihood_f, [50, 1, 1], args=(X,-1), full_output=True, disp=False)
P1=result[0]
(result[1], result[0])
Out[167]:
(4418.599495886474, array([ 59.61196161, 0.28650831, 1.15141838]))
In [168]:
#Try a different set of start paramters, the fit is still not good and the parameter estimate is still way off
result=so.fmin(likelihood_f, [50, 0.5, 0], args=(X,-1), full_output=True, disp=False)
P1=result[0]
(result[1], result[0])
Out[168]:
(4417.1495981801972,
array([ 6.24809397e+01, 2.91877405e-01, 6.63343536e-04]))
In [169]:
#In this case we need a loop to get it right
result=[]
for i in range(40, 120): #in fact (80, 120) should probably be enough
_=so.fmin(likelihood_f, [i, 0.5, 0], args=(X,-1), full_output=True, disp=False)
result.append((_[1], _[0]))
In [170]:
#get the MLE
P2=sorted(result, key=lambda x: x[0])[0][1]
sorted(result, key=lambda x: x[0])[0]
Out[170]:
(4399.780263084549,
array([ 9.37289361e+01, 3.84587087e-01, 3.36856705e-04]))
In [171]:
#Which one is visually better?
plt.hist(X, bins=20, normed=True)
plt.plot(range(260), ss.nbinom.pmf(range(260), np.round(P1[0]), P1[1], np.round(P1[2])), 'g-')
plt.plot(range(260), ss.nbinom.pmf(range(260), np.round(P2[0]), P2[1], np.round(P2[2])), 'r-')
Out[171]:
[<matplotlib.lines.Line2D at 0x109776c10>]
I know this thread is quite old, but current readers may want to look at this repo which is made for this purpose: https://github.com/gokceneraslan/fit_nbinom
There's also an implementation here, though part of a larger package: https://github.com/ernstlab/ChromTime/blob/master/optimize.py
I stumbled across this thread, and found an answer for anyone else wondering.
If you simply need the n, p parameterisation used by scipy.stats.nbinom you can convert the mean and variance estimates:
mu = np.mean(sample)
sigma_sqr = np.var(sample)
n = mu**2 / (sigma_sqr - mu)
p = mu / sigma_sqr
If you the dispersionparameter you can use a negative binomial regression model from statsmodels with just an interaction term. This will find the dispersionparameter alpha using MLE.
# Data processing
import pandas as pd
import numpy as np
# Analysis models
import statsmodels.formula.api as smf
from scipy.stats import nbinom
def convert_params(mu, alpha):
"""
Convert mean/dispersion parameterization of a negative binomial to the ones scipy supports
Parameters
----------
mu : float
Mean of NB distribution.
alpha : float
Overdispersion parameter used for variance calculation.
See https://en.wikipedia.org/wiki/Negative_binomial_distribution#Alternative_formulations
"""
var = mu + alpha * mu ** 2
p = mu / var
r = mu ** 2 / (var - mu)
return r, p
# Generate sample data
n = 2
p = 0.9
sample = nbinom.rvs(n=n, p=p, size=10000)
# Estimate parameters
## Mean estimates expectation parameter for negative binomial distribution
mu = np.mean(sample)
## Dispersion parameter from nb model with only interaction term
nbfit = smf.negativebinomial("nbdata ~ 1", data=pd.DataFrame({"nbdata": sample})).fit()
alpha = nbfit.params[1] # Dispersion parameter
# Convert parameters to n, p parameterization
n_est, p_est = convert_params(mu, alpha)
# Check that estimates are close to the true values:
print("""
{:<3} {:<3}
True parameters: {:<3} {:<3}
Estimates : {:<3} {:<3}""".format('n', 'p', n, p,
np.round(n_est, 2), np.round(p_est, 2)))
The following code generates a sample of size 100 from trunctated normal distributions with different intervals. Is there any effecient(vectorised) way of doing this?
from scipy.stats import truncnorm
import numpy as np
sample=[]
a_s=np.random.uniform(0,1,size=100)
b_s=a_s+0.2
for i in range(100):
sample.append(truncnorm.rvs(a_s[i], b_s[i], size=100))
print sample
One day in the not so distant future, all NumPy/SciPy functions will broadcast all their arguments, and you will be able to do truncnorm.rvs(a_s, b_s, size=100), but since we are not there yet, you could manually generate your random samples from a uniform distribution and the CDF and PPF of a normal distribution:
import numpy as np
from scipy.stats import truncnorm, norm
a_s = np.random.uniform(0, 1, size=100)
b_s = a_s + 0.2
cdf_start = norm.cdf(a_s)
cdf_stop = norm.cdf(b_s)
cdf_samples = np.random.uniform(0, 1, size=(100, 100))
cdf_samples *= (cdf_stop - cdf_start)[:, None]
cdf_samples += cdf_start[:, None]
truncnorm_samples = norm.ppf(cdf_samples)