skew normal distribution in scipy

From the Wikipedia description,

from scipy import linspacefrom scipy import pi,sqrt,expfrom scipy.special import erffrom pylab import plot,showdef pdf(x):    return 1/sqrt(2*pi) * exp(-x**2/2)def cdf(x):    return (1 + erf(x/sqrt(2))) / 2def skew(x,e=0,w=1,a=0):    t = (x-e) / w    return 2 / w * pdf(t) * cdf(a*t)    # You can of course use the scipy.stats.norm versions    # return 2 * norm.pdf(t) * norm.cdf(a*t)n = 2**10e = 1.0 # locationw = 2.0 # scalex = linspace(-10,10,n) for a in range(-3,4):    p = skew(x,e,w,a)    plot(x,p)show()

If you want to find the scale, location, and shape parameters from a dataset use scipy.optimize.leastsq, for example using e=1.0,w=2.0 and a=1.0,

fzz = skew(x,e,w,a) + norm.rvs(0,0.04,size=n) # fuzzy datadef optm(l,x):    return skew(x,l[0],l[1],l[2]) - fzzprint leastsq(optm,[0.5,0.5,0.5],(x,))

should give you something like,

(array([ 1.05206154,  1.96929465,  0.94590444]), 1)

The accepted answer is more or less outdated, because a skewnorm function is now implemented in scipy. So the code can be written a lot shorter:

 from scipy.stats import skewnorm import numpy as np from matplotlib import pyplot as plt X = np.linspace(min(your_data), max(your_data)) plt.plot(X, skewnorm.pdf(X, *skewnorm.fit(your_data))