How do I make plots of a 1-dimensional Gaussian distribution function using the mean and standard deviation parameter values (μ, σ) = (−1, 1), (0, 2), and (2, 3)?
I'm new to programming, using Python.
Thank you in advance!
1
-
Possible duplicate of python pylab plot normal distributionTrevor Boyd Smith– Trevor Boyd Smith2019-10-24 22:12:00 +00:00Commented Oct 24, 2019 at 22:12
Add a comment
|
5 Answers
With the excellent matplotlib and numpy packages
from matplotlib import pyplot as mp
import numpy as np
def gaussian(x, mu, sig):
return (
1.0 / (np.sqrt(2.0 * np.pi) * sig) * np.exp(-np.power((x - mu) / sig, 2.0) / 2)
)
x_values = np.linspace(-3, 3, 120)
for mu, sig in [(-1, 1), (0, 2), (2, 3)]:
mp.plot(x_values, gaussian(x_values, mu, sig))
mp.show()
will produce something like
4 Comments
Andrew Mao
Your gaussian PDF is wrong - you need to scale by (\sqrt(2\pi)\sigma)^(-1). Additionally, x*x is much faster than pow(x, 2).
sinapan
shouldn't the x-axis be from -3 to 3?
danodonovan
@sinapan yes it should (updated). The old X values where counters, not the values of X (ie a mistake).
Stefnotch
@AndrewMao Does your comment still apply?
The correct form, based on the original syntax, and correctly normalized is:
def gaussian(x, mu, sig):
return 1./(np.sqrt(2.*np.pi)*sig)*np.exp(-np.power((x - mu)/sig, 2.)/2)
3 Comments
Kyle McDonald
This is the only answer with normalization that matches scipy. Needed to add a couple "np", and the decimal marks are superfluous
1/(np.sqrt(2*np.pi)*sig)*np.exp(-np.power((x - mu)/sig, 2)/2)
GreenhouseVeg
Finally someone who remembered π in the denominator!
Tom Pohl
Instead of
np.power(x, 2) one could use np.square(x) (which might be a bit faster due to specialization).you can read this tutorial for how to use functions of statistical distributions in python. https://docs.scipy.org/doc/scipy/tutorial/stats.html
from scipy.stats import norm
import matplotlib.pyplot as plt
import numpy as np
#initialize a normal distribution with frozen in mean=-1, std. dev.= 1
rv = norm(loc = -1., scale = 1.0)
rv1 = norm(loc = 0., scale = 2.0)
rv2 = norm(loc = 2., scale = 3.0)
x = np.arange(-10, 10, .1)
#plot the pdfs of these normal distributions
plt.plot(x, rv.pdf(x), x, rv1.pdf(x), x, rv2.pdf(x))
Comments
In addition to previous answers, I recommend to first calculate the ratio in the exponent, then taking the square:
def gaussian(x,x0,sigma):
return np.exp(-np.power((x - x0)/sigma, 2.)/2.)
That way, you can also calculate the gaussian of very small or very large numbers:
In: gaussian(1e-12,5e-12,3e-12)
Out: 0.64118038842995462
Comments
You are missing a parantheses in the denominator of your gaussian() function. As it is right now you divide by 2 and multiply with the variance (sig^2). But that is not true and as you can see of your plots the greater variance the more narrow the gaussian is - which is wrong, it should be opposit.
So just change the gaussian() function to:
def gaussian(x, mu, sig):
return np.exp(-np.power(x - mu, 2.) / (2 * np.power(sig, 2.)))
1 Comment
Eugene
This formula is wrong because if you integrate it from minus infinity to infinity you will get sqrt(2)*sqrt(pi) that isn't right. The right formula is 1/sqrt(2*pi)*exp(-x^2/2). Although, in this form, its mean is 0 and variance is 1, you can shift and scale this gaussian as you like
