Alternative for scipy.stats.norm.pdf?

问题

Does anyone know of an alternative for scipy.stats.norm.pdf()? I'm hosting my python site on Google App Engine and Google doesn't support SciPy.

I've tried this function, but that didn't return the same results as scipy:

def normpdf(x, mu, sigma):
    u = (x-mu)/abs(sigma)
    y = (1/(sqrt(2*pi)*abs(sigma)))*exp(-u*u/2)
    return y

For example:

print scipy.stats.norm.pdf(20, 20, 10)
print normpdf(20, 20, 10)

print scipy.stats.norm.pdf(15, 20, 10)
print normpdf(15, 20, 10)

print scipy.stats.norm.pdf(10, 20, 10)
print normpdf(10, 20, 10)

Returns these values:

0.0398942280401
0.0398942280401

0.0352065326764
0.0146762663174

0.0241970724519
0.0146762663174

回答1:

You got tricked by pythons integer division arithmetics! Here is some working code:

from __future__ import division

import scipy.stats
from numpy import *

def normpdf(x, mu, sigma):
    u = (x-mu)/abs(sigma)
    y = (1/(sqrt(2*pi)*abs(sigma)))*exp(-u*u/2)
    return y


print scipy.stats.norm.pdf(20, 20, 10)
print normpdf(20, 20, 10)

print scipy.stats.norm.pdf(15, 20, 10)
print normpdf(15, 20, 10)

print scipy.stats.norm.pdf(10, 20, 10)
print normpdf(10, 20, 10)

Note the first line! Otherwise, you could convert each input variable to a float, e.g. by multiplying by 1.

回答2:

The division by 2 inside of the exp is being interpreted as integer division whenever u evaluates to an int. To prevent this you can ensure that u always evaluates to a float by manually casting it:

def normpdf(x, mu=0, sigma=1):
    u = float((x-mu) / abs(sigma))
    y = exp(-u*u/2) / (sqrt(2*pi) * abs(sigma))
    return y

(I also provided default arguments for mu and sigma, you could remove those if you wanted)

来源：https://stackoverflow.com/questions/8669235/alternative-for-scipy-stats-norm-pdf

标签

python

math

scipy

normal-distribution