如何计算累积正态分布？

I am looking for a function in Numpy or Scipy (or any rigorous Python library) that will give me the cumulative normal distribution function in Python.

Here’s an example:

>>> from scipy.stats import norm
>>> norm.cdf(1.96)
0.9750021048517795
>>> norm.cdf(-1.96)
0.024997895148220435

In other words, approximately 95% of the standard normal interval lies within two standard deviations, centered on a standard mean of zero.

If you need the inverse CDF:

>>> norm.ppf(norm.cdf(1.96))
array(1.9599999999999991)

It may be too late to answer the question but since Google still leads people here, I decide to write my solution here.

That is, since Python 2.7, the math library has integrated the error function math.erf(x)

The erf() function can be used to compute traditional statistical functions such as the cumulative standard normal distribution:

from math import *
def phi(x):
    #'Cumulative distribution function for the standard normal distribution'
    return (1.0 + erf(x / sqrt(2.0))) / 2.0

Ref:

https://docs.python.org/2/library/math.html

https://docs.python.org/3/library/math.html

How are the Error Function and Standard Normal distribution function related?

Adapted from here http://mail.python.org/pipermail/python-list/2000-June/039873.html

from math import *
def erfcc(x):
    """Complementary error function."""
    z = abs(x)
    t = 1. / (1. + 0.5*z)
    r = t * exp(-z*z-1.26551223+t*(1.00002368+t*(.37409196+
        t*(.09678418+t*(-.18628806+t*(.27886807+
        t*(-1.13520398+t*(1.48851587+t*(-.82215223+
        t*.17087277)))))))))
    if (x >= 0.):
        return r
    else:
        return 2. - r

def ncdf(x):
    return 1. - 0.5*erfcc(x/(2**0.5))

To build upon Unknown’s example, the Python equivalent of the function normdist() implemented in a lot of libraries would be:

def normcdf(x, mu, sigma):
    t = x-mu;
    y = 0.5*erfcc(-t/(sigma*sqrt(2.0)));
    if y>1.0:
        y = 1.0;
    return y

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

def normdist(x, mu, sigma, f):
    if f:
        y = normcdf(x,mu,sigma)
    else:
        y = normpdf(x,mu,sigma)
    return y

Starting Python 3.8, the standard library provides the NormalDist object as part of the statistics module.

It can be used to get the cumulative distribution function ( – probability that a random sample X will be less than or equal to x) for a given mean (mu) and standard deviation (sigma):

from statistics import NormalDist

NormalDist(mu=0, sigma=1).cdf(1.96)
# 0.9750021048517796

Which can be simplified for the standard normal distribution (mu = 0 and sigma = 1):

NormalDist().cdf(1.96)
# 0.9750021048517796

NormalDist().cdf(-1.96)
# 0.024997895148220428

Alex’s answer shows you a solution for standard normal distribution (mean = 0, standard deviation = 1). If you have normal distribution with mean and std (which is sqr(var)) and you want to calculate:

from scipy.stats import norm

# cdf(x < val)
print norm.cdf(val, m, s)

# cdf(x > val)
print 1 - norm.cdf(val, m, s)

# cdf(v1 < x < v2)
print norm.cdf(v2, m, s) - norm.cdf(v1, m, s)

Read more about cdf here and scipy implementation of normal distribution with many formulas here.

Taken from above:

from scipy.stats import norm
>>> norm.cdf(1.96)
0.9750021048517795
>>> norm.cdf(-1.96)
0.024997895148220435

For a two-tailed test:

Import numpy as np
z = 1.96
p_value = 2 * norm.cdf(-np.abs(z))
0.04999579029644087

Simple like this:

import math
def my_cdf(x):
    return 0.5*(1+math.erf(x/math.sqrt(2)))

I found the formula in this page https://www.danielsoper.com/statcalc/formulas.aspx?id=55

As Google gives this answer for the search netlogo pdf, here’s the netlogo version of the above python code

    ;; Normal distribution cumulative density function
    to-report normcdf [x mu sigma]
        let t x - mu
        let y 0.5 * erfcc [ - t / ( sigma * sqrt 2.0)]
        if ( y > 1.0 ) [ set y 1.0 ]
        report y
    end

    ;; Normal distribution probability density function
    to-report normpdf [x mu sigma]
        let u = (x - mu) / abs sigma
        let y = 1 / ( sqrt [2 * pi] * abs sigma ) * exp ( - u * u / 2.0)
        report y
    end

    ;; Complementary error function
    to-report erfcc [x]
        let z abs x
        let t 1.0 / (1.0 + 0.5 * z)
        let r t *  exp ( - z * z -1.26551223 + t * (1.00002368 + t * (0.37409196 +
            t * (0.09678418 + t * (-0.18628806 + t * (.27886807 +
            t * (-1.13520398 +t * (1.48851587 +t * (-0.82215223 +
            t * .17087277 )))))))))
        ifelse (x >= 0) [ report r ] [report 2.0 - r]
    end