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How to calculate a Normal Distribution percent point function in python

How do I do the equivalent of scipy.stats.norm.ppf without using Scipy. I have python's Math module has erf built in but I cannot seem to recreate the function.

PS: I cannot just use scipy because Heroku does not allow you to install it and using alternate buildpacks breaches the 300Mb maximum slug size limit.

like image 221
SARose Avatar asked Dec 27 '16 03:12

SARose


1 Answers

There's not a simple way to use erf to implement norm.ppf because norm.ppf is related to the inverse of erf. Instead, here's a pure Python implementation of the code from scipy. You should find that the function ndtri returns exactly the same value as norm.ppf:

import math

s2pi = 2.50662827463100050242E0

P0 = [
    -5.99633501014107895267E1,
    9.80010754185999661536E1,
    -5.66762857469070293439E1,
    1.39312609387279679503E1,
    -1.23916583867381258016E0,
]

Q0 = [
    1,
    1.95448858338141759834E0,
    4.67627912898881538453E0,
    8.63602421390890590575E1,
    -2.25462687854119370527E2,
    2.00260212380060660359E2,
    -8.20372256168333339912E1,
    1.59056225126211695515E1,
    -1.18331621121330003142E0,
]

P1 = [
    4.05544892305962419923E0,
    3.15251094599893866154E1,
    5.71628192246421288162E1,
    4.40805073893200834700E1,
    1.46849561928858024014E1,
    2.18663306850790267539E0,
    -1.40256079171354495875E-1,
    -3.50424626827848203418E-2,
    -8.57456785154685413611E-4,
]

Q1 = [
    1,
    1.57799883256466749731E1,
    4.53907635128879210584E1,
    4.13172038254672030440E1,
    1.50425385692907503408E1,
    2.50464946208309415979E0,
    -1.42182922854787788574E-1,
    -3.80806407691578277194E-2,
    -9.33259480895457427372E-4,
]

P2 = [
    3.23774891776946035970E0,
    6.91522889068984211695E0,
    3.93881025292474443415E0,
    1.33303460815807542389E0,
    2.01485389549179081538E-1,
    1.23716634817820021358E-2,
    3.01581553508235416007E-4,
    2.65806974686737550832E-6,
    6.23974539184983293730E-9,
]

Q2 = [
    1,
    6.02427039364742014255E0,
    3.67983563856160859403E0,
    1.37702099489081330271E0,
    2.16236993594496635890E-1,
    1.34204006088543189037E-2,
    3.28014464682127739104E-4,
    2.89247864745380683936E-6,
    6.79019408009981274425E-9,
]

def ndtri(y0):
    if y0 <= 0 or y0 >= 1:
        raise ValueError("ndtri(x) needs 0 < x < 1")
    negate = True
    y = y0
    if y > 1.0 - 0.13533528323661269189:
        y = 1.0 - y
        negate = False

    if y > 0.13533528323661269189:
        y = y - 0.5
        y2 = y * y
        x = y + y * (y2 * polevl(y2, P0) / polevl(y2, Q0))
        x = x * s2pi
        return x

    x = math.sqrt(-2.0 * math.log(y))
    x0 = x - math.log(x) / x

    z = 1.0 / x
    if x < 8.0:
        x1 = z * polevl(z, P1) / polevl(z, Q1)
    else:
        x1 = z * polevl(z, P2) / polevl(z, Q2)
    x = x0 - x1
    if negate:
        x = -x
    return x

def polevl(x, coef):
    accum = 0
    for c in coef:
        accum = x * accum + c
    return accum
like image 76
K. A. Buhr Avatar answered Sep 23 '22 00:09

K. A. Buhr