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.
2 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
The function ppf is the inverse of y = (1+erf(x/sqrt(2))/2. So we need to solve this equation for x, given y between 0 and 1. Here is a code doing this by the bisection method. I imported SciPy function to illustrate that the result is the same.
from math import erf, sqrt
from scipy.stats import norm # only for comparison
y = 0.123
z = 2*y-1
a = 0
while erf(a) > z or erf(a+1) < z: # looking for initial bracket of size 1
if erf(a) > z:
a -= 1
else:
a += 1
b = a+1 # found a bracket, proceed to refine it
while b-a > 1e-15: # 1e-15 ought to be enough precision
c = (a+b)/2.0 # bisection method
if erf(c) > z:
b = c
else:
a = c
print sqrt(2)*(a+b)/2.0 # this is the answer
print norm.ppf(y) # SciPy for comparison
Left for you to do:
- preliminary bound checks (y must be between 0 and 1)
- scaling and shifting if other mean / variance are desired; the code is for standard normal distribution (mean 0, variance 1).
