# How to generate geometric random numbers?

Discussion in 'Python' started by MyInfoStation, Jul 23, 2006.

1. ### MyInfoStationGuest

Hi all,

I am a newbie to Python and would like to genereate some numbers
according to geometric distribution. However, the Python Random package
seems do not have implemented functionality. I am wondering is there
exist any other libraries that can do this job?

Thanks a lot,

Da

MyInfoStation, Jul 23, 2006

2. ### Gerhard FiedlerGuest

The usual way is to generate standard random numbers (linear distribution)
and then apply whatever transformation you need to generate the desired
distribution.

Gerhard

Gerhard Fiedler, Jul 23, 2006

3. ### Robert KernGuest

That only works if there is such a transformation.

The geometric distribution and many others have been implemented in numpy:

http://www.scipy.org/NumPy

In [1]: from numpy import random

In [2]: random.geometric(0.5, size=100)
Out[2]:
array([1, 5, 2, 3, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 2, 2, 3, 3, 1, 1, 1, 1, 1,
2, 2, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 3, 1,
4, 1, 1, 1, 2, 1, 2, 3, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 6, 1, 1, 3, 2,
1, 1, 2, 1, 1, 7, 2, 1, 1, 2, 1, 1, 2, 4, 1, 2, 1, 4, 2, 1, 1, 2, 1,
4, 2, 1, 1, 3, 1, 3, 1])

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco

Robert Kern, Jul 23, 2006
4. ### MyInfoStationGuest

Thanks a lot. I will try it out.

But I am still surprised because the default Random package in Python
can generate so few discrete random distritbuions, while it can
generate quite a few continuous distribution, including some not very
common one.

Da

MyInfoStation, Jul 24, 2006
5. ### Paul RubinGuest

It looks pretty simple to transform the uniform distribution to the
geometric distribution. The formula for its cdf is pretty simple:

cdf(p,n) = (1-p)**(n-1)*p

For fixed p, if the cdf is c, we get (unless I made an error),

n = log(c, 1-p) - 1

So choose a uniform point c in the unit interval, run it through that
formula, and round up to the nearest integer.

See http://en.wikipedia.org/wiki/Geometric_distribution

Paul Rubin, Jul 24, 2006
6. ### Paul RubinGuest

I meant n = log(c/p, 1-p) - 1
sorry.

Paul Rubin, Jul 24, 2006
7. ### Robert KernGuest

import random
from math import ceil, log

def geometric(p):
""" Geometric distribution per Devroye, Luc. _Non-Uniform Random Variate
Generation_, 1986, p 500. http://cg.scs.carleton.ca/~luc/rnbookindex.html
"""

# p should be in (0.0, 1.0].
if p <= 0.0 or p > 1.0:
raise ValueError("p must be in the interval (0.0, 1.0]")
elif p == 1.0:
# If p is exactly 1.0, then the only possible generated value is 1.
# Recognizing this case early means that we can avoid a log(0.0) later.
# The exact floating point comparison should be fine. log(eps) works just
# dandy.
return 1

# random() returns a number in [0, 1). The log() function does not
# like 0.
U = 1.0 - random.random()

# Find the corresponding geometric variate by inverting the uniform variate.
G = int(ceil(log(U) / log(1.0 - p)))
return G

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco

Robert Kern, Jul 24, 2006
8. ### Paul RubinGuest

I usually owuld write that as int(ceil(log(U, 1.0 - p))).

Paul Rubin, Jul 24, 2006
9. ### Robert KernGuest

Knock yourself out. I was cribbing from my C implementation in numpy.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco

Robert Kern, Jul 24, 2006
10. ### Paul RubinGuest

Oh cool, I thought you were pasting from a Python implementation. No prob.

Paul Rubin, Jul 24, 2006