How to generate geometric random numbers?

[MyInfoStation]

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

 

[Gerhard Fiedler]

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?

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

Gerhard

 

[Robert Kern]

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

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
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

 

[MyInfoStation]

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

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
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

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

 

[Paul Rubin]

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.

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
for more about the distribution.

 

[Paul Rubin]

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

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

 

[Robert Kern]

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

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
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

 

[Paul Rubin]

G = int(ceil(log(U) / log(1.0 - p)))

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

 

[Robert Kern]

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

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
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

 

[Paul Rubin]

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

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