Python/Numpy/Scipy: Draw Poisson random values with different lambda
Asked Answered
F

1

6

My problem is to extract in the most efficient way N Poisson random values (RV) each with a different mean/rate Lam. Basically the size(RV) == size(Lam).

Here it is a naive (very slow) implementation:

import numpy as NP

def multi_rate_poisson(Lam):
    rv = NP.zeros(NP.size(Lam))
    for i,lam in enumerate(Lam):
        rv[i] = NP.random.poisson(lam=lam, size=1)
    return rv

That, on my laptop, with 1e6 samples gives:

Lam = NP.random.rand(1e6) + 1
timeit multi_poisson(Lam)
1 loops, best of 3: 4.82 s per loop

Is it possible to improve from this?

Filmore answered 21/4, 2013 at 18:18 Comment(0)
S
3

Although the docstrings don't document this functionality, the source indicates it is possible to pass an array to the numpy.random.poisson function.

>>> import numpy
>>> # 1 dimension array of 1M random var's uniformly distributed between 1 and 2
>>> numpyarray = numpy.random.rand(1e6) + 1 
>>> # pass to poisson
>>> poissonarray = numpy.random.poisson(lam=numpyarray)
>>> poissonarray
array([4, 2, 3, ..., 1, 0, 0])

The poisson random variable returns discrete multiples of one, and approximates a bell curve as lambda grows beyond one.

>>> import matplotlib.pyplot
>>> count, bins, ignored = matplotlib.pyplot.hist(
            numpy.random.poisson(
                    lam=numpy.random.rand(1e6) + 10), 
                    14, normed=True)
>>> matplotlib.pyplot.show()

This method of passing the array to the poisson generator appears to be quite efficient.

>>> timeit.Timer("numpy.random.poisson(lam=numpy.random.rand(1e6) + 1)",
                 'import numpy').repeat(3,1)
[0.13525915145874023, 0.12136101722717285, 0.12127304077148438]
Susannasusannah answered 17/9, 2013 at 5:40 Comment(0)

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