[SciPy-user] Simple combinatorics with Numpy

Mico Filós elmico.filos@gmail....
Wed Oct 22 05:57:11 CDT 2008


Thanks a lot for your answers.

The functions in sample.py do perfectly the job.

The motivation for this unordered sampling with replacement was perhaps a
little unclear in my first post. I need to count the different ways that a mean
sample of size 3 can occur. The probability of a given value for the sample
mean depends on how many *ordered* samples give rise to that given value. For
instance,

   {1,1,1} = [(1,1,1)]                          -> 1
   {1,1,3} = [(1,1,3), (1,3,1), (3,1,1)]   -> 5/3
   {1,3,5} = [(1,3,5), (3,1,5), (1,5,3), (5,1,3), (5,3,1), (3,5,1)]   -> 8/3
   {1,3,3} = [(1,3,3), (3,1,3), (3,3,1)]   -> 7/3
   {3,3,3} = [(3,3,3)]                     -> 3 ...

where {} and () denote the unordered and ordered samples, respectively. The
number of different orderings associated with a particular sample is given by a
multinomial distribution: if there are 3 places and 4 different numbers
repeated k1, k2, k3, k4 times, there are c = 3!/(k1!k2!k3!k4!) ordered samples.

Of course, I could also have proceeded the brute force way, enumerating all 3^4
ordered samples, computing their average, and counting the number of times each
different value occurs.

Sorry for being so wordy.


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