How to efficiently generate a random integer within a range from an arbitrary probability distribution

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Paolo Binetti
Paolo Binetti 2017-1-7
评论： the cyclist 2017-1-8
I need to generate a random integer within a range from an arbitrary probability distribution, within a loop of 100000 iterations. My implementation works, but I am not sure it is mathematically clean, and it takes forever:
pdf = [ 0.9 0.3 0.003 0.1 0.07 0.0005 0.003 0.15 0.009 0.08 ]; % discrete prob distrib function
cdf = cumsum(pdf); % cumulative distribution function
cdf = cdf / max(cdf); cdf(1) = 0; % normalization
index = ceil(interp1(cdf, [1:numel(pdf)], rand(1)))
Notice that the pdf above is just an example: my actual case is a vector of about 500 numbers.
Here is a different solution, which seems mathematically cleaner, but does not work for my overall problem, and is just as slow as above:
pdf = [ 0.9 0.3 0.003 0.1 0.07 0.0005 0.003 0.15 0.009 0.08 ]; % discrete prob distrib function
cdf = cumsum(pdf); % cumulative distribution function
cdf = cdf - min(cdf); cdf = cdf / max(cdf); % normalization
index = round(interp1(cdf, [1:numel(pdf)], rand(1)))
Is there a more efficient/correct way to do this?

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采纳的回答

the cyclist
the cyclist 2017-1-8
编辑：the cyclist 2017-1-8
I think that
index = sum(rand()>cdf)+1;
will be much faster than using interp1 as you do, and will give the same result.
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the cyclist
the cyclist 2017-1-8
Regarding speed ...
I get roughly equivalent results between my solution and randsample. Note that both of these solutions can be vectorized:
index = sum(rand(N,1)>cdf,2)';
and
index = randsample(1:numel(pdf),N,true,pdf);
These will both generate N = 100,000 samples in a few milliseconds.
I did not try to vectorize your solution, but as it currently stands, it took 5 seconds to generate that many.

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更多回答（1 个）

the cyclist
the cyclist 2017-1-8
Do you have the Statistics and Machine Learning Toolbox? If so, you can do this with the randsample command.

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