Randomly generating integers with prior probabilities

Question

I'm after a nice method to draw from a range of integers given known prior probabilities.

Say I wanted an 80% chance of drawing 1, a 15% chance of drawing 2, and a 5% chance of drawing 3...

I'm obviously thinking about this wrong, since neither

[~,I] = max([.80 .15 .05] .* rand(1,3))

nor

[~,I] = max([.80 .15 .05] + rand(1,3))

appears to achieve this...(rand samples from a uniform distribution)

Does anyone have a suggestion that will extend to any range of integers?

** Additional Comment **

I realise now for rational probabilities it's trivial just to create a vector comprised of these options proportionally (80% 1s, 15% 2s, and 5% 3s), and then just randperm the contents of the vector and choose the first element over and over...

This method is bad if the probabilities do not behave nicely...

In R, there's the `sample` and `sample.int` functions. e.g. `sample(3,100,replace=TRUE,p=c(.8,.15,.05))` will generate 100 values from the distribution in your question. — Glen_b, Aug 13 '17 at 00:56
Thanks @Glen_b. I need to implement a solution in MATLAB...and so it would be great to understand how this function in R is achieving this... — NBland, Aug 13 '17 at 03:39
R is open source; you can read the code but I don't think you'll find it as enlightening in this case as the indicated duplicates. The [table](https://stats.stackexchange.com/a/68041/805) method would probably serve you well. It's quite fast -- you take a longish array and fill it with values in the right proportions so that you are almost sampling with the required probabilities; there's normally a few "leftover" cells at the end which take you to a second step of generation that take you up to the exact probabilities you need (and that step can use any convenient other method) — Glen_b, Aug 13 '17 at 05:39

Semoi · Answer 1 · 2017-08-13T08:32:46.693

Why don't you split your interval into three pieces of unequal length and assign the corresponding integer. Here the Matlab code:

%% define percentage
perc = [80, 95, 100];

%% number of random integers
nMax = 1e3;

%% generate integers
tic
vec = randi([1, 100],1,nMax);
t   = toc;
disp(['Generated ' num2str(nMax) ' random integers in ' num2str(t) 's'])

%% get indices
tic
% idx(numel(perc)).dat = NaN; % only needed if you have many intervals
for i = 1:numel(perc)
   idxVec(i).dat      = find(vec <= perc(i));
   vec(idxVec(i).dat) = NaN;
end
t = toc;
disp(['Indexed ' num2str(nMax) ' random integers in ' num2str(t) 's'])

%% plotting:
figure(1);
for i = 1:numel(perc)
    plot(idxVec(i).dat, i*ones(1,numel(idxVec(i).dat)), '.');
    hold on;
end
hold off

I'm not sure how big your database is. However, for my standard the code is fast:

Generated 10 000 000 random integers in 0.21746s
Indexed 10 000 000 random integers in 0.35026s

PS: You should remove the time taking (tic, toc) and the figure. Especially the plotting will take time for a large database.

This is reasonable for my given example, but it becomes a bit clunky when I want to choose from a larger range of values. Thanks for your answer. :) — NBland, Aug 13 '17 at 03:36

Randomly generating integers with prior probabilities

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