How to pick a value according to its probability

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Steven
Steven 2011-12-7
评论： Real User 2023-4-28
Hi,
Let's say
P = [0.1 0.3 0.4 0.2]
X = [1 2 5 9]
where P(n) is the probability to select the X(n) element. I wish to make a function that select a "random" element of X according to its probability, like
f = myfun(P,X)
>> f = 2 (occurs around 30%)
thx a lot
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Sean de Wolski
Sean de Wolski 2011-12-7
f = X(find(rand<cumsum(P),1,'first'))
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Walter Roberson
Walter Roberson 2011-12-7
The answers in the other thread took care in case cumsum(P) < 1 as can happen due to round-off error.

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更多回答（3 个）

Jonathan
Jonathan 2018-9-3
编辑：Jonathan 2018-9-3
The accepted answer is not doing any sanity check, and is sensitive to rounding errors. You should use randsample instead.
To sample n points from X, with replacement, and probabilities P:
randsample( X, n, true, P )
This can also be used with a custom RandStream (see documentation). Be aware that this function does NOT check for negative values in P; check manually if needed.
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Steven Lord
Steven Lord 2020-4-16
You can use discretize (which didn't exist when this question was asked originally) to do this. Generate uniform random numbers, bin them using bins whose widths are given by P, and for each bin return the corresponding element of X.
P = [0.1 0.3 0.4 0.2];
X = [1 2 5 9];
values = discretize(rand(1, 1e4), cumsum([0 P]), X);
histogram(values, 'Normalization', 'probability')
The probabilities shown in the histogram should agree pretty closely with the values in P.
Mendi
Mendi 2021-7-9
The fastest one (100ns-200ns):
function [idx] = get_random_choice(p)
% Random choice with probability
% Example: get_random_choice([0.2,0.7,0.1])
N=length(p); idx=1; cump=0;
r=rand;
while(idx<N)
cump=cump+p(idx);
if(cump>r),break,else,idx=idx+1;end
end
end

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