How to pick a value according to its probability
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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
4 Commenti
Risposta accettata
Sean de Wolski
il 7 Dic 2011
f = X(find(rand<cumsum(P),1,'first'))
1 Commento
Walter Roberson
il 7 Dic 2011
The answers in the other thread took care in case cumsum(P) < 1 as can happen due to round-off error.
Più risposte (3)
Jonathan
il 3 Set 2018
Modificato: Jonathan
il 3 Set 2018
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.
4 Commenti
Steven Lord
il 16 Apr 2020
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.
0 Commenti
Mendi
il 9 Lug 2021
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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