bucketing values
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Hi
I have a vector of values from 0 to 1 representing probabilities
A = [0.8756 0.1185 0.0059]
How can i generate a random number using rand() and map it to a value in A without looping. I can;t see how i can do this with vectorisation or with the built in functions.
My aim is generate a random numbers so 87% of the time I choose A(1), 11% A(2), etc...
Thanks
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Respuesta aceptada
Teja Muppirala
el 21 de Jul. de 2011
If you have the Statistics Toolbox installed, there is the built-in RANDSAMPLE command to do this.
A = [0.8756 0.1185 0.0059];
X = randsample(numel(A),5000,true, A); % X gets 5000 samples
tabulate(X)
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Más respuestas (5)
Daniel Shub
el 21 de Jul. de 2011
Assuming your array A is short ...
A = [0.8756 0.1185 0.0059]
x = rand(1e3, 1);
y(x < A(1)) = 'a';
y(x >= A(1) & x < A(2)) = 'b';
y(x >= A(2)) = 'c';
2 comentarios
Fangjun Jiang
el 21 de Jul. de 2011
I think all the answers provided are along the same line. In your case, the A needs to be [0.87,0.98,1].
Sean de Wolski
el 21 de Jul. de 2011
Like this?
A = [0.8756 0.1185 0.0059]; %percentiles (must sum to 1
R = rand(1,1000); %random data
B = [11 17 19]; %sample data to extract A% of the time
[junk,Bin] = histc(R,[-inf cumsum(A)]); %find the bin
C = B(Bin) %extract
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Fangjun Jiang
el 21 de Jul. de 2011
Something like this:
A = [0.8756 0.1185 0.0059];
RandNum=rand(1000,1);
Index=2*ones(size(RandNum));
Index(RandNum<A(1))=1;
Index(RandNum>1-A(3))=3;
sum(Index==1)
sum(Index==2)
sum(Index==3)
ans =
892
ans =
102
ans =
6
close enough?
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