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Selecting a random number with some probability

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Aftab Ahmed Khan
Aftab Ahmed Khan el 25 de Feb. de 2015
Respondida: Steven Lord el 16 de Jun. de 2022
Hello Everyone, I am using this one line of code to generate a single value either to be 1 or 2 with equal probability but my question is that how can i select the value to be 1 with 60% probability and the value to be 2 with 40% probability ? Thank you.

Respuesta aceptada

Torsten el 25 de Feb. de 2015
if x<0.6
Best wishes

Más respuestas (5)

pankaj singh
pankaj singh el 10 de Mzo. de 2018
The simplest technique is to use inbuilt Matlab function 'randscr'.
Suppose you want to generate M by N matrix of W, X, Y, and Z with probabilities i,j,k, and l. Then use
out = randsrc(M,N,[W,X,Y,Z;i,j,k,l]); % i+j+k+l = 1;
In your case, as you want a single value to be generated, your M x N = 1 x 1 matrix; the values are 1 with 60% probability (i.e. 0.6) and 2 with 40% (i.e. 0.4) probability, therefore use this;
out = randsrc(1,1,[1,2;0.6,0.4]);
Note that the above is just an example. You can create any matrix size with any number of values. The sum of probabilities must be equal to 1.
  3 comentarios
Ana Gabriela Arteaga
Ana Gabriela Arteaga el 11 de Jun. de 2020
same problem.
Jonathan Ford
Jonathan Ford el 11 de Jul. de 2021
I'm not sure what is causing your error, but you could try writing your own randsrc function, something like this:
function X = myrandsrc(M, N, A)
X = reshape(A(1,sum(A(2,:) < rand(M*N,1)*ones(1,size(A,2)),2)+1),M,N);
X = myrandsrc(4,5,[1 2 3 4; 0.4 0.7 0.9 1])
will return something like:
X =
1 2 3 2 2
2 1 4 3 2
3 2 2 3 2
2 2 2 3 2
Note that for this implementation, you need to use the cumulative probability distribution in the second row of A, so with the above call you will get ~40% of 1, 30% of 2, 20% of 3 and 10% of 4.
Below is the function separated out into multiple lines, to better explain how it works:
function X = myrandsrc(M, N, A)
% the number of elements to chose from
sz = size(A,2);
% generate some random numbers
r = rand(M*N,1)*ones(1,sz);
% determine the correct elements
r = sum(A(2,:) < r,2)+1;
% select the correct elements
X = A(1,r);
% Reshape into M x N matrix
X = reshape(X,M,N);

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Daniel el 25 de Feb. de 2015
You can use rand, which gives uniform distribution and look if the number is below or above 0.6.
if(0.6 <= rand()){
select = 1;
} else {
select = 2;
That should give 60/40 chances. There are more elegant ways to do that though.
  2 comentarios
Trung Khoa Le
Trung Khoa Le el 17 de En. de 2019
Could you please give me a bit explanation why this way makes sense or some documentation that I can read to gain some intuition? Thanks
Luciano Anastassiou
Luciano Anastassiou el 22 de Mayo de 2019
Hi Trung Khoa Le,
The intuition is simply that "rand" generates a random number between 0 and 1. Then when you apply the "if" constraint, you are telling the system to only give out the result "select = 1" when that random number is below 0.6. Otherwise it will give out "select = 2".
If you repeated this 1000s of times, it would give out "select = 1" 60% of the time, because 60% of those random numbers between 0 and 1 will be below 0.6, and the other 40% of the time it will give out "select = 2".

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Jos (10584)
Jos (10584) el 25 de Feb. de 2015
Editada: Jos (10584) el 25 de Feb. de 2015
For two values it is simple
VAL = [10 20] % 2 values
P = .8 % probabbility of selecting first value
Ndraw = 20 % number of draws
R = rand(Ndraw,1) < P
SEL = VAL(2 - R) % use as index into VAL
For more complicated cases you might be interested inTake a look at my RANDP function, which picks random values with relative probabilities.

Steven Lord
Steven Lord el 16 de Jun. de 2022
Another approach is to use the discretize function to discretize a uniform random number between 0 and 1 as generated by the rand function. Because edges is [0, 0.6, 1] any values in uniform that are in the range [0, 0.6) will be mapped to 1 in oneOrTwo and any values in uniform in the range [0.6, 1] will be mapped to 2.
probabilities = [0.6 0.4];
edges = cumsum([0 probabilities])
edges = 1×3
0 0.6000 1.0000
uniform = rand(1, 1e4);
oneOrTwo = discretize(uniform, edges);
We can check using histogram that the generated numbers have the right distribution (or close to it.)
histogram(oneOrTwo, 'Normalization', 'probability')
% Draw lines at the desired probabilities
yline(probabilities, 'r:')
Those look to be in pretty good agreement with the desired probabilities to me.

Ka Mirul
Ka Mirul el 14 de Nov. de 2017
Editada: KSSV el 16 de Jun. de 2022
I have create a video about generating random number in MATLAB
Hope that will help you

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