Generating a random number from array based on requirements

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Matlab Beginner
Matlab Beginner 2021-2-24
评论： Walter Roberson 2021-2-24
I am using the code below to allow me to select a single "random" element from an array with each element having its own weighting.
A=[1,2,3,4];
p=[10 20 30 40];
c=cumsum(p);
[~,r]=histc(rand(1,1),[0 c/c(end)]);
r=A(r);
Suppose my requirements now change, and only elements 3 and 4 can be selected from A. Without having to redo the percentages (i.e. changing p array), how can I modify the inputs to the histc function to achieve this?
Thank you

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回答（1 个）

Rik
Rik 2021-2-24
Ran in:
A=[1,2,3,4];
p=[10 20 30 40];
c=cumsum(p);
isAllowed=[3,4];
while true
[~,r]=histc(rand(1,1),[0 c/c(end)]);
if ismember(r,isAllowed),break;end
end
disp(r)
4
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Rik
Rik 2021-2-24
You could use something like the mod function to wrap the results so they will always index isAllowed, however, that will completely break the distribution. So instead of changing p you would skew the results so much that you're effectively ignoring it.
Walter Roberson
Walter Roberson 2021-2-24
Ran in:
A = [1, 2, 3, 4];
p = [10 20 30 40];
isAllowed=[3,4];
c = cumsum(p);
N = 100;
[~,r]=histc(rand(1,N),[0 c/c(end)]);
r = A(r(ismember(r,isAllowed)));
stem(r)
This is the rejection process. Notice that the size of the returned data is only 70% of nominal. If you need a fixed number of samples output then you have to go back and ask for more. This can be pretty expensive -- for example you might be asking to accept only items with a 5% probability, and then to get 100 outputs on average you would need 100/0.05 = 2000 tries.

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