Conditional Random number generation
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wang syr
am 2 Mär. 2017
Kommentiert: Bruno Luong
am 12 Aug. 2023
Hello there, For example; If I want to generate 5 random integer numbers with a sum of 20, how can I do that?
" ... example = ceil(10*rand(100, 5)) ... "
Akzeptierte Antwort
Roger Stafford
am 4 Mär. 2017
function R = randfixedsumint(m,n,S);
% This generates an m by n array R. Each row will sum to S, and
% all elements are all non-negative integers. The probabilities
% of each possible set of row elements are all equal.
% RAS - Mar. 4, 2017
if ceil(m)~=m|ceil(n)~=n|ceil(S)~=S|m<1|n<1|S<0
error('Improper arguments')
else
P = ones(S+1,n);
for in = n-1:-1:1
P(:,in) = cumsum(P(:,in+1));
end
R = zeros(m,n);
for im = 1:m
s = S;
for in = 1:n
R(im,in) = sum(P(s+1,in)*rand<=P(1:s,in));
s = s-R(im,in);
end
end
end
return
6 Kommentare
Yu Takahashi
am 9 Feb. 2021
Bearbeitet: Walter Roberson
am 10 Feb. 2021
Wondering whether it is possible to specify the max and min of the devided value? i.e., something like what you kindly provided in the randfixedsum function, thanks!
Ref
Bruno Luong
am 12 Aug. 2023
@Walter Roberson "Wondering whether it is possible to specify the max and min of the devided value?"
Weitere Antworten (2)
Walter Roberson
am 2 Mär. 2017
9 Kommentare
Walter Roberson
am 4 Mär. 2017
Ah. I don't think I know how to implement your suggestion, though, at least not without generating all of the possible choices that sum to 20 and then picking one at random.
John D'Errico's https://www.mathworks.com/matlabcentral/fileexchange/12009-partitions-of-an-integer can calculate all of the possible partitions; a question is whether we can avoid having to take that step.
Walter Roberson
am 4 Mär. 2017
https://en.wikipedia.org/wiki/Partition_(number_theory)#Restricted_part_size_or_number_of_parts talks about restricted partitioning briefly, and ties it to change making problems, which does indeed sound equivalent to the approach I was taking. Those are in turn tied to knapsack problems.
Bruno Luong
am 10 Aug. 2020
m = 5;
n = 3;
s = 10;
This will generate uniform distribution with sum criteria
% generate non-negative integer random (m x n) array row-sum to s
[~,r] = maxk(rand(m,s+n-1),n-1,2);
z = zeros(m,1);
r = diff([z, sort(r,2), (s+n)+z],1,2)-1;
1 Kommentar
Bimal Ghimire
am 4 Okt. 2020
While generating conditional random numbers, how can we generate random numbers that has a limit of some maximum value and have certain specified sum value?
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