random number generation for known sum
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i what do generate some 15 random numbers that sum for each separate value of say, 60 , 70, 40, 65. How would i go abouts this?
sum = [60 70 40 65];
n=15;
b = [ ];
for i = 1:1:n,
a = randi([1,??]);
b = [b a];
end
采纳的回答
Azzi Abdelmalek
2013-8-27
n=60;
m=1:n;
a=m(sort(randperm(60,15)));
b=diff(a);
b(end+1)=60-sum(b)
sum(b)
13 个评论
harley
2013-8-27
Walter Roberson
2013-8-27
You must be using an older MATLAB that does not have that enhancement to randperm(). Try
t = randperm(n);
a=m(sort(t(1:15));
harley
2013-8-27
Arnab Pal
2018-11-4
I want to generate the same, but I need Non-integer numbers. Is it possible?
Bruno Luong
2018-11-4
编辑：Bruno Luong
2018-11-4
@Arnab
sumtarget = 60;
n = 5;
x = diff([0,sort(randperm(sumtarget+n-1,n-1)),sumtarget+n])-1
Bruno Luong
2018-11-4
Ah sorry some how I read non-negative.
Tejas
2020-7-13
Is there an easy way to control the maximum value each number in 'b' can have? Say, each value in 'b' must be less than or equal to 7, and still the 15 numbers in 'b' should add up to 60.
Walter Roberson
2020-7-13
Roger's FEX contribution, the a and b parameters are lower bound and upper bound. You can use zeros for the lower bound if that is appropriate for your situation.
Walter Roberson
2020-7-13
Bruno Luong
2020-7-13
编辑：Bruno Luong
2020-7-13
The easiest way is perhaps using Roger FEX function, then do some sort of "integering" the float solution
sumatarget = 60
n = 15;
ub = 7;
x = floor([0; cumsum(randfixedsum(n,1,sumatarget,0,ub))]);
x(end) = sumatarget; % prevent floating point error
r = diff(x)
The distribution might be not perfectly uniform but possibly close enough and suitable for what ever you want to do with it.
Tejas
2020-7-14
Integering the solution from Roger's function works very well for me. I do not require the numbers to be perfectly uniform. Thanks!
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