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

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Azzi Abdelmalek
Azzi Abdelmalek 2013년 8월 27일

1 개 추천

n=60;
m=1:n;
a=m(sort(randperm(60,15)));
b=diff(a);
b(end+1)=60-sum(b)
sum(b)

댓글 수: 13

thanks Azzi, i get an error
Error in ==> Untitled at 3
a=m(sort(randperm(60,15)));
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
harley 2013년 8월 27일
thanks
Arnab Pal
Arnab Pal 2018년 11월 4일
I want to generate the same, but I need Non-integer numbers. Is it possible?
Bruno Luong
Bruno Luong 2018년 11월 4일
편집: Bruno Luong 2018년 11월 4일
@Arnab
sumtarget = 60;
n = 5;
x = diff([0,sort(randperm(sumtarget+n-1,n-1)),sumtarget+n])-1
Arnab Pal
Arnab Pal 2018년 11월 4일
편집: Arnab Pal 2018년 11월 4일
Sir, It is generating the integer numbers only.
Bruno Luong
Bruno Luong 2018년 11월 4일
Ah sorry some how I read non-negative.
For random non-negative floating numbers, you need using Roger Stafford FEX
Tejas
Tejas 2020년 7월 13일
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
Walter Roberson 2020년 7월 13일
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.
Tejas
Tejas 2020년 7월 13일
편집: Tejas 2020년 7월 13일
I want the numbers to be integers, as in the original question. Roger's FEX contribution seems to work with real numbers.
Bruno Luong
Bruno Luong 2020년 7월 13일
편집: Bruno Luong 2020년 7월 13일
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
Tejas 2020년 7월 14일
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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2013년 8월 27일

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2020년 7월 14일

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