How to create a random row vector whose sum of elements and number of elements is at my desire?

For example I want a row vector(1,5) and its total sum should be 20. Is there any function which randomly returns a 1*5 vector with 20 as its sum of elements?

[Jos (10584)]

Do the elements need to be integers?

[Vivek]

yes jos

Answer 1

[Andrei Bobrov]

n = 5;
m = 20;

a = rand(n,1);
out = round(a/sum(a)*m);
out(1) = out(1) - sum(out) + m;

[Vivek]

This also working. Thank you all for multiple answers

[Roger Stafford]

I hope you won't mind my criticizing your solution, Andrei, but I have noticed that a small proportion of the time, this algorithm comes up with a -1 at out(1) in the case m = 20 and n = 5. Here is an example:

 m = 20;
 a = [0.03;0.23;0.22;0.56;0.65]; % <-- Suppose 'rand' gives this
 out = round(a/sum(a)*m);
 out(1) = out(1) - sum(out) + m;
 out' =  -1     3     3     7     8

For n = 6 it is even possible to come with -2 in that position:

 m = 20;
 a = [0.045;0.451;0.351;0.251;0.751;0.151]; % <-- or this
 out = round(a/sum(a)*m);
 out(1) = out(1) - sum(out) + m;
 out' =  -2     5     4     3     8     2

It appears to be caused by unlucky roundings in which the sum of fractional numbers is just m at one point but after rounding their sum is greater than m with the first element having rounded down to zero. Your adjustment

 out(1)=out(1)-sum(out)+m

will then produce a negative value at out(1).

[Andrei Bobrov]

Hi Roger! I fully agree with you, for the condition when all positive numbers in vector out , it is need to replace round at floor .

And your solution very nice.

Answer 2

[Roger Stafford]

If only positive integers are allowed (>0) then do:

 m = 20; % Choose the desired sum
 n = 5; % Choose the number of integers
 v = diff([0,sort(randperm(m-1,n-1)),m]);

If zeros are permitted, then do:

 m = 20; % Choose the desired sum
 n = 5; % Choose the number of integers
 v = diff([0,sort(randperm(m+n-1,n-1)),m+n])-ones(1,n);;

Note: These algorithms randomly select among all possible solutions with uniform probability.

Note 2: If you have the older version of 'randperm' that has only one input argument, you can substitute

 p = randperm(n);
 p = p(1:k);

for

 p = randperm(n,k);

Note 3; There is no limitation such as n<=15 on the size of n in randperm(n). It is very different from perm.

[Andrei Bobrov]

+1

[Roger Stafford]

I neglected to point out, Vivek, that the total number of possible solutions to your problem is implicit in these two expressions. If zeros are not permitted, the total is (m-1)!/(n-1)!/(m-n)!, namely the number of possible combinations of n-1 things chosen out of m-1 things. With zeros permitted it is (m+n-1)!/(n-1)!/m!, the number of combinations of n-1 things out of m+n-1 things. The probabilities are equal for all possibilities simply because the combinations produced by doing the two-argument 'randperm' followed by a 'sort' should all be of equal probability (provided Mathworks' random number generator is working properly.)

[Vivek]

Thanks Roger

Answer 3

[Azzi Abdelmalek]

Edit

n=20;
m=21;
p=fix(m/n);
v=randi(p,1,n);
a=m-sum(v);
x=zeros(n,ceil(a/n));
x(1:a)=1;
v=v+sum(x',1)

[Vivek]

sorry I didn't mention. Numbers must be integers. Also if we use perm then it is limited to 15 elements. So any other idea?

[Azzi Abdelmalek]

v=randi(4,1,5)
a=20-sum(v)
x=zeros(5,ceil(a/5));
x(1:a)=1
v=v+sum(x',1)

[Vivek]

Ya Azzi. Working fine. Thank you.