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

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Vivek
Vivek 2013-11-30
评论： Andrei Bobrov 2013-12-3
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?

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Andrei Bobrov
Andrei Bobrov 2013-11-30
n = 5;
m = 20;
a = rand(n,1);
out = round(a/sum(a)*m);
out(1) = out(1) - sum(out) + m;
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Roger Stafford
Roger Stafford 2013-12-2
编辑：Roger Stafford 2013-12-2
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
Andrei Bobrov 2013-12-3
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.

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Roger Stafford
Roger Stafford 2013-11-30
编辑：Roger Stafford 2013-11-30
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.
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Roger Stafford
Roger Stafford 2013-12-1
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.)

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Azzi Abdelmalek
Azzi Abdelmalek 2013-11-30
编辑：Azzi Abdelmalek 2013-11-30
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)
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