random number generation for known sum

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

 採用された回答

Azzi Abdelmalek
Azzi Abdelmalek 2013 年 8 月 27 日

1 投票

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

13 件のコメント

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