Borrar filtros
Borrar filtros

any logic to do this programming on random number

2 visualizaciones (últimos 30 días)
joy
joy el 11 de Jun. de 2014
Editada: John D'Errico el 11 de Jun. de 2014
Hi,
I have generated 1000 random numbers from a normal distribution with std deviation=0.10 and mean=0.36
like
r=0.36+0.10*randn(1,1000)
now, I want to select 10 numbers from this random numbers, the summation of which will be in between 3.55 to 3.65..
I need 30 sets of such 10 numbers which sum will lie between this..3.55 to 3.65
any logic to do so
thanks
  2 comentarios
Geoff Hayes
Geoff Hayes el 11 de Jun. de 2014
Editada: Geoff Hayes el 11 de Jun. de 2014
Note that the above equation is not quite correct (or your standard deviation and mean are reversed). If std is 0.1 and the mean is 0.36, then
r = 0.36+0.1*randn(1,1000);
Try std(r) and mean(r) to verify this.
joy
joy el 11 de Jun. de 2014
yes...u r right..thank u

Iniciar sesión para comentar.

Respuesta aceptada

Geoff Hayes
Geoff Hayes el 11 de Jun. de 2014
From the set of 1000 random numbers, you could do something like the following
% generate the random numbers
r = 0.36+0.1*randn(1,1000);
% pre-allocate memory for the 30 sets of 10 numbers
sets = zeros(10,30);
% generate each set
for i=1:30
% randomly choose 10 indices from list of random numbers
n = length(r);
idcs = randi(n,10,1);
% re-select the set of 10 indices if that set fails one of the three tests
while length(unique(idcs))~=10 || sum(r(idcs))<3.55 || sum(r(idcs))>3.66
idcs = randi(n,10,1);
end
% save the set of data corresponding to these indices
sets(:,i) = r(idcs);
% remove the 10 elements from r so that they are not picked again
r(idcs) = [];
end
Of course, the problem with the above is the while loop - if more sets are chosen, then it may become more difficult to find 10 such elements that satisfy the sum requirements and so the code may (eventually) become stuck in the loop.

Más respuestas (1)

Sean de Wolski
Sean de Wolski el 11 de Jun. de 2014
Editada: Sean de Wolski el 11 de Jun. de 2014
I would take a different approach. First generate the 30 numbers who you must sum to:
xsum = rand(1,30)+3.55;
Now generate the 10 random numbers that sum to xsum using Roger's randfixedsum
Example:
%%Desired sums
xsum = rand(1,30)+3.55;
%%build random numbers
rn = zeros(10,30);
for ii = 1:30
rn(:,ii) = randfixedsum(10,1,xsum(ii),0,xsum(ii));
end
%%verify
assert(norm(sum(rn)-xsum)<10^-14)
  3 comentarios
Sean de Wolski
Sean de Wolski el 11 de Jun. de 2014
Ahh! Good catch.
Sean de Wolski
Sean de Wolski el 11 de Jun. de 2014
I'll have to think about it a little more.

Iniciar sesión para comentar.

Categorías

Más información sobre Random Number Generation en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by