any logic to do this programming on random number
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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 个评论
Geoff Hayes
2014-6-11
编辑：Geoff Hayes
2014-6-11
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
2014-6-11
采纳的回答
Geoff Hayes
2014-6-11
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.
更多回答（1 个）
Sean de Wolski
2014-6-11
编辑：Sean de Wolski
2014-6-11
I would take a different approach. First generate the 30 numbers who you must sum to:
xsum = rand(1,30)+3.55;
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 个评论
John D'Errico
2014-6-11
编辑：John D'Errico
2014-6-11
A flaw here is the numbers won't be at all normally distributed. Given the bound constraint on the mean, they won't be normally distributed anyway, but this gives up on any semblance of normality.
Sean de Wolski
2014-6-11
Ahh! Good catch.
Sean de Wolski
2014-6-11
I'll have to think about it a little more.
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