serveral normrnd without a loop

Hey there :)

I would like to create a vector that is made of several normal distributions without using a loop.

Here is the code with the loop:

% my values so far
sig=0.43;
mu=[-3.9, -1.5, 0.3, 2.1]; 
length=[1, 5, 9, 5];

% loop
req_zeros=max(length);
x=nan(numel(mu),req_zeros);
for k=1:numel(mu)
   x(k,:)=[normrnd(mu(k),sig,[1,length(k)]), zeros(1,req_zeros-length(k))];
end

% the vector im looking for is:
x=x(x~=0).';

Can anyone tell me how to get to x without using a loop at all?

I'm very thankful for any help!

Answer 1

[per isakson]

What about this?

>> cell2mat( arrayfun( @(m,l) normrnd( m, sig, [1,l] ),  mu, len, 'uni',false ) )
ans =
  Columns 1 through 7
      -4.1537      -1.6196      -1.3182      -2.2182      -1.2972      -2.0215      0.32846
  Columns 8 through 14
      0.58051      0.44064      0.76553      0.73261      0.02011      0.41053     -0.10608
  Columns 15 through 20
     -0.26837       2.4977          2.1       2.0764       2.4918       2.3557

I renamed your length to len because length is a Matlab function. Sorry for the lower case L.

No, the numbers are in a different order.

The script below produces the same result as your script

rng('default')
mx  = max( len );
cac = arrayfun( @(m,l) [normrnd(m,sig,[1,l]),zeros(1,mx-l)],  mu, len, 'uni',false );
y = reshape( cell2mat(reshape(cac,[],1)), 1,[] );
y(y==0)=[]; % There is a vanishingly small chance that normrnd() returns the value zero.

[xena42]

Thank you so much!

I came up with this now, which should do pretty much the same, if i understand your code correctly:

(unsing len instead of length as you suggested)

mu=repmat(mu,max(len),1);
sig=sig*ones(size(mu));
logicCut=cumsum(repmat(len, max(len),1)) <= len.^2; % true for numbers I'n interested in
x=normrnd(mu,sig,size(mu));
x=x(logicCut);    % cuts away unwanted numbers