Unable to perform assignment because the left and right sides have a different number of elements

when i try to run the follwing code, is shows the error as

Unable to perform assignment because the left and right sides have a different number of elements.

in the line "y(j)=x(j)+sigma.*randn(size(j));"

Help me out to rectify this..

Code:

Var=numel(x);
 nMu=ceil(mu*nVar);
 j=randsample(nVar,nMu);
    if numel(sigma)>1
        sigma = sigma(j);
    end
    y=x;
    y(j)=x(j)+sigma.*randn(size(j));

Thanks in advance..

[KALYAN ACHARJYA]

Is this complete code?

• x is not defined?
• mu?
• j indexing without define it?

[Walter Roberson]

I predict that sigma is empty.

Answer 1

[KSSV]

This error happens when you try to save more number of elements than initialized.

Example:

A = zeros(1,3) ; 
A(1) = rand(1,2)  % this will give error 

In the above you need to save only one element in A(1) but you are saving two, so error pops out.

A = zeros(1,3) ;
k = [];    % empty matrix
A(1) = []   % error

The above also throws error. I included this example keeping @Walter Roberson sir's comment.

Check your code and rethink on it.

Answer 2

[Walter Roberson]

Oh wait... is it possible that x is a row vector? If so then with vector j, no matter whether j is a row vector or column vector, x(j) would be a row vector because x is a row vector. Vector being indexed by vector gives orientation the same as the source vector, not the orientation of the indexing vector.

randsample() in that form returns a column vector, so j is a column vector.

x(j)+sigma.*randn(size(j))
^^^^ ^^^^^  ^^^^^^^^^^^^^^
row  scalar column

and row + column gives a 2D array . You would get an nMu x nMu output array, not a 1 x nMu output array.

y(j) = reshape(x(j),[],1) + reshape(sigma.*randn(size(j)), [], 1);

Or if you are sure that x is a row vector, then

y(j) = x(j).' + sigma.*randn(size(j));

[sastra university]

Thank you so much for the answer...