- x is not defined?
- mu?
- j indexing without define it?
Unable to perform assignment because the left and right sides have a different number of elements
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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..
2 commentaires
KALYAN ACHARJYA
le 5 Fév 2021
Is this complete code?
Réponses (2)
KSSV
le 5 Fév 2021
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
Check your code and rethink on it.
0 commentaires
Walter Roberson
le 5 Fév 2021
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));
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