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

Using it as a part of Genetic Algorithm optimization. The code was working fine in its original form. But when tried to optimize by using my objective function, I got into this error

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

Error in Mutate (line 25)

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

Error in ga (line 144)

popm(k).Position=Mutate(p.Position,mu,VarMin,VarMax);

function y=Mutate(x,mu,VarMin,VarMax)

    nVar=numel(x);
    
    nmu=ceil(mu*nVar);
    
    j=randsample(nVar,nmu);
    
    sigma=0.1*(VarMax-VarMin);
    
    y=x;
    y(j)=x(j)+sigma*randn(size(j));
    
    y=max(y,VarMin);
    y=min(y,VarMax);

end

[Walter Roberson]

use

dbstop if error

I suspect that varmin or varmax are empty

Answer 1

[KSSV]

This line:

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

will create error..because, you are trying to save more number of elements then it is initialized. Try:

 y(:,:,j)=x(j)+sigma*randn(size(j));

or

 y{j}=x(j)+sigma*randn(size(j));

You have to check the dimensions/ size of the RHS and intialize the LHS to store values.

[Muhammad Irfan]

When using y(:,:,j)=x(j)+sigma*randn(size(j));

[Muhammad Irfan]

When using y{j}=x(j)+sigma*randn(size(j));

[KSSV]

You ahve to initialize y, and then use.

As you have y = x. y is of same size as x and it is array. When you try the baove it will throw errors. Show us the input dimensions, we can help you further.

[Muhammad Irfan]

%% Problem Definition

CostFunction=@(x) obj_fun(x);     % Cost Function

nVar=81;             % Number of Decision Variables

VarSize=[1 nVar];   % Decision Variables Matrix Size

for i=1:81
VarMin(i)= 0.05;         % Lower Bound of Variables
VarMax(i)= 1.1;% Upper Bound of Variables
end

%% GA Parameters

MaxIt=100;     % Maximum Number of Iterations

nPop=100;       % Population Size

pc=0.7;                 % Crossover Percentage
nc=2*round(pc*nPop/2);  % Number of Offsprings (also Parnets)
gamma=0.4;              % Extra Range Factor for Crossover

pm=0.3;                 % Mutation Percentage
nm=round(pm*nPop);      % Number of Mutants
mu=0.1;         % Mutation Rate

[Muhammad Irfan]

function y=Mutate(x,mu,VarMin,VarMax)

    nVar=numel(x);
    
    nmu=ceil(mu*nVar);
    
    j=randsample(nVar,nmu);
    
    sigma=0.1*(VarMax-VarMin);
    y=zeros(size(x));
    y=x;
   y(j)=x(j)+sigma.*randn(size(j));
    
    y=max(y,VarMin);
    y=min(y,VarMax);

end

Got this error after this:

Matrix dimensions must agree.

Error in Mutate (line 25)

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

Error in ga (line 148)

popm(k).Position=Mutate(p.Position,mu,VarMin,VarMax);

[Walter Roberson]

You have not shown your call to ga. As outside observers, we do not know that the VarMin and VarMax that you show the initialization for, are the same ones that are passed to Mutate .

Again I recommend dbstop if error and examine the variables when the code stops.

Answer 2

[Walter Roberson]

Okay, so your problem is that your x is not the orientation you expect.

When you index a vector of values at a vector of indices, then the shape of the result depends upon the shape of the original vector, not the shape of the index.

>> A = [3 7 9]
A =
     3     7     9

>> A([1 3])
ans =
     3     9

>> A([1 3].')
ans =
     3     9

>> B = A.'
B =
     3
     7
     9

>> B([1 3])
ans =
     3
     9

>> B([1 3].')
ans =
     3
     9

Meanwhile, you forced a specific orientation onto your j and so your randn(size(j)) has a specific orientation.

Suppose for example that your x is a 9 x 1 column vector and your randn(size(j)) is a 1 x 9 row vector then when you index x(j) you get back a column vector, and since R2016b, column vector + row vector gives you back a full-sized vector instead of giving you an error message.

[Muhammad Irfan]

I checked it.

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

y=1x81 double

j=9x1 double

x=1x81 double

sigma=1x81 double

randn(size(j))=9x1 double

So, How to get rid of error??/

[Walter Roberson]

    j = randsample(nVar,nmu) .';

Then j would be 1 x 9 and size(j) would be [1, 9] so randn(size(j)) would be 1 x 9

However, now that I see that sigma is 1 x 81, you need to modify

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

to

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

[Muhammad Irfan]

After updation of following statement,

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

I got error, shown below:

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

Error in Mutate (line 28)

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

Error in ga (line 148)

popm(k).Position=Mutate(p.Position,mu,VarMin,VarMax);

Error in main (line 14)

ga;

[Muhammad Aqib]

Were you able to solve the problem?

[Walter Roberson]

j is 9 x 1. randn(size(j)) would be 9 x 1. sigma(j) would be 1 x 9 because sigma is a row vector. So you have a 1 x 9 .* a 9 x 1, which is going to give you a 9 x 9 result.

Reminder that when you index a vector at a vector, the shape depends on the shape of the vector being indexed, not upon the shape of the indexing array.