Indexing with parentheses '()' must appear as the last operation of a valid indexing expression

des=[0.1 0.5 0.5];
L = length(des);
x = zeros(1,L)
etaa= 0.04;
w=zeros(1,L);
% w2=zeros(1,1);
e=zeros(1,1);
for k = 1:1500
    x = [randi(50)/50, x(1:end-1)]
    x = transpose(x);
    y_fix = des * x;

    y_adapt = w*x;
  
   %%%%% error update%%%%
     e(k)=y_fix-y_adapt; 
     
   %%%%% weights update%%% 
for j= 1:L
     w(j)(k+1)=w(j)(k)+etaa * x(j)* e(k);      %MARKED
 
end
end

Answer 1

[Walter Roberson]

You declared

w=zeros(1,L);

Indexing that with a single subscript would lead you to a single double precision location.

What is your expectation, then, of what

w(j)(k)

might mean? The k'th location inside the single double precision number?

If you are creating two dimensional arrays, then you should initialize w as two dimensional rather than as a vector, and you would use two subscripts, such as w(j,k)

[Ali Mukhtar]

no .. not two dimesional array

for j= 1:L
     w(j)(k+1)=w(j)(k)+etaa * x(j)* e(k);      %MARKED
end

j is the location of elememts which go till "L" . K is the iteration on which code is running . in above term i just want to update elements of "w" array for next term (K+1)

[Walter Roberson]

If you need to keep a record of what all of the values were, whether for plotting or later calculations, then you need to use one of:

• a 2D array (most efficient, especially if pre-allocated w = zeros(L, 1500)
• a cell array (less efficient)
• a non-scalar struct (I am not sure how this would compare in efficiency to a cell array)
• a table (less efficient than any of the above)

If you do not need to keep a record, then just leave out the (k+1) and (k) indexing, like

     w(j) = w(j) + etaa * x(j)* e(k);      %MARKED

In some situations, you will find you need to know what the value for the previous generation was, such as if the code involved

     w(j,k+1) = w(j,k) + etaa * x(j) * e(k);
     x(j) = sqrt(w(j,k));

here, the w(j,k) in the x(j) line is the value of w(j) before it was was updated in the w(j,k+1) line. In such a case, where you need to know the old value to complete a sequence of iterations, then you cannot just get rid of the second index:

     w(j) = w(j) + etaa * x(j) * e(k);
     x(j) = sqrt(w(j));                   %NOT THE SAME

here the w(j) is the updated value, whereas in the original it was the value before being updated.

In order to deal with this kind of situation, you would not need to keep the full 2D array. Instead you can use something like

     oldw = w;
     for j = etc
         w(j) = oldw(j) + etaa * x(j) * e(k);
         x(j) = sqrt(oldw(j));
     end

[Ali Mukhtar]

do you know about horzcat Error???

[Walter Roberson]

Yes. horzcat() errors occur when you try to use [] to put together values that have different numbers of rows. For example,

    x = [rand(50)/50, 1:10]

would try to use [] to put together a 50 x 50 random array with a 1 x 10 array; this would be an error.

[Ali Mukhtar]

how could we solve it ... ???

x = [randi(50)/50, x(1:end-1)] its giving me error in this equation ... im not getting to solve it

[Walter Roberson]

x = zeros(1,L)

x starts out as a row vector.

x = [randi(50)/50, x(1:end-1)]

in MATLAB, when you use a single vector to index into a vector, then the result of the indexing does not depend upon the shape of the vector you indexed with, and instead depends upon the shape of the vector being indexed. So because x is a row vector, the result of x(1:end-1) is a row vector. [] between a scalar (which has one row) and a row vector (which has one row) works fine, giving you a row vector.

x = transpose(x);

You make the row vector into a column vector.

Next iteration, you encounter

x = [randi(50)/50, x(1:end-1)]

in MATLAB, when you use a single vector to index into a vector, then the result of the indexing does not depend upon the shape of the vector you indexed with, and instead depends upon the shape of the vector being indexed. So because x is now a column vector, the result of x(1:end-1) is a column vector. [] between a scalar (which has one row) and a column vector (which has multiple rows) fails.

You need to decide whether x is a row vector or a column vector, and make the code consistent for it. Do not keep changing its orientation.

[Ali Mukhtar]

thanks alot your tips are always life saving ...;-)