how to control the length of indexed array element?

m=[1 2 3 4];
m_d=0.5.*ones(1,length(m));
for i=1:length(m)
    m_xx=m(i)(1:(length(m(i))*m_d(1))); %error is here
end

I want to control an indexed array element length for my project and I really can't get the correct syntax for this

[Tommy]

Can you explain a bit more? What should m_xx equal after each iteration?

[Mohamed Salah]

m_xx should be equal to m(i) (from 1 to the last bit of m(i)) multiplied by factor m_d(1)

Answer 1

[per isakson]

"(from 1 to the last bit of m(i)) "

Here is my shot in the dark

%%
m   = uint8([1,2,3,4]);    
len = length(m);
m_d = 0.5*ones(1,len);
m_xx = cell(1,len);
for jj = 1 : len
    m_xx{jj} = bitget( m(jj), [1:8], 'uint8' );
end
%%
m_xx{1}
m_xx{4}

outputs

ans =
  1×8 uint8 row vector
   1   0   0   0   0   0   0   0
>> m_xx{4}
ans =
  1×8 uint8 row vector
   0   0   1   0   0   0   0   0

Comments

• bitget only takes whole numbers (I chose uint8 to keep the output short)
• cannot [not meaningsful to] multiply an integer with 0.5
• I stored the result in a cell array, but there are other alternatives

In reponse to comment

Now I understand your question, I think.

Mehmed Saad have listed problems with your puzzling statement.

Your statement

   m_xx = a(1:(length(a)*.5));

works because the length(a) is an even number.

Try

%%
a=[1 2 3 4];
b=[3 4 5 6];
c=[1 2 4 5]; 
m=[a b c];      % concatenates a, b and c into a row of length 12. That's less useful. 
%%
m   = { a, b, c };    % cell array. Comma is more readable than space  
len = length( m );
m_xx = cell( 1, len );      % pre-allocate cell array for holding the results
for jj = 1 : length( m )    % i is the imaginary unit
    m_xx{jj} = m{jj}( 1 : length(m{jj})*0.5 ); 
end

and

>> m_xx{:}
ans =
     1     2
ans =
     3     4
ans =
     1     2
     

Comment

The statement in the for-loop looks a lot like your statement. However, you missed

• that m and m_xx need to be cell arrays and
• the difference between () and {} when using cell arrays.

See Access Data in Cell Array

[Mohamed Salah]

I didn't mean that really

look suppose that

a=[1 2 3 4];
b=[3 4 5 6];
c=[1 2 4 5]; 
m=[a b c];

So I want to get m(1) half length -for example-

so I'd write

m_xx=m(1)(1:(length(m(1))*0.5) %and here comes the error 

an working example without the for loop I want to make is to take every element of m directly like this

m_xx=a(1:(length(a)*.5)); 

i want m_xx=[1 2] from the first half of a

[Tommy]

Ah. See if this gets you where you're going:

a=[1 2 3 4];
b=[3 4 5 6];
c=[1 2 4 5];
m={a;b;c}; % a, b, and c can be different lengths
for i=1:numel(m)
    m_xx=m{i}(1:end/2)
end

[per isakson]

I added to my answer

Answer 2

[Mehmed Saad]

1. you cannot feed 0 as array index it will give error
2. for loop runs from 0 to length of m which will make it run for 5 iterations instead of 4
3. m(i) will give you ith index. 1:length(m(i)) will be 1 as length(m(i)) will always be 1
4. in order to access m's index from 1 to i, you have to replace all that code with m(1:i). also start for loop from 1 and not from 0

[Mohamed Salah]

sorry that was a typo when I was writing the code here but that's not the error

the error appears in the line I commented

[Mohamed Salah]

()-indexing must appear last in an

index expression.

That's the error

[Mehmed Saad]

suppose i=1

see how this line behaves

m(i) (1:(length(m(i)) *m_d(1)))

• m(1) is 1
• length(m(1)) is 1
• m_d(1) is 0.5
• (length(m(1)) *m_d(1)) is 0.5
• (1:(length(m(1)) *m_d(1))) is 1:0.5 which will be empty
• There is no operator between m(i) and (1:(length(m(i)) *m_d(1))) resulting in error

Now try to correct it in command window.. When you get the Right answer,then replace 1 with i

[Mohamed Salah]

length of m in my code is 4000 so that's not the problem

in your last element of the the list you said no operator so the error appears so that's exactly my question I don't know the operator to make m_xx half the length of m(1)

[Mohamed Salah]

that's the comment I wrote on the answer below that maybe help to clarify things

look suppose that

a=[1 2 3 4];
b=[3 4 5 6];
c=[1 2 4 5]; 
m=[a b c];

So I want to get m(1) half length -for example-

so I'd write

m_xx=m(1)(1:(length(m(1))*0.5) %and here comes the error 

an working example without the for loop I want to make is to take every element of m directly like this

m_xx=a(1:(length(a)*.5)); 

i want m_xx=[1 2] from the first half of a