Could anyone help me how to write the matrix with respect to diagnol in the for loop.

If i run the following code

N_UE=[2 4 6 8 10];
N_SC=[12 14 16 18 20];
for t= 1:length(N_UE)
for r= 1:length(N_UE)
G=rand(N_UE(t),N_SC(r))
C=[diag(1:N_UE(t)),diag(1:N_SC(r))]
end
end

it results in Error using horzcat Dimensions of matrices being concatenated are not consistent.

Error in (line 6) C=[diag(1:N_UE(t)),diag(1:N_SC(r))]

Could anyone help me to solve it.

[Birdman]

What is the purpose of line 6?

[Prabha Kumaresan]

I need to have diagnol representation of matrix C.I tried with the above one but it gives error.As G gives random selection of (N_UE(t),N_SC(r)),in the same way i want to represent C as a diagnol representation of (N_UE(t),N_SC(r)).Could you please help me on this.

[Stephen23]

"Could anyone help me to solve it."

You are trying to concatenate two matrices of different sizes, which is an error. There a multiple possible ways to "solve" that: padding, subsampling, interpolating, etc. So far you have told us nothing useful about what this code should do, nor how you imagine the output of two differently sized matrices should be concatenated together.

[Birdman]

diag command builds SQUARE MATRIX and you try to concantenate two different square matrices. What is the purpose?

[Prabha Kumaresan]

If m=5
n= 10
the command [diag(1:m),diag(m+1:n)] works 
But in my code the values of m and n keep
on changing with respect to 5 different 
sets of scenario.
m=[2 4 6 8 10];
n=[12 14 16 18 20];
for t= 1:length(m)
for r= 1:length(n)
[diag(1:m(t)),diag(m+1(t):n(r))]
end
end
Could you please help me how to write the 
diagonal matrix with respect to 5 different 
sets of scenario.

[KSSV]

Why don't you give an matrix example...you are expecting, instead of code with errors?

[Prabha Kumaresan]

C1=[1     0     1     0     1     0     1     0     1     0     1     0;
    0     2     0     2     0     2     0     2     0     2     0     2]

C2=[1     0     0     0     1     0     0     0     1     0     0     0     1     0;
    0     2     0     0     0     2     0     0     0     2     0     0     0     2;
    0     0     3     0     0     0     3     0     0     0     3     0     0     0;
    0     0     0     4     0     0     0     4     0     0     0     4     0     0]

Answer 1

[Rik]

I'll add my solution as a separate answer to avoid confusion.

m=[2 4 6 8 10];
n=[12 14 16 18 20];
iwant = cell(length(m),1) ;
for t= 1:length(m)
    n_=ceil(n(t)/m(t));%round up to nearest multiple
    C=repmat(diag(1:m(t)),1,n_);
    iwant{t}=C(:,1:n(t));%crop back to only needed cols
end

For me this just executes as expected. The fourth run yields an 8x18 matrix:

1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0
0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 2
0 0 3 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0
0 0 0 4 0 0 0 0 0 0 0 4 0 0 0 0 0 0
0 0 0 0 5 0 0 0 0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0
0 0 0 0 0 0 7 0 0 0 0 0 0 0 7 0 0 0
0 0 0 0 0 0 0 8 0 0 0 0 0 0 0 8 0 0

[jaah navi]

I have attached the image of the fourth run. If you see in the first image it is showing 24 columns instead of 18.In the second image it was showing [8x18 double].Could you please check on it.

[Rik]

You were the one removing the semi-colons. This code first makes a multiple of diag matrices and the trims it down to the needed number of cols. It even says so in the comments.

[jaah navi]

Yes.I have removed the semicolons to see the result of each and every command during every run in the command window.Could you please tell me which command line in the code performs the trim.

[Walter Roberson]

The line that says "%crop back to only needed cols" is the one that does the trim.

[jaah navi]

ok.Thanks.

[Rik]

As KSSV told you, thanks is accepting the answer. This is a hobby, I do this for free. Clicking accept on the answer that solved your problem best is the easiest way to show a bit of gratitude, as well as moving it to the top of the page, which makes it easier for other people to find the answer.

[jaah navi]

As in this page i couldn't find accepting the answer and so i replied thanks.

[Walter Roberson]

Only the person who asked the question can accept the answer (at least for the first week, after which someone with a high enough reputation can accept it.)

[Rik]

You should find the button to accept an answer right next to or below the profile picture of the person supplying the answer.

[Walter Roberson]

Rik, it was not jaah who asked the Question so jaah would not be able to Accept the answer.

Answer 2

[KSSV]

k = [1 2] ;

C = diag(k) ;

C1 = repmat(C,1,6) ;
k = [1 2 3 4] ;
C = diag(k) ;
C2 = repmat(C,1,3) ;

C21 = diag([1 2]) ;
C22 = zeros(2) ;
C212 = [C21 ; C22] ;
C2 = [C2 C212] ;

[jaah navi]

thanks for your response. But how can I use it in my code using for loop

m=[2 4 6 8 10];
n=[12 14 16 18 20];
for t= 1:length(m)
for r= 1:length(n)
------------
------------
end
end
for first run the matrix size should be (2,12)
for second run it should be (4,14)......
for last run it should be (10,20)

[KSSV]

N = 3 ;
m=[2 4 6 8 10];
n=[12 14 16 18 20];
iwant = cell(length(m),1) ;
for t= 1:length(m)
    k = [m(t) n(t)] ;
    C = diag(k) ;
    iwant{t} = repmat(C,1,N) ;
end

[Rik]

@KSSV I interpreted it slightly differently:

m=[2 4 6 8 10];
n=[12 14 16 18 20];
iwant = cell(length(m),1) ;
for t= 1:length(m)
    n_=ceil(n(t)/m(t));%round up to nearest multiple
    C=repmat(diag(1:m(t)),1,n_);
    iwant{t}=C(:,1:n(t));%crop back to only needed cols
end

[jaah navi]

Ok.but i want to have the result for (2,12) to be

[1   0   1   0   1   0   1   0   1   0   1   0
0   2   0   2   0   2   0   2   0   2   0   2]

(4,14) to be  [1   0   0   0   1   0   0   0   1   0   0   0   1   0; 0   2   0   0   0   2   0   0   0   2   0   0   0   2; 0   0   3   0   0   0   3   0   0   0   3   0   0   0; 0   0   0   4   0   0   0   4   0   0   0   4   0   0]

[jaah navi]

Thanks a lot.It worked.

[KSSV]

Thanks is accepting the answer.