can someone check the code i have problem

Hello!

I have a matrixR(6x30) contains one value by row,the sum of R colomns is given by A, each value of A has a corresponding element in a vector B :

A = [7;5;7;5;5;7]

B = [4;6;4;6;6;4]

in order to transform each value in B to a column vector ,i generated a matrix C where the number of columns is the max value of B as follow

 C =
 
 0     1     2     3     4     0
 0     1     2     3     4     5
 0     1     2     3     4     0
 0     1     2     3     4     5
 0     1     2     3     4     5
 0     1     2     3     4     0

After this I returned to matrix R and finding the index of the value corresonding to A, and start fill the the next elements until finish the elements of each vector of C,

this is my code for generating C

 k = max(B);
 C = zeros(size(B,1),k);
 for i = 1:size(B,1)
     ins = B(i);
     compteur = 0;
     for j =1:ins
         C(i,j) = compteur;
         compteur = compteur +1;
     end
 end

and to fill elements of R, starting to the first nonzero until the number of element existing in each corresponding value in C,

 m = 6;
 n = 30;
 L = 2000;
 Req = zeros(m,n);
 for i = 1:m
     for j =1:n
         if R(i,j) ~= 0
             start = find(R(i,:)~=0,1);  %find index of first value
             End = start+size(C,2); % number of columns to fill 
             end2 = size(C,2);
             for jj=1:jend2
                 gg = (L*C(1,jj))/B(i); % value in each column
                 Req(i,start) = gg;
                 start = start + 1;
             end
         end
     end
 end

the problem with this code is the number of element exceed the needed interval, but i want to fill just the number of element from the first value until the number of element in C without considering last zeros.

any suggestion is appreciated .

[Dyuman Joshi]

What exactly are you trying to do? Are you trying to obtain the elements of R from given conditions? It would be helpful in knowing what your desired output is.

The C matrix you have shown above is not same as the output from the code, I assume that is a typo?

B = [4;6;4;6;6;4];
k = max(B);
C = zeros(size(B,1),k);
for i = 1:size(B,1)
    ins = B(i);
    compteur = 0;
    for j =1:ins
        C(i,j) = compteur;
        compteur = compteur +1;
    end
end
C

C = 6×6

0 1 2 3 0 0 0 1 2 3 4 5 0 1 2 3 0 0 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 0 0

Also, it would be helpful if you could attach or provide the data for R.

You can also improve your code by vectorizing it at some points

B = [4;6;4;6;6;4];
k = max(B);
C = zeros(size(B,1),k);
for i = 1:size(B,1)
    ins = B(i);
    C(i,1:ins)=0:ins-1;
end
C

C = 6×6

0 1 2 3 0 0 0 1 2 3 4 5 0 1 2 3 0 0 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 0 0

[Cedrik]

@Dyuman Joshi hello!

R is a matrix of (6x30) , in each row i have just one value , I want to generate a new matrix Rreq with the same dimension of R. so i will find the index of each value in R, after this I will multiply this value by the element of vector C and divided by the value in B.

for example ,i will fill the first row of Rreq, the first value in R is "7" its index is 3. so i will start from the column 3 of Req and fill just the number of element of the first vector C [0 1 2 3 0 0], each time i multiply (7*0)/4 , (7*1)/4, (7*2)/4 ,(7,3)/4, here 4 is the first element of B. i want to fill just the 4 elements

0     0     0     0     0     0     0     1.75     3.5     5.25     0     0     0     0     0     0     0  0     0     0     0     0     0     0     0     0     0     0     0     0

Also i won't exceed the limits of my matrix, for e.g if the index of the value in R is at the last column , i will consider the value just for this column , i will not add other ones.

Answer 1

[Dyuman Joshi]

You can do that without defining C (If I understand what you want to do correctly)

B = [4;6;4;6;6;4];
m = 6;
n = 30;
L = 2000;
Req = zeros(m,n);
for idx=1:m
        %since you have only 1 non-zero per row
        first=find(R(idx,:))
        %use the value as an index
        last=min(n,first+B(idx)-1)
        %modify Req accordingly
        Req(idx,first:last)=R(idx,first)*(0:(last-first))/B(idx)
end

[Cedrik]

Exactly! thank you very much!