# can someone check the code i have problem

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Cedrik on 25 Jan 2023
Commented: Cedrik on 26 Jan 2023
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 .
##### 2 CommentsShowHide 1 older comment
Cedrik on 26 Jan 2023
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.

Dyuman Joshi on 26 Jan 2023
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 on 26 Jan 2023
Exactly! thank you very much!