How to sum previous columns before specific element ?

Hello! I have a row vector E(1,50), and I have a binary matrix A (20,50). In each row of A there is only "1". I want to add values in previous columns of E before each "1" existing in A. As example E(1,6) and A(3,6)

E = [2 3 5 4 1 8 ] 
A = [0 1 0 0 0 0 ;...
     0 0 0 1 0 0 ;...
     0 0 0 0 0 1] 

I have a value k in each time I will verify if it is upper to the sum of previous element of E. For this example A(1,2) =1, so the condition is "if k- (2+3)>0, then the two columns in E."

If it is the case I will fill matrix B same dimension of A.

Now for element B(1,2)=1.

For A(2,4)=1, I will take the sum of 4 previous columns in E (4+5+3+2).

etc...

How can I do this? Thanks in advance.

[the cyclist]

What you are trying to do is not clear to me.

Can I suggest that you make an example of input and desired output? (Maybe you could make a smaller example, such as E = 1x5 vector, and A is a 3x5 binary matrix?)

Don't describe what you want. Show us an example of what you want.

[Maria]

@the cyclist okay, i gave more details above.

[Torsten]

So the aim is to create the matrix B that has the same rows as A if the condition holds and will have a zero row if not ?

And what about the value for k ? Is it constant throughout the process or does it change with the row in question ?

[Maria]

@Torsten yes "k" is constant and each time i will compare it with the sum of previous columns of E.

Thank you for your reply.

Answer 1

[Image Analyst]

Try this:

E = [2 3 5 4 1 8 ];
A = [0 1 0 0 0 0 ;...
    0 0 0 1 0 0 ;...
    0 0 0 0 0 1];
[rows, columns] = size(A);
% Initialize B
B = zeros(size(A));
for row = 1 : rows
    % Find the first column where A is 1 for this row.
    first1column = find(A(row, :), 1, 'first');
    % Sum the values of E from column 1 up until this first1column and
    % assign to B in that row and column
    B(row, first1column) = sum(E(1:first1column));
end
% Display B in command window
B

B = 3×6

0 5 0 0 0 0 0 0 0 14 0 0 0 0 0 0 0 23

[Maria]

@Image Analyst thank you, i will adopt this solution.