Help with column-wise matrix manipulation without for loop

I have the matrix (which is basically the matrix representation of several lines)

A = [10     5    10  % first two columns are the gradients in x-direction and the last column is the y-intercept
    0     4     3
    2     0     8
    0     6     5];

I want to find the intersection between the lines starting from bottom going up (for each column seperately) when the gradients are non-zero, e.g., the last row of first column has zero gradient so I ignore it and move up to third row, but the second row has zero gradient as well, so I calculate its intersection with the first row:

The first row elements are always non-zero and since there is no "previous row" before the first row, the intersection of the those lines by default are set to zero. I have achieved my target using for loop, but I was hoping to avoid for loops if possible (I will have more columns than just 3). Here is what I have done:

for k = 1:2
    
    B = A(A(1:end,k)~=0,[k,n]); % for each column, I am finding the matrix that does not include zero gradients
    [r,~] = size(B);
    c = 2:r;
    if isempty(c)
        x{k} = 0;
    else
        x{k} = [0; (B(c,end)-B(c-1,end))./(B(c-1,1)-B(c,1))];
    end
    
end

which gives me my desired results:

>> x{1}

ans =

         0
   -0.2500

>> x{2}

ans =

     0
    -7
    -1

Notice how the dimensions of x{1} and x{2} are different, and thus I had to use curly brackets. Avoiding the for loop can save me great deal of computational time. Thanks in advance! Let me know if you have any question.

Answer 1

[Shojiro SHIBAYAMA]

Honestly, I didn't get what you want do precisely, but the following code may help you. IMO, for loop in MATLAB doesn't prolong the computation time.

dAn_p = A(2:end,:)-A(1:(end-1),:); % 'next' and 'previous' matrix from A.
tmp = cell2mat(arrayfun(@(ix)-dAn_p(:,end)./dAn_p(:,ix), 1:3,'un',0)); % take the gradient
tmp(A(2:end,:)==0|A(1:(end-1),:)==0)=0 % set elements zeros where any corresponding elements in A are zeros.

[J AI]

Basically, the first two columns decide the computations to be done, however, they are seperate entitites and do not affect each other. Now this is how the computation works:

The basic idea: Intersection is only calculated between the current line and the preceding non-zero gradient line. The calculations are done from bottom to up.

The first column: the very last row has zero, so I move up and find non-zero in 3rd, however, 2nd has zero, so I move up again and find non-zero in the 1st. So I calculate the intersection between the two lines with gradients 2 and 10 and y-intercepts 8 and 10 respectively. However, since there are no more lines prior to the first row, I set its intersection to zero. Total of two intersecting points for column 1 and thus x{1} = [0; -0.25].

The second column: Starting from bottom, last row is non-zero but 3rd is zero, so the intersection is calculated between last and 2nd (non-zero gradient). Similarly, intersection between 2nd and 1st row (non-zero gradient) is calculated. By default, first row's intersection is set to zero. Total of three intersecting points for column 2 and thus x{2} = [0; -7; -1].

I hope I could make myself clear about the problem.

By the way, I didn't understand the code you provided. Could you kindly explain?

Thank you for your effort!