adding next values in an array

hi all, I have a question

I have a vector like this [1;2;3;-5;-6;2;3], when it's negative i would like to add in the next value and get something like that [1;2;3;-5;-11;-9;-6]. Which means when it's negative it starts adding the next value!

the thing is that i have an array with 48x258 dimensions, so the example above would be for each column!

thanks a lot!

[Alexandra Harkai]

What would you like to do when the last element of a column is negative?

[Nikolas Spiliopoulos]

it won't be :) But i need to do something else when the sum of the values becomes positive

[Alexandra Harkai]

In this case -5 is negative so the next element will be -6+(-5)=-11. Then -11 is negative so the next element will be 2+(-11)=-9. Which is negative so the next becomes 3+(-9)=-6 which is negative so would you want to do anything with it?

I could be totally wrong understanding your logic.

[Nikolas Spiliopoulos]

not really, i just need the logic for n elements,

in my real case it wont be my last element negative. This is just an example to understand the logic. thanks!

Answer 1

[Guillaume]

 M = [1 2 3 -5 -6 2 3]'  %demo data
 M + cumsum([zeros(1, size(M, 2)); M(1:end-1, :)] .* (cumsum(M<0, 1)>1), 1)

Will do what you want on any sized matrix.

Explanation of above code:

1. M<0 finds sign of M
2. cumsum(M<0, 1) will be 1 or more at the first negative value in each column
3. cumsum(M<0, 1)>1 will be one after the first negative value all the way down
4. [zeros(1, size(M, 2)); M(1:end-1, :)] is M shifted down by one row
5. 4.*5 filters the shifted M so that all positive values before the first negative values are 0 AND the first negative value is 0
6. cumsum of 6 + M is the desired result

[Nikolas Spiliopoulos]

thanks for the answer!! however, is there any way to do it with i and j (using for)? cos I would like to check if the sum value is negative or positive. If positive i need to stop the procedure until the next negative value! is it possible? thanks!

Answer 2

[Alexandra Harkai]

Considering the last element on a column will not be added to any other elements, you can loop through the whole thing:

function [a] = addNegatives(a)
  for col = 1:size(a,2)   % for each column
    for k = 1:size(a, 1)-1  % don't do it for the last element
      if a(k, col) < 0
        a(k+1, col) = a(k, col) + a(k+1, col);
      end
    end
  end
end