adding next values in an array

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Nikolas Spiliopoulos
编辑： Nikolas Spiliopoulos 2017-2-17
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!
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Alexandra Harkai
Alexandra Harkai 2017-2-17
编辑：Alexandra Harkai 2017-2-17
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
Nikolas Spiliopoulos 2017-2-17
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!

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Guillaume
Guillaume 2017-2-17
编辑：Guillaume 2017-2-17
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
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Nikolas Spiliopoulos
Nikolas Spiliopoulos 2017-2-17
编辑：Nikolas Spiliopoulos 2017-2-17
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!

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Alexandra Harkai
Alexandra Harkai 2017-2-17
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

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