# Adding values to cells as the indices increase

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Holmbrero el 22 de Dic. de 2020
Comentada: Holmbrero el 4 de En. de 2021
Hi!
I have a n x m cell array and i want to add a multiple of a constant as n and m increases.
(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3) ---- add constant to row nr 2.
(3,1) (3,2) (3,3) ---- add constant x2 to row nr 3.
I would also like to add a constant to columns as they increase.
Any suggestions?
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Holmbrero el 22 de Dic. de 2020
I'll try to elaborate.
(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)
I want to add a constant to row nr 2 ((2,1) (2,2) (2,3) ) and a multiple of the same constant to row nr 3 ((3,1) (3,2) (3,3) )
I also want to add another constant to column nr 1 ((1,2) (2,2) (2,3)) and a multiple of the same constant to columnt 3 ((1,3) (2,3) (3,3)).
I could just add them manually but i need the solution to work on a arbitrary nxm cell array.
Did that make my question easier to interpret?
Regards
Aaron Fredrick el 23 de Dic. de 2020
Yes I understand your question but, just to blear could you tell as the matrix increases each time,
a constant is added to 2nd row and a multiple of that to the 3rd row, and another constant to 1st column and a multiple of that to the 3rd column
OR
a constant is added to 2nd row and a multiple of that to the last row, and another constant to 1st column and a multiple of that to the last column
And, is the the multiplication of the constant is same for each time?

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Gaurav Garg el 28 de Dic. de 2020
Hi Holmbrero,
I understand that you wish to add multiples of constants to each row. I am providing with a pseudo-code to help you do that -
for i=1:size(c,1)
for j=1:size(c,2)
c{i,j} = c{i,j} + constant * i
end
end
Assuming that 'c' is your cell array, the above code would add each cell element with a constant, depending on the row number.
You can similarly replicate the code to add each element with constant, depending on column number, like this -
for i=1:size(c,1)
for j=1:size(c,2)
c{i,j} = c{i,j} + constant * j
end
end
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Holmbrero el 4 de En. de 2021
Hi!
I solved it by creating two new matrices with the added constants and then simply added these to the cell array.
The problem was somehow sort of hard to describe but i think your solution is also appliciable.

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