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I have a matrix lets call it AAA (n by 3 matrix) it looks a bit like the first one below called AAA - the important bit is coloumn 1 which contains numbers between say 1 and 20 such numbers can repeat.
1 100 102 4 55 58 11 339 341 3 55 56 2 50 53 11 123 127 14 55 59
I have a second matrix lets call it BBB (20 by 2 coloumn 2 is all zero's for the moment)where I have populated only the first coloumn with the numbers 1 through 20.
1 0 2 0 3 0 4 0 ... 20 0
I want to loop through matrix AAA and populate coloumn 2 in matrix BBB with the number in coloumn 2 from matrix AAA next to its appropirate number. At each update I will sum up coloumn 2 in matrix BBB (which i can obviously handle). So in the above example after the first loop matrix BBB will have the number 100 in row 1 coloumn 2. After the second loop it will have the number 55 in coloumn 2 of row 4 after the next loop it will have 339 in coloumn 2 of row 11. As mentioned I will do something after each loop or update but that bit I can handle.
So it will start with something like this
[rows cols] = size(AAA) for i = 1:rows "then not sure if i should be using something like a string compare or a map.objects or what the most appropriate way to continue is"
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If elements of column 1 of AAA are really positive integers, you can create BBB as a vector and index the element in BBB with elements from column 1 of AAA. Example:
[rows cols] = size(AAA) ; BBB = zeros(max(AAA(:,1)), 1) ; for k = 1 : rows id = AAA(k,1) ; BBB(id) = BBB(id) + AAA(k,2) ; end
After running this with the AAA that you provided, you get
BBB = 100 50 55 55 0 0 0 0 0 0 462 0 0 55
In this code, the line
BBB = zeros(max(AAA(:,1)), 1) ;
preallocates memory (and initializes it to 0's) for a number of elements given by the max value present in column 1 of AAA.
Then, the lines
id = AAA(k,1) ; BBB(id) = BBB(id) + AAA(k,2) ;
uses element (k,1) of AAA as an index in BBB. It adds AAA(k,2) to the value present at this index in BBB. In this solution, the fact that you use elements of the first column of AAA as indices in BBB makes it so there is no need to have these values stored in a first column of BBB. To illustrate, it is enough that row 11 of BBB is 462 to get this sum: you can get is as BBB(11); there is no need to have B(11,:)=[11, 462].
I can update this solution if you really need to store elements of column 1 of AAA in column 1 of a 2D BBB. You would need this if elements of column 1 of AAA could not be used as indices, but it would complicate a little the approach, because you would have to find relevant rows in BBB using a test and logical indexing or FIND.