Sum elements of corresponding equal elements

for i=1:length(C)-1 
    for j=i+1:length(C)
        if C(i,1)==C(j,1)
              C(i,3)=C(i,2)+C(j,2);
        end
    end
end

C is a 36x2 matrix and I want to do the following: I want to check the 1st column for equal values. Let's say we find the (5,1), the (18,1) and the (21,1) elements in the column to be equal. Then I would like to sum the corresponding (5,2)+(18,2)+(21,2) elements in the second column and make each sum appear in the corresponding 3rd column, i.e. in the positiions (5,3), (18,3) and (21,3).

The above seems to almost work. I think it is because after it has searched and found an equality, it gives up on a 2nd potential equality and that's my problem!

Answer 1

[Akira Agata]

How about the following solution?

% Create sample 36-by-3 array C
rng('default');                             % for reproducability
C = [randi(10,36,1),rand(36,1),nan(36,1)];  % Sample array C

% Apply findgroups and splitapply functions to calcurate sum for each group
[group,ID] = findgroups(C(:,1));
val = splitapply(@sum,C(:,2),group);

% Arrange the result to the 3rd column of C
[~,loc] = ismember(C(:,1),ID);
C(:,3) = val(loc);

[tandemuse]

thank you for your reply.

This almost seemed to work however it doesn't for the matrix I've attached.

e.g. there are 7 instances of the value 0.6, and each is equal to 1 so we should get a sum of 7.

However we get a sum of 5 or 2.

[tandemuse]

I can't really tell why it doesn't work for this particular case.

I've noticed however that if the values are integers instead of decimals it does work.

I mean that instead of having 0.1 0.2 etc you have 1 2 etc.

Maybe there is some problem with findgroups and decimals?

[Walter Roberson]

Reminder that findgroups() uses exact comparisons .

>> [C(4),C(14),C(4)-C(14)]
ans =
                       0.3                       0.3     -5.55111512312578e-17

You will need to use something like

[ID, ~, group] = uniquetol(C(:,1));

[tandemuse]

YES! Thank you!

That is EXACTLY it. For future reference to others who might face the same problem, what is happening is that the command findgroups views the two "0.3" as distinct elements and doesn't put them in the same group.

The way I solved it is that I simply rounded the 1st column of C before the operation.

Here is the code:

C(:,1)=round(C(:,1),2);
groups = findgroups(C(:,1));
groups_sum = splitapply(@sum,C(:,2),groups);
pz=[unique(C(:,1)) groups_sum]