Sum elements of corresponding equal elements
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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!
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Akira Agata
le 17 Oct 2020
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);
4 commentaires
Walter Roberson
le 18 Oct 2020
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));
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