How to solve : Subscripted assignment dimension mismatch error
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Hello,
I want to calculate mean of rows with multiple criteria. Below is the for loop which I am using, however I get error :Subscripted assignment dimension mismatch error. It may be due to when it looks for no. of rows to calculate mean, if it finds only one row, then the calculated mean is scalar? If so I want to keep the original row, instead of calculating the mean, Could you please help in this regard how to adjust the script?
Where X is a 5467-by-513 matrix , id is a 143-by-1 vector and wkno is a 44-by-1 vector
for ii=1:size(id,1);
for jj=1:size(wkno,1);
tst(ii,jj)= X(:,1)==id(ii,1) & X(:,2)==wkno(jj,1);
M(ii,jj)=mean(X(tst,:));
end
end
Many thanks
0 Comments
Accepted Answer
Thorsten
on 12 Oct 2015
Edited: Thorsten
on 12 Oct 2015
tst is a vector, so you cannot store it in a single scalar tst(ii,jj). Same for M(ii,jj). This may work:
M = nan(numel(id), numel(wkno), size(X,2));
for ii=1:size(id,1);
for jj=1:size(wkno,1);
tst = X(:,1)==id(ii,1) & X(:,2)==wkno(jj,1);
if sum(tst) == 1
M(ii,jj,:) = X(tst,:))
elseif sum(tst) > 1 % could be 0 if nothing is matching...
M(ii,jj,:)=mean(X(tst,:));
end
end
end
3 Comments
Thorsten
on 12 Oct 2015
It depends on your data what makes senses. You could sum across the 3rd dimension, for example:
imshow(sum(M, 3), []);
More Answers (1)
Walter Roberson
on 12 Oct 2015
It would be the other way around. When there is only one match then the mean calculated would be a scalar, and that is the case that would work. When there are multiple matches then the mean would be a vector with as many entries as the number of columns, and you cannot store a vector into M(ii,jj) as that is a single array location.
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