" Array dimensions must match for binary array op."

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IM
IM el 5 de Jun. de 2020
Comentada: Walter Roberson el 6 de Jun. de 2020
Dear experts,
I need your help;)
I got an Error :" Array dimensions must match for binary array op."
I have 1x14 cell ThAll : { 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double 4-D double}.
Each "4-D double" is a different size matrix . I need to find mean value matrix across all of these 4-D double matrices.
At first , I wanted to make all matrices in one ThAll cell array in the same size with NaN, then add all of them and find one mean value matrix across all matrices. I do it in next way:
for k = 1:length(ThAll) % Im not sure if Im making the same size for all matrices right; I think in this part is error
m = size(ThAll{k},1);
ThAll{k}(m+1, : , :, :) = NaN;
end
sum = ThAll{1}; % sum of all matrices
for i = 2:length(ThAll)
sum = sum + ThAll{i};
end
meanThAll = sum ./ length(ThAll); % one mean matrix across all matrices in cell array
Can someone help with this Error :" Array dimensions must match for binary array op" in my case ?
A lot thanks in advance!
  2 comentarios
Walter Roberson
Walter Roberson el 5 de Jun. de 2020
Which MATLAB release are you using? In particular, are you using R2016b or later?
IM
IM el 5 de Jun. de 2020
Im using R2020a

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Respuestas (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov el 5 de Jun. de 2020
Here is one of the easy solutions:
for jj=1:numel(ThAll)
TH_all=ThAll{jj};
for kk=1:4
TH_ALL(jj, kk)=mean2(TH_all(:,:,kk));
end
end
  4 comentarios
IM
IM el 6 de Jun. de 2020
Hey Walter! Yes, the error "Brace indexing is not supported for variables of this type" occurring on TH_all=ThAll{jj};
After :
for jj=1:numel(ThAll)
TH_all=ThAll{jj};
for kk=1:4
TH_ALL(jj, kk)=mean2(TH_all(:,:,kk));
end
end
ThAll is no more 1x14 cell array . ThAll became simply "4-D double".
Walter Roberson
Walter Roberson el 6 de Jun. de 2020
That does not make sense unless it happened before the code you posted.

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