How can I fix this error - "Matrix dimensions must agree" "x{end,1}.AppName == aa1{selA,1}.Type"
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if Sign == 1
if appm==0
tt.AppName=aa1{selA,1}.Type;
tt.status='[On]';
tt.dataOn=i-7;
x{end+1,1}=tt;
clear tt;
appm=1;
end
SignStr = '[On]';
iCount = iCount + 1;
if ~isempty(r{selA,1})
r{selA,1} = r{selA,1} + 1;
else
r{selA,1} = 1;
end
else
if isempty(x)~=1 && x{end,1}.AppName == aa1{selA,1}.Type
% AppName=aa1{selA,1}.Type;
tt.AppName=aa1{selA,1}.Type;
tt.status='[Off]';
tt.dataOn=i-7;
x{end+1,1}=tt;
clear tt;
end
1 commentaire
Guillaume
le 24 Déc 2018
Please use proper indenting in your code as it makes it much more readable.
Comments are missing from your code.
Réponses (2)
Image Analyst
le 24 Déc 2018
Evidently selA is not a single scalar number so you're not selecting a single cell, but an array of cells.
2 commentaires
Image Analyst
le 24 Déc 2018
Well, just a guess since we can't debug it. But here is a solution guaranteed to solve it: Click here
It also helps to break things into simple, temporary variables:
v1 = x{end,1}.AppName
v2 = aa1{selA,1}.Type
theyMatch = v1 == v2
My current guess is that they are strings and he should be using strcmpi(v1, v2) instead of using ==.
theyMatch = strcmpi(v1, v2);
if theyMatch && ............
Guillaume
le 24 Déc 2018
Modifié(e) : Guillaume
le 24 Déc 2018
x{end,1}.AppName and aa1{selA,1}.Type are not the same size and neither of them is scalar in the non-scalar dimension of the other. As the error message tells you "Matrix dimensions must agree" since == does an element by element comparison.
if ~isempty(x) && strcmp(x{end,1}.AppName), aa1{selA,1}.Type) %return true if both char vectors are the same
Otherwise, it may be that isequal may work, or it may be that you've got a bug somewhere that makes the two matrices a different size.
Note that:
if isempty(x)~=1
is simpler as
if ~isempty(x)
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