How can I fix this error - "Matrix dimensions must agree" "x{end,1}.AppName == aa1{selA,1}.Type"

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

[Guillaume]

Please use proper indenting in your code as it makes it much more readable.

Comments are missing from your code.

Answer 1

[Image Analyst]

Evidently selA is not a single scalar number so you're not selecting a single cell, but an array of cells.

[Guillaume]

Hum, no. If SelA wasn't scalar the error would be Expected one output from a curly brace or dot indexing expression, but there were xxx results.

aa1{selA,1}.Type would error even before matlab can carry out the comparison.

[Image Analyst]

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 && ............

Answer 2

[Guillaume]

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 you are comparing char vectors, the function to use is strcmp:

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)