How to solve 'Error using vertcat, Dimensions of arrays being concatenated are not consistent.'? But they have the same dimensions.

Hi Everyone,

This error occurs whenever I have changed a value in my table to 0 or above 10. Whenever I have values between 1 and 9 the code works. What could be the reason for this and any suggestion on how can I make it work for 0?

Thanks,

[Sindar]

Can you include your code?

[Thi Ng]

Yes, sorry the code still looks messy.. My input would be a table 4x1 or a value, that looks like this in the work space [1;2;3;4].

All numbers from 1-9 work but not 0. I tried to change the if-statements but it doesn't seem to work.

Thanks for your help.

To correct my comment from above. It does work for values above 9 now. 0 is still causing a problem.

%Selections
t1 = M1_6_3; %for M_6
t2 = M1_3_3; % for M_3


if Selection(1,1) == 'M_6'
     % All Options for A1, A2, etc. 
for M1_6_A1 = [ 0 0 0 ];
    if t1(1,1) == 0
    elseif t1(1,1) == 1
        M1_6_A1 = [ 0 0 1 ];
    elseif t1(1,1) == 2
        M1_6_A1 = [ 0.1 0.5 0.8 ];
    elseif t1(1,1) == 3
        M1_6_A1 = [ 0 0.9 1 ];
    elseif t1(1,1) == 4
        M1_6_A1 = [ 0.2 0.8 0.8 ];
    elseif t1(1,1) == 5
        M1_6_A1 = [ 0.2 0.8 0.3 ];
    elseif t1(1,1) == 6
        M1_6_A1 = [ 0.7 0.9 0.2 ];
    elseif t1(1,1) == 7
        M1_6_A1 = [ 0.9 1 0 ];
    elseif t1(1,1) == 8
        M1_6_A1 = [ 0.9 0.5 0 ];
    elseif t1(1,1) == 9
        M1_6_A1 = [ 1 0 0 ];
    else
        M1_6_A1 = [ 1 1 1 ];
    end
 
 
end

[Thi Ng]

I have found a way around it to replace all 0 with 10.

t1{:,:}(t1{:,:}==0) = 10;

Then defined an additional elseif statement for 10, which works.

I am still happy to hear an advice on why 0 was causing that problem.

Thanks for your help!

[Russell Marki]

in your for loop you are defining M1_6_A1 to be 0 then 0 then 0, if M1_6_A1 is not defined later on it will remain a scalar.

[Thi Ng]

Thanks, do you have an example how that would look like?

I tried changing the code for a test run to:

for M1_6_A1 = [ 1 1 0 ];
    if t1(1,1) == 0
    end

It still gives me the same error. I think I didn't understand it correctly.

[Walter Roberson]

M1_6_A1 = [ 0 0 0 ];
    if t1(1,1) == 0
    elseif t1(1,1) == 1
        M1_6_A1 = [ 0 0 1 ];
    elseif t1(1,1) == 2
        M1_6_A1 = [ 0.1 0.5 0.8 ];
    elseif t1(1,1) == 3
        M1_6_A1 = [ 0 0.9 1 ];
    elseif t1(1,1) == 4
        M1_6_A1 = [ 0.2 0.8 0.8 ];
    elseif t1(1,1) == 5
        M1_6_A1 = [ 0.2 0.8 0.3 ];
    elseif t1(1,1) == 6
        M1_6_A1 = [ 0.7 0.9 0.2 ];
    elseif t1(1,1) == 7
        M1_6_A1 = [ 0.9 1 0 ];
    elseif t1(1,1) == 8
        M1_6_A1 = [ 0.9 0.5 0 ];
    elseif t1(1,1) == 9
        M1_6_A1 = [ 1 0 0 ];
    else
        M1_6_A1 = [ 1 1 1 ];
    end
 

No for loop needed.

You could rewrite as

   
    if t1(1,1) == 0
         M1_6_A1 = [ 0 0 0 ];
    elseif t1(1,1) == 1
        M1_6_A1 = [ 0 0 1 ];
    elseif t1(1,1) == 2
        M1_6_A1 = [ 0.1 0.5 0.8 ];
    elseif t1(1,1) == 3
        M1_6_A1 = [ 0 0.9 1 ];
    elseif t1(1,1) == 4
        M1_6_A1 = [ 0.2 0.8 0.8 ];
    elseif t1(1,1) == 5
        M1_6_A1 = [ 0.2 0.8 0.3 ];
    elseif t1(1,1) == 6
        M1_6_A1 = [ 0.7 0.9 0.2 ];
    elseif t1(1,1) == 7
        M1_6_A1 = [ 0.9 1 0 ];
    elseif t1(1,1) == 8
        M1_6_A1 = [ 0.9 0.5 0 ];
    elseif t1(1,1) == 9
        M1_6_A1 = [ 1 0 0 ];
    else
        M1_6_A1 = [ 1 1 1 ];
    end

[Thi Ng]

Thank you very much! Works perfectly.

[Thi Ng]

I would mark the answers as accepted, but they are all in the comments. Many thanks!

Answer 1

[Walter Roberson]

M1_6_A1 = [ 0 0 0 ];
    if t1(1,1) == 0
    elseif t1(1,1) == 1
        M1_6_A1 = [ 0 0 1 ];
    elseif t1(1,1) == 2
        M1_6_A1 = [ 0.1 0.5 0.8 ];
    elseif t1(1,1) == 3
        M1_6_A1 = [ 0 0.9 1 ];
    elseif t1(1,1) == 4
        M1_6_A1 = [ 0.2 0.8 0.8 ];
    elseif t1(1,1) == 5
        M1_6_A1 = [ 0.2 0.8 0.3 ];
    elseif t1(1,1) == 6
        M1_6_A1 = [ 0.7 0.9 0.2 ];
    elseif t1(1,1) == 7
        M1_6_A1 = [ 0.9 1 0 ];
    elseif t1(1,1) == 8
        M1_6_A1 = [ 0.9 0.5 0 ];
    elseif t1(1,1) == 9
        M1_6_A1 = [ 1 0 0 ];
    else
        M1_6_A1 = [ 1 1 1 ];
    end
 

No for loop needed.

You could rewrite as

   
    if t1(1,1) == 0
         M1_6_A1 = [ 0 0 0 ];
    elseif t1(1,1) == 1
        M1_6_A1 = [ 0 0 1 ];
    elseif t1(1,1) == 2
        M1_6_A1 = [ 0.1 0.5 0.8 ];
    elseif t1(1,1) == 3
        M1_6_A1 = [ 0 0.9 1 ];
    elseif t1(1,1) == 4
        M1_6_A1 = [ 0.2 0.8 0.8 ];
    elseif t1(1,1) == 5
        M1_6_A1 = [ 0.2 0.8 0.3 ];
    elseif t1(1,1) == 6
        M1_6_A1 = [ 0.7 0.9 0.2 ];
    elseif t1(1,1) == 7
        M1_6_A1 = [ 0.9 1 0 ];
    elseif t1(1,1) == 8
        M1_6_A1 = [ 0.9 0.5 0 ];
    elseif t1(1,1) == 9
        M1_6_A1 = [ 1 0 0 ];
    else
        M1_6_A1 = [ 1 1 1 ];
    end