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

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Thi Ng 2020-4-22

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回答： Walter Roberson 2020-4-22

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,

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Thi Ng 2020-4-22

Thank you very much! Works perfectly.

Thi Ng 2020-4-22

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

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采纳的回答

Walter Roberson 2020-4-22

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https://ww2.mathworks.cn/matlabcentral/answers/519715-how-to-solve-error-using-vertcat-dimensions-of-arrays-being-concatenated-are-not-consistent-but#answer_427504

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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

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