Z must be a matrix, not a scalar or vector.

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Alexandra Roxana 2023-3-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1921345-z-must-be-a-matrix-not-a-scalar-or-vector

评论： Star Strider 2023-3-21

采纳的回答： Star Strider

在 MATLAB Online 中打开

I want to have a 3D plot with a solution vector but I have the error that Z must be a matrix.

a=0;
b=4;
c=0;
d=6;
g=1;
dxs=0.2;
dxf=0.25;
dy=0.5;
NNAB=(2*g)/dxs+(b-a-2*g)/dxf-1;
NNAD=(d-c)/dy+1;
NTN=NNAD*NNAB;
sol=rand(3*NTN-4*NNAB,1);
omega2=sol(2*NTN-2*NNAB+1:3*NTN-4*NNAB);
figure(2)
[X,Y]=meshgrid(a:b,c:d);
surf(X,Y,omega2)
Error using surf
Z must be a matrix, not a scalar or vector.

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Rik 2023-3-1

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What exactly is your question?

size(omega2)
ans = 1×2
   187     1

To use surf, the final variable should contain a height for each index of X and Y. This is clearly not the case. Why exactly did you pick these functions?

Cameron 2023-3-1

Look at your X,Y, and omega2 variables. The size of all of them should be the same when using the surf function.

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Star Strider 2023-3-1

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https://ww2.mathworks.cn/matlabcentral/answers/1921345-z-must-be-a-matrix-not-a-scalar-or-vector#answer_1183010

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It would be helpful to have the actual data rather than a ‘proxy problem’. The ‘Z’ vector should have the dame number of elements as the matrices. If so, it is then straightforward to interpolate them to form matrices, demonstrated here.

a=0;

b=4;

c=0;

d=6;

g=1;

dxs=0.2;

dxf=0.25;

dy=0.5;

NNAB=(2*g)/dxs+(b-a-2*g)/dxf-1;

NNAD=(d-c)/dy+1;

NTN=NNAD*NNAB;

sol=rand(3*NTN-4*NNAB,1);

omega2=sol(2*NTN-2*NNAB+1:3*NTN-4*NNAB);

omega2 = omega2(randperm(numel(omega2),35)); % Random Subset With The Same Number Of Elements As The Matrices

figure(2)

[X,Y]=meshgrid(a:b,c:d);

F = scatteredInterpolant(X(:),Y(:),omega2); % Create 'scatteredInterpolant' Object

omega2 = F(X,Y); % Interpolate

surf(X,Y,omega2)

colormap(turbo)

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Star Strider 2023-3-16

编辑：Star Strider 2023-3-16

在 MATLAB Online 中打开

You can use the entire vector if you are willing to make some compromises.

This is the only way I can think of to produce the result you apparently want using the entire vector —

a=0;

b=4;

c=0;

d=6;

g=1;

dxs=0.2;

dxf=0.25;

dy=0.5;

NNAB=(2*g)/dxs+(b-a-2*g)/dxf-1;

NNAD=(d-c)/dy+1;

NTN=NNAD*NNAB;

sol=rand(3*NTN-4*NNAB,1);

omega2=sol(2*NTN-2*NNAB+1:3*NTN-4*NNAB);

fctr = factor(numel(omega2)); % Prime Factors

a_b = linspace(a, b, fctr(1)); % Create Vector

c_d = linspace(c, d, fctr(2)); % Create Vector

[X,Y]=meshgrid(a_b,c_d); % Create Matrices

Omega2 = griddata(X(:), Y(:), omega2, X, Y); % Interpolate To Calculate Matrix

figure(2)

surf(X,Y,Omega2)

xlabel('a:b')

ylabel('c:d')

zlabel('omega2')

The online Run feature is currently down for scheduled maintenance (according to the pop-up notice that appears when I try to run this), so I ran it on MATLAB Online to be certain it worked. It does. I cannot determine if it produces the resullt you want.

EDIT — (16 Mar 2023 at 18:48)

Ran code.

Star Strider 2023-3-18

在 MATLAB Online 中打开

The only problem is if the number is prime, since prime numbers can only be integer-divided by themselves and 1. For the others, the factor function still works.

If the length of ‘omega2’ is a prime, it may be necessary to truncate it to a non-prime length and then use the appropriate factoring techniques to calculate the vectors. This approach assumes that a prime length is allowed, and sets the length of one vector to the prime number and the other vector to 1.

The immediate problem appears to be when there are more than two prime factors to the length. That simply requires some choices, for example —

a=0;
b=4;
c=0;
d=6;
omega2 = 1:randi(1E+4);
omega2Len = numel(omega2)
omega2Len = 3852
fctr = factor(numel(omega2));                   % Prime Factors
primeFactors = fctr
primeFactors = 1×5
     2     2     3     3   107
if numel(primeFactors) > 2                      % More Than Two Prime Factors
    max2 = maxk(primeFactors,2);                % Select Two Highest Factors
    Len_a_b = prod(max2);
    Len_c_d = numel(omega2)/Len_a_b;
elseif numel(primeFactors) == 1                 % Length Is Prime
    Len_a_b = primeFactors;
    Len_c_d = 1
else                                            % Only Two Prime Factors
    Len_a_b = primeFactors(1);
    Len_c_d = primeFactors(2);
end
VectrLens = [Len_a_b Len_c_d]
VectrLens = 1×2
   321    12
a_b = linspace(a, b, Len_a_b);                  % Create Vector
c_d = linspace(c, d, Len_c_d);                  % Create Vector
Check = numel(a_b) * numel(c_d)
Check = 3852

This is one option, however there are others. It all depends on how you want to deal with those situations. Experiment with this approach to get the result you want.