Unable to perform assignment

Question

JRC 2024-4-18

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2108761-unable-to-perform-assignment

评论： JRC 2024-4-18

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

I have code that compiles for t=4. For t greater than 4 the following error message appears:

"Unable to perform assignment because the left and right sides have a different number of elements"

The code is:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

t=5; 
syms u10 u9 u8 u7 u6 u5 u4 u3 u2 u1; 
syms x10 x9 x8 x7 x6 x5 x4 x3 x2 x1; 
syms y10 y9 y8 y7 y6 y5 y4 y3 y2 y1;
a12 =0.8; 
a13=0.05;
b21 =0.6; 
u=[];x=[];y=[];
for j=1:1:t+1
    u = [u; sym(['u' num2str(j)])];
    x = [x; sym(['x' num2str(j)])];
    y = [y; sym(['y' num2str(j)])];
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
J(t+1)=0;
for N=t:-1:1
    sis1 = x(N)+a12*y(N)-a13*x(N)*y(N);
    sis2 = y(N)-a12*y(N)+b21*u(N)+a13*x(N)*y(N);
    Func =u(N)^2+3*y(N); 
    JJ(N) = subs(Func + J(N+1), {x(N+1),y(N+1)},{sis1,sis2});
    JJ1(N)=simplify(JJ(N));
    b1(N) = diff(JJ1(N), u(N));
    [p1(N)] = solve(b1(N), u(N));
    [p(N)] = simplify(p1(N));
    J1(N) = subs(JJ(N), {u(N)}, {p(N)});
    J(N) = simplify(J1(N));
end
Unable to perform assignment because value of type 'sym' is not convertible to 'double'.

Caused by:
    Error using mupadengine/feval2char
    Unable to convert expression containing symbolic variables into double array. Apply 'subs' function first to substitute values for variables.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Thanks.

2 个评论
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Dyuman Joshi 2024-4-18

在 MATLAB Online 中打开

Dynamically defining variables is not a good idea, as you have done here -

syms u10 u9 u8 u7 u6 u5 u4 u3 u2 u1; 
syms x10 x9 x8 x7 x6 x5 x4 x3 x2 x1; 
syms y10 y9 y8 y7 y6 y5 y4 y3 y2 y1;

and here -

u=[];x=[];y=[];
for j=1:1:t+1
    u = [u; sym(['u' num2str(j)])];
    x = [x; sym(['x' num2str(j)])];
    y = [y; sym(['y' num2str(j)])];
end

A better method is to define a symbolic array like this, so that you can define all of them in a single command (as I have done below) -

syms u [1 10]

u

u =

And use indexing to access the individual variables -

u(1)

ans =

%Define them collectively in a single command

syms u x y [1 10]

As for the error occuring in the 2nd for loop, it is not clear to me what the objective is or what you want to do, thus I can not suggest anything other than correcting the size of preallocated variable.

JRC 2024-4-18

Thank you for the symbolic command that simplifies the expression, but did not solve the problem. The problem is in looping since I cannot compile for t equal to 5, for example.

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Answer 1

Walter Roberson 2024-4-18

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2108761-unable-to-perform-assignment#answer_1443976

移动：Walter Roberson 2024-4-18

在 MATLAB Online 中打开

t=5; 
syms u10 u9 u8 u7 u6 u5 u4 u3 u2 u1; 
syms x10 x9 x8 x7 x6 x5 x4 x3 x2 x1; 
syms y10 y9 y8 y7 y6 y5 y4 y3 y2 y1;
a12 =0.8; 
a13=0.05;
b21 =0.6; 
u=[];x=[];y=[];
for j=1:1:t+1
    u = [u; sym(['u' num2str(j)])];
    x = [x; sym(['x' num2str(j)])];
    y = [y; sym(['y' num2str(j)])];
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
J(t+1) = sym(0);
P1 = zeros(1,t,'sym');
for N=t:-1:1
    fprintf('N = %d\n', N);
    sis1 = x(N)+a12*y(N)-a13*x(N)*y(N);
    sis2 = y(N)-a12*y(N)+b21*u(N)+a13*x(N)*y(N);
    Func =u(N)^2+3*y(N); 
    JJ(N) = subs(Func + J(N+1), {x(N+1),y(N+1)},{sis1,sis2});
    JJ1(N)=simplify(JJ(N));
    b1(N) = diff(JJ1(N), u(N));
    sol = solve(b1(N), u(N));
    if isempty(sol)
        fprintf('warning: for N = %d there are no solutions\n', N);
        p1(N) = sym(NaN);
    else
        if numel(sol) > 1
            fprintf('warning: for N = %d, there are %d solutions\n', N, numel(sol));
        else
            fprintf('warning: for N = %d, there is the expected single solution\n', N);
        end
        p1(N) = sol(1);
    end
    [p(N)] = simplify(p1(N));
    J1(N) = subs(JJ(N), {u(N)}, {p(N)});
    J(N) = simplify(J1(N));
end
N = 5
warning: for N = 5, there is the expected single solution
N = 4
warning: for N = 4, there is the expected single solution
N = 3
warning: for N = 3, there is the expected single solution
N = 2
warning: for N = 2, there is the expected single solution
N = 1
warning: for N = 1, there are 5 solutions