Saving vector result from each loop if solution is correct.

Question

Deokjae Jeong 2021-10-31

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1575978-saving-vector-result-from-each-loop-if-solution-is-correct

评论： Star Strider 2021-10-31

Let's say DMP.m uses fsolve and output a solution vector, named xx. It can be `Equation solved', 'No solution found', or 'Equation solved but inaccuracy possible.'

I want to store the vector if it tells `Equation solved'.

So for example, my code is

for r=1:10
    DMP
    if exitflag==1
        xxx(r,:)=xx 
    end
end
Unrecognized function or variable 'r'.

But Matlab gives me the error message: Unrecognized function or variable 'r'.

Could you help me the coding?

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Answer 1

Star Strider 2021-10-31

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https://ww2.mathworks.cn/matlabcentral/answers/1575978-saving-vector-result-from-each-loop-if-solution-is-correct#answer_820868

Use the exitflag output from fsolve to determine whether a solution was found.

.

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Star Strider 2021-10-31

在 MATLAB Online 中打开

I am not certain what is throwing that error. I created my own version of ‘DMP’ to test it, and it appears to work correctly as originally written (with my ‘DMP’ code, at least) —

for r=1:10
    [xx,exitflag] = DMP(r);
    if exitflag==1
        xxx(r,:)=xx;
    end
end
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 0.6180
fv = 3.2817e-08
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 1.0000
fv = 1.0658e-13
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 1.3028
fv = 1.7319e-13
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 1.5616
fv = 2.2162e-08
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 1.7913
fv = 2.7710e-10
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 2.0000
fv = 7.5939e-13
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 2.1926
fv = 3.4635e-11
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 2.3723
fv = 5.3424e-08
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 2.5414
fv = 7.0818e-09
exitflag = 1
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
xx = 2.7016
fv = 6.7687e-08
exitflag = 1
xxx
xxx = 10×1
    0.6180
    1.0000
    1.3028
    1.5616
    1.7913
    2.0000
    2.1926
    2.3723
    2.5414
    2.7016
function [xx,exitflag] = DMP(r)
[xx,fv,exitflag] = fsolve(@(x)x.^2+x - r, rand)
end

I only see the posted code and nothing else, so what is not posted may be causing the problem.

.

Star Strider 2021-10-31

在 MATLAB Online 中打开

Note that ‘DPM’ is a script, not a function, so it will not return anything. I changed it to be a function here.

Try this version (it takes too long to run it here, so I ran it offline on my computer) —

for r=1:10
    [xx,exitflag] = DMP;
    if exitflag==1
        xxx(r,:)=xx;
    end
end
xxx
function F = equations(x,beta,w,p,z,b,d)
u=x(1);
v=x(2);
theta=x(3);
eta=x(4);
a=x(5);
lambda=x(6);
c=x(7);
F(1)=((abs(lambda)+b)*(1-u)*u)/((abs(lambda)+b)*(u-1)+100*a*u)-v;
F(2)=p-(p*c)/(beta*(100*a/(1+theta)))*(1-beta*(1-d)*(1-abs(lambda)))-w;
F(3)=abs(eta)*p+(1-abs(eta))*z+abs(eta)*theta*p*c*(1-d)-w;
F(4)=v/u-theta;
F(5)=-.3697544+.0341093/abs(eta)+7.533068/c-theta;
F(6)=-.0410965+3.417642*abs(d)-abs(lambda);
F(7)=66.2881*abs(eta)-1061.607*lambda+693.8852*a+3.338711*p-4.223466*w-c;
F(8)=-.0162202/c/eta+.0009784*c+.0016295/eta+1.396799*lambda-.0047*p+.0056439*w-a;
%F(9)=.0523244+2.903846*v-.6319993/c-.0003297/eta/eta+.0000121/eta/eta/eta-u
F(10)=(beta*100*a*(p-eta*p+eta*z-z))/(beta*100*a*eta*theta*p*(1-d)+p*(1+theta)*(1-beta*(1-d)*(1-lambda)))-c;
end
function [xx,exitflag] = DMP
beta=0.998418435031256; %monthly (yearly=0.981185442086183)
w=21.42592; %Period 4
p=28.1777866666667; %Period 4
z=0.0113878916464317;  %Period 4
b=0.0119785793333333; %Period 3
d=0.0169109966666667; %Period 4
optimoptions('fsolve');
options = optimoptions('fsolve');
%options.Algorithm = 'trust-region'
options.Algorithm = 'Levenberg-Marquardt';
options.MaxFunctionEvaluations = 1000000;
options.MaxIterations = 100000;
options.StepTolerance = 1e-20;
options.FunctionTolerance = 1e-5;
options.OptimalityTolerance: 1e-5;
st = 0.01;
ed = 0.04;
u_ini = (ed-st).*rand(1,1) + st;
st = 0.005;
ed = 0.017;
v_ini = (ed-st).*rand(1,1) + st;
theta_ini=v_ini/u_ini;
st = 0.02;
ed = 0.06;
a_ini = (ed-st).*rand(1,1) + st;
st = 0.06;
ed = 0.20;
eta_ini = (ed-st).*rand(1,1) + st;
st = 0.01;
ed = 0.03;
lambda_ini = (ed-st).*rand(1,1) + st;
st = 12;
ed = 30;
c_ini = (ed-st).*rand(1,1) + st;
x0=[u_ini,v_ini,theta_ini,eta_ini,a_ini,lambda_ini,c_ini];
[x fval exitflag]=fsolve(@equations,x0,options,beta,w,p,z,b,d);
u=x(1);
v=x(2);
theta=x(3);
eta=x(4);
a=x(5);
lambda=x(6);
c=x(7);
xx=[a,eta,lambda,c,u,v];
end

It ran without error, and produced (in one run) —

xxx =

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

0 0 0 0 0 0

0.0413 0.1399 0.0165 24.0175 0.0418 0.0078

0.0399 0.0820 0.0167 18.9749 0.0229 0.0101

The zeros indicate a run in which the solution was not found. The random values are the reason that it will not always converge on a solution. This varies between runs.

.

Deokjae Jeong 2021-10-31