Saving vector result from each loop if solution is correct.
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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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Star Strider
2021-10-31
0 个投票
14 个评论
Deokjae Jeong
2021-10-31
Star Strider
2021-10-31
My pleasure!
.
Deokjae Jeong
2021-10-31
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
xxx
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.
.
Deokjae Jeong
2021-10-31
Star Strider
2021-10-31
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Deokjae Jeong
2021-10-31
Star Strider
2021-10-31
I am not certain what MATLAB version/release is being used, however in older releases, ‘i’ and ‘j’ when used alone were considered the imaginary operator 
, so here ‘i’ is not a real number. Either substitute it for another variable name or use ‘ii’. Then it will not be confused with the imaginary operator.
, so here ‘i’ is not a real number. Either substitute it for another variable name or use ‘ii’. Then it will not be confused with the imaginary operator. .
Deokjae Jeong
2021-10-31
编辑：Deokjae Jeong
2021-10-31
Star Strider
2021-10-31
I suspect that ‘k’ may be being used int the ‘DMP’ function, however is not being passed to it as an argument, and that throws the error. Since I have never seen the ‘DMP’ function (except the one I wrote to test the code that worked correctly) or the reference to it in the posted code (note that the ‘DMP’ call alone as presented in the posted code will run it, however will not pass anything to it or return anything from it), that is only a guess.
My test code sends ‘r’ to it as an argument and returns the necessary information from it in the function call to it. It shows the correct way to call ‘DMP’ with its arguments, and return values from it to the calling script, so they can be used approporiately.
.
Deokjae Jeong
2021-10-31
Star Strider
2021-10-31
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
0.0399 0.0820 0.0167 18.9749 0.0229 0.0101
0.0399 0.0820 0.0167 18.9749 0.0229 0.0101
0.0399 0.0820 0.0167 18.9749 0.0229 0.0101
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
Star Strider
2021-10-31
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
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