How to store answers of variables from for loop in structures

Question

GAGANDEEP KAUR 2020-11-25

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/661888-how-to-store-answers-of-variables-from-for-loop-in-structures

编辑： Walter Roberson 2025-3-28

I need to solve some equations for symbolic variables for multiple set of data so for loop is used, while storing the output I am facing an issue in creating suitable structures. Here are the code statements. If anyone can please help.

syms a b c d e

x1=[0.5096 0.5092 0.5087 ];

x2=[0.0963 0.0964 0.0965 ];

x3=[0.3941 0.3944 0.3948];

T=[394.15 399.15 404.15 ];

N=length(x1);

K1=zeros(1,N);K2=zeros(1,N);K3=zeros(1,N);K4=zeros(1,N);K5=zeros(1,N);

A=zeros(1,N);B=zeros(1,N);C=zeros(1,N);E=zeros(1,N);D=zeros(1,N);x1n=zeros(1,N);x2n=zeros(1,N);x3n=zeros(1,N);

X1=zeros(1,N);X2=zeros(1,N);X3=zeros(1,N);X4=zeros(1,N);X5=zeros(1,N);X6=zeros(1,N);X7=zeros(1,N);

X12=zeros(1,N);X16=zeros(1,N);

for i=1:N

Sout=struct();

K1(i)=exp((-8.549)+(6692/T(i)))

K2(i)=9.34*10^-7

K3(i)=exp((16.93565)+((1250)/T(i))+(-2.575*log(T(i))))

%Equilibrium constant for reaction 4(Polyiodide formation a)

K4(i)=exp((-936.28)+((40216.27)/T(i))+(151.983*(log(T(i))))+(-0.1675*(T(i))))

K5(i)=exp((1044.78)+(-45171.42/T(i))+(-165.20*log(T(i)))+(0.1511*(T(i))))

eqns=[(b*b)/((x3(i)-b)*(a-b))==K5(i),(a*(a-b))/((x1(i)-a)^4)*(x2(i)-a)==K4(i),(c/((x1(i)-c)^5)*(x2(i)-c))==K1(i),(e^2)/((x1(i)-e)^2)==K2(i),(d^2)/((x1(i)-d)*(x2(i)-d))==K3(i)]

S=vpasolve(eqns,[a,b,c,d,e]);

mask = S.a < 0 |S.b<0| S.c < 0 | S.d < 0 | S.e < 0 ;

S.a(mask)=[];

S.b(mask)=[];

S.c(mask)=[];

S.d(mask)=[];

S.e(mask)=[];

Sout(i).a=min(S.a)

Sout(i).b=min(S.b)

Sout(i).c=min(S.c)

Sout(i).d=min(S.d)

Sout(i).e=min(S.e)

A(i)=(Sout(i).a)

B(i)=(Sout(i).b)

C(i)=(Sout(i).c)

D(i)=(Sout(i).d)

E(i)=(Sout(i).e)

x1n(i)=x1(i)-A(i)-D(i)-E(i)-C(i);

x2n(i)=x2(i)-A(i)+B(i)-C(i)-D(i);

x3n(i)=x3(i)-B(i);

X1(i)=x1n(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X2(i)=x2n(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X3(i)=x3n(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X4(i)=A(i)+D(i)+E(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X5(i)=C(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X6(i)=A(i)-B(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X7(i)=B(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X12(i)=E(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

X16(i)=C(i)+D(i)/(x1n(i)+x2n(i)+x3n(i)+A(i)+B(i)+C(i)+D(i)+E(i));

end

K1 = 1×3

1.0e+03 * 4.5793 0 0

K2 = 1×3

1.0e-06 * 0.9340 0 0

K3 = 1×3

111.8264 0 0

K4 = 1×3

1.0e+03 * 3.2760 0 0

K5 = 1×3

10.7550 0 0

eqns =

Sout = struct with fields:

a: 0.45644511101436436229583126272584

Sout = struct with fields:

a: 0.45644511101436436229583126272584 b: 0.32634425642891616456663020396749 - 0.0085063316052863842253167082483287i

Sout = struct with fields:

a: 0.45644511101436436229583126272584 b: 0.32634425642891616456663020396749 - 0.0085063316052863842253167082483287i c: 0.40499614171701609798318306855559 - 0.064873041847274051977377108187709i

Sout = struct with fields:

a: 0.45644511101436436229583126272584 b: 0.32634425642891616456663020396749 - 0.0085063316052863842253167082483287i c: 0.40499614171701609798318306855559 - 0.064873041847274051977377108187709i d: 0.096100276263303125736310619771794

Sout = struct with fields:

a: 0.45644511101436436229583126272584 b: 0.32634425642891616456663020396749 - 0.0085063316052863842253167082483287i c: 0.40499614171701609798318306855559 - 0.064873041847274051977377108187709i d: 0.096100276263303125736310619771794 e: 0.00049202066312225458567174360052425

A = 1×3

0.4564 0 0

B =

0.3263 - 0.0085i 0.0000 + 0.0000i 0.0000 + 0.0000i

C =

0.4050 - 0.0649i 0.0000 + 0.0000i 0.0000 + 0.0000i

D = 1×3

0.0961 0 0

E = 1×3

1.0e-03 * 0.4920 0 0

K1 = 1×3

1.0e+03 * 4.5793 3.7020 0

K2 = 1×3

1.0e-06 * 0.9340 0.9340 0

K3 = 1×3

111.8264 104.0385 0

K4 = 1×3

1.0e+03 * 3.2760 2.6828 0

K5 = 1×3

10.7550 11.9893 0

eqns =

Sout = 1x2 struct array with fields:

a

Sout = 1x2 struct array with fields:

a b

Sout = 1x2 struct array with fields:

a b c

Sout = 1x2 struct array with fields:

a b c d

Sout = 1x2 struct array with fields:

a b c d e

A = 1×3

0.4564 0.4546 0

B =

0.3263 - 0.0085i 0.3299 - 0.0088i 0.0000 + 0.0000i

C =

0.4050 - 0.0649i 0.4003 - 0.0670i 0.0000 + 0.0000i

D = 1×3

0.0961 0.0962 0

E = 1×3

1.0e-03 * 0.4920 0.4916 0

K1 = 1×3

1.0e+03 * 4.5793 3.7020 3.0085

K2 = 1×3

1.0e-06 * 0.9340 0.9340 0.9340

K3 = 1×3

111.8264 104.0385 96.9274

K4 = 1×3

1.0e+03 * 3.2760 2.6828 2.2141

K5 = 1×3

10.7550 11.9893 13.2375

eqns =