Output for while loop

Question

Mateusz Brzezinski 2020-1-29

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/502569-output-for-while-loop

评论： Mateusz Brzezinski 2020-1-31

Hello,

I wrote a script that gives roots based on the Newton Raphson iteration. However, it is looped for changing one factor (in this case A0). I would like to store all outputs (uA and uB) and correlate it with changing values of A0 so in text file I will have sth. like this:

A0=

0.01

uA=

0.06

uB=

0.83

A0=

0.02

uA=

0.09

uB=

0.82

and so on...

How I should do this?

This is my simple script:

diary Logs
format long
A0Start=0.000;
A0End=1.000;
A0Step=0.01;
for h=A0Start:A0Step:A0End
    A0=h;
    prec=1e-10;
    uA=0.2;
    uB=0.8;
    B0=0.550;
    KA=64.54860696;
    KB=7.717484273;
    KAB=96.50371794;
    x=[uA;uB];
    dev=1;
    indx=0;
    while dev>prec
         f=[(A0*x(1)^2)+(((A0/(2*KA))^(1/2))*x(1))+(KAB/2)*(((A0*B0)/(KA*KB))^(1/2))*x(1)*x(2)-A0; (B0*x(2)^2)+(((B0/(2*KB))^(1/2))*x(2))+(KAB/2)*(((A0*B0)/(KA*KB))^(1/2))*x(1)*x(2)-B0];
         J=[2*A0*x(1)+(A0/(2*KA))^(1/2)+(KAB*x(2)*((A0*B0)/(KA*KB))^(1/2))/2, (KAB*x(1)*((A0*B0)/(KA*KB))^(1/2))/2; (KAB*x(2)*((A0*B0)/(KA*KB))^(1/2))/2, 2*B0*x(2)+(B0/(2*KB))^(1/2)+(KAB*x(1)*((A0*B0)/(KA*KB))^(1/2))/2];
         delx=-inv(J)*f; 
         x=x+delx;
         dev=max(abs(f));
         indx=indx+1;
    end
    disp 'for A0';
    disp (h)
    %disp 'last additive';
    %disp(delx)
    %disp 'iterations';
    %disp (indx)
    disp 'uA and uB found as';
    disp (x)
end
diary off

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Geoff Hayes 2020-1-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/502569-output-for-while-loop#answer_412673

在 MATLAB Online 中打开

Mateuz - try storing the data in an array which you can then (for example) write to file after the loop has completed. For example,

format long
A0Start=0.000;
A0End=1.000;
A0Step=0.01;
myData = zeros(length(A0Start:A0Step:A0End),3);
k = 1;
for h=A0Start:A0Step:A0End
    A0=h;
    prec=1e-10;
    uA=0.2;
    uB=0.8;
    B0=0.550;
    KA=64.54860696;
    KB=7.717484273;
    KAB=96.50371794;
    x=[uA;uB];
    dev=1;
    indx=0;
    while dev>prec
         f=[(A0*x(1)^2)+(((A0/(2*KA))^(1/2))*x(1))+(KAB/2)*(((A0*B0)/(KA*KB))^(1/2))*x(1)*x(2)-A0; (B0*x(2)^2)+(((B0/(2*KB))^(1/2))*x(2))+(KAB/2)*(((A0*B0)/(KA*KB))^(1/2))*x(1)*x(2)-B0];
         J=[2*A0*x(1)+(A0/(2*KA))^(1/2)+(KAB*x(2)*((A0*B0)/(KA*KB))^(1/2))/2, (KAB*x(1)*((A0*B0)/(KA*KB))^(1/2))/2; (KAB*x(2)*((A0*B0)/(KA*KB))^(1/2))/2, 2*B0*x(2)+(B0/(2*KB))^(1/2)+(KAB*x(1)*((A0*B0)/(KA*KB))^(1/2))/2];
         delx=-inv(J)*f; 
         x=x+delx;
         dev=max(abs(f));
         indx=indx+1;
    end
    myData(k,1) = A0;  % or h as they are the same
    myData(k,2:3) = x; % assuming that x is a 1x2 array
    k = k + 1;
end