estimating parameters by lsqcurvfit

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mallela ankamma rao
评论： mallela ankamma rao 2022-4-29
Good afternoon sir
sir please tell me how to draw graph for infected data vs predicted for this lscurvefit code.
i wrote code like this
plot(time,infected ,time,Cv) but it shows error sir.
code is
data = [ 1 25
2 75
3 227
4 296
5 258
6 236
7 192
8 126
9 71
10 28
11 11
12 7];
time = data(1:12,1);
infected= data(1:12,2);
beta0 = 1;
lamdaa0= 0.019;
lamdas0= 0.0715;
etas0 = 0.03;
etaq0 = 0.04;
gammaa0 = 0.2;
gammaq0 = 0.13;
gammah0 = 0.07;
mua0 = 0.0001;
muh0 = 0.0002;
lb =[0,0,0,0,0,0,0,0,0,0]; ub = [1,1,1,1,1,1,1,1,1,1];
B0 = [beta0; lamdaa0; lamdas0; etas0; etaq0; gammaa0; gammaq0; gammah0; mua0; muh0 ];
options=optimset('MaxFunEvals', 10000, 'MaxIter', 10000, 'TolFun', 0.0001, 'TolX',0.0001,'Display','on');
[B,resnorm,RESIDUAL,exitflag,OUTPUT,LAMBDA,Jacobian] = lsqcurvefit(@diff1,B0,time,infected,lb,ub,options);
disp(B)
%plot(time,infected,time,C)
function C = diff1(B,time)
x0 = [1217378052,120,3,2,1,1,0,0];
[t,Cv] = ode45(@DifEq, time, x0);
function dC = DifEq(t, x)
N = 1390000000;
pi = 700 ;
zetaa = 0.1;
zetas = 0.2;
zetaq = 0.3;
zetah = 0.3;
omega = 0.2;
theta = 0.5;
mu = 0.0000425;
beta = B(1);
lamdaa= B(2);
lamdas = B(3);
etas = B(4);
etaq = B(5);
gammaa = B(6);
gammaq = B(7);
gammah = B(8);
mua = B(9);
muh = B(10);
xdot = zeros(7,1);
xdot(1) = pi -beta*(zetaa*x(3) +zetas*x(4) +zetaq*x(5)+zetah*x(6))*(x(1)/N) -mu*x(1);
xdot(2) = beta*(zetaa*x(3) +zetas*x(4) +zetaq*x(5)+zetah*x(6))*(x(1)/N) -(omega+mu)*x(2);
xdot(3) = theta*omega*x(2)-(lamdaa+gammaa+mua+mu)*x(3);
xdot(4) = (1-theta)*omega*E-(lamdas+etas+mu)*x(4);
xdot(5) = lamdaa*Ia+lamdas*Is-(etaq+gammaq+mu)*x(5);
xdot(6) = etas*Is+etaq*Q- (gammah+muh+mu)*x(6);
xdot(7) = gammaa*x(4) + gammaq*x(5) + gammah*x(6)
end
C = Cv(:,2);
plot(time,infected ,time,Cv)
end
Error is
Unrecognized function or variable 'infected'.
Error in lscurvfit6>diff1 (line 72)
plot(time,infected ,time,Cv)
Error in lsqcurvefit (line 225)
initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});
Error in lscurvfit6 (line 19)
[B,resnorm,RESIDUAL,exitflag,OUTPUT,LAMBDA,Jacobian] = lsqcurvefit(@diff1,B0,time,infected,lb,ub,options);
Caused by:
Failure in initial objective function evaluation. LSQCURVEFIT cannot continue.

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回答（2 个）

Torsten
Torsten 2022-4-29
编辑：Torsten 2022-4-29
Although this has been answered several times already, here we go again:
data = [ 1 25
2 75
3 227
4 296
5 258
6 236
7 192
8 126
9 71
10 28
11 11
12 7];
time = data(1:12,1);
infected= data(1:12,2);
beta0 = 1;
lamdaa0= 0.019;
lamdas0= 0.0715;
etas0 = 0.03;
etaq0 = 0.04;
gammaa0 = 0.2;
gammaq0 = 0.13;
gammah0 = 0.07;
mua0 = 0.0001;
muh0 = 0.0002;
lb =[0,0,0,0,0,0,0,0,0,0]; ub = [1,1,1,1,1,1,1,1,1,1];
B0 = [beta0; lamdaa0; lamdas0; etas0; etaq0; gammaa0; gammaq0; gammah0; mua0; muh0 ];
options=optimset('MaxFunEvals', 10000, 'MaxIter', 10000, 'TolFun', 0.0001, 'TolX',0.0001,'Display','on');
[B,resnorm,RESIDUAL,exitflag,OUTPUT,LAMBDA,Jacobian] = lsqcurvefit(@diff1,B0,time,infected,lb,ub,options);
C = diff1(B,time);
plot(time,[infected,C])
function C = diff1(B,time)
x0 = [1217378052,120,3,2,1,1,0,0];
[t,Cv] = ode45(@DifEq, time, x0);
function dC = DifEq(t, x)
N = 1390000000;
pi = 700 ;
zetaa = 0.1;
zetas = 0.2;
zetaq = 0.3;
zetah = 0.3;
omega = 0.2;
theta = 0.5;
mu = 0.0000425;
beta = B(1);
lamdaa= B(2);
lamdas = B(3);
etas = B(4);
etaq = B(5);
gammaa = B(6);
gammaq = B(7);
gammah = B(8);
mua = B(9);
muh = B(10);
xdot = zeros(7,1);
xdot(1) = pi -beta*(zetaa*x(3) +zetas*x(4) +zetaq*x(5)+zetah*x(6))*(x(1)/N) -mu*x(1);
xdot(2) = beta*(zetaa*x(3) +zetas*x(4) +zetaq*x(5)+zetah*x(6))*(x(1)/N) -(omega+mu)*x(2);
xdot(3) = theta*omega*x(2)-(lamdaa+gammaa+mua+mu)*x(3);
xdot(4) = (1-theta)*omega*E-(lamdas+etas+mu)*x(4);
xdot(5) = lamdaa*Ia+lamdas*Is-(etaq+gammaq+mu)*x(5);
xdot(6) = etas*Is+etaq*Q- (gammah+muh+mu)*x(6);
xdot(7) = gammaa*x(4) + gammaq*x(5) + gammah*x(6)
end
C = Cv(:,2);
end
Walter Roberson
Walter Roberson 2022-4-29
infected= data(1:12,2);
You do that assignment once, before any function calls.
You do call
[B,resnorm,RESIDUAL,exitflag,OUTPUT,LAMBDA,Jacobian] = lsqcurvefit(@diff1,B0,time,infected,lb,ub,options);
but there, infected is going to become the ydata for curve fitting purposes; that copy of infected is not going to be passed to any lower level under the name infected
plot(time,infected ,time,Cv)
You are inside function diff1 . infected is not a defined variable there.
Perhaps you were thinking of somehow passing it in to function diff1() so that they could be used for plotting purposes -- that could be done by parameterizing the @diff1 call.
You do some funny things with time vs t in the plot(), but that should be okay... unless, that is, the ode*() call stopped early. It would be better to plot t,Cv than time,Cv in case the full time span does not get covered.
  1 个评论
mallela ankamma rao
mallela ankamma rao 2022-4-29
thanks for your reply sir
please run the program and give one line code for graph because i am confused about graph sir
Thank you sir

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