I am confused by the error: "Error evaluating inline model function"

Question

Jinglei 2022-12-31

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1886712-i-am-confused-by-the-error-error-evaluating-inline-model-function

评论： Jinglei 2023-1-19

myfunc=inline('fitting(c,x)','c','x');
c0=[1.5,1.5];
c=nlinfit(x,y,myfunc,c0)
function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[~,ye] = ode45(@eq2, xspan, y0);
ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));
end 
function dy=eq2(x,y)
global c
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=0.9728.*(2.5-(1 + 0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1 + 0.2181))./(1 + 0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);
dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);
end

When I run the program, the error came out as below.

Error using nlinfit (line 207)

Error evaluating inline model function.

Error in youhuachuli1228 (line 65)

c=nlinfit(x,y,myfunc,c0)

Caused by:

Error using inlineeval (line 14)

Error in inline expression ==> fitting(c,x)

Undefined function 'fitting' for input arguments of type 'double'.

I think the problem is that the function 'fitting(c,x)' is in type of syms instead of 'double'. But I don't know how to fix it. Does anyone have any ideas?

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

Walter Roberson 2022-12-31

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1886712-i-am-confused-by-the-error-error-evaluating-inline-model-function#answer_1138942

When you use inline() the expression is evaluated in the base workspace. Any function mentioned must be resolved from the base workspace. That effectively prohibits local functions and nested functions. 'fitting' would have to have its own fitting.m file instead of defined in the same file.

The function fitting is not symbolic at all.

You also failed to initialize the global variable c.

Do not use inline() unless someone imposes using it as a job requirement (for example a homework example showing how inline breaks and so should be avoided)

global is not recommended.

https://www.mathworks.com/help/matlab/math/parameterizing-functions.html

54 个评论
显示 52更早的评论隐藏 52更早的评论

Torsten 2022-12-31

编辑：Torsten 2022-12-31

在 MATLAB Online 中打开

Your fitting function will return NaN and the solver will quit.

x = [0

5

20

30

50

80

100

120

150

];

y = [0

0.17917

0.44143

0.68912

0.76926

0.78383

0.7984

1.06795

0.90768

];

myfunc=@fitting;%inline('fitting(c,x)','c','x');

c0=[1.5,1.5];

c=nlinfit(x,y,myfunc,c0)

c = 1×2

0.1720 0.8251

%c = lsqcurvefit(myfunc,c0,x,y)

ytotal = fitting(c,x);

hold on

plot(x,ytotal)

plot(x,y,'o')

hold off

function ytotal = fitting(c,x)

xspan = x;

y0 = zeros(2,1);

[~,ye] = ode45(@(x,y)eq2(x,y,c), xspan, y0);

ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));

end

function dy=eq2(x,y,c)

%y(1)=CE,y(2)=LE

dy=zeros(2,1);

dy(1)=0.9728.*(2.5-(1+0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);

dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);

end

Torsten 2023-1-1

编辑：Torsten 2023-1-2

在 MATLAB Online 中打开

Since the code above works, I will no longer comment on this.

Use the code below and you won't get any syntax errors.

x = [0;5;20;30;50;80;100;120;150];

xs = 0:0.1:150;

y = [0;0.17917;0.44143;0.68912;0.76926;0.78383;0.7984;1.06795;0.90768];

myfunc=@fitting;

c0=[1.5,1.5];

ytotal0 = fitting(c0,xs);

c=nlinfit(x,y,myfunc,c0)

c = 1×2

0.1720 0.8251

xs = 0:0.1:150;

ytotal = fitting(c,xs);

%c = lsqcurvefit(myfunc,c0,x,y)

hold on

plot(x,y,'o')

plot(xs,ytotal0)

plot(xs,ytotal)

hold off

function ytotal = fitting(c,x)

xspan = x;

y0 = zeros(2,1);

[~,ye] = ode45(@(x,y)eq2(x,y,c), xspan, y0);

ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));

end

function dy=eq2(x,y,c)

%y(1)=CE,y(2)=LE

dy=zeros(2,1);

dy(1)=0.9728.*(2.5-(1+0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);

dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);

end

Walter Roberson 2023-1-1

编辑：Walter Roberson 2023-1-1

在 MATLAB Online 中打开

You can try a number of different starting points and hope for a better answer. If you do, then you will get ode failures, and nlinfit failures. You can get a better fit, in theory, but in practice it is still quite far from where you would hope.

x = [0;5;20;30;50;80;100;120;150];

y = [0;0.17917;0.44143;0.68912;0.76926;0.78383;0.7984;1.06795;0.90768];

myfunc=@fitting;

rng(123456);

best_r = inf;

best_c = [-inf, -inf];

for iters = 1 : 50

c0 = [1.5 1.5] + randn(1,2)/5;

try

[c, r] =nlinfit(x,y,myfunc,c0);

nr = norm(r);

if nr < best_r

fprintf('improved on iteration %d\n', iters);

best_r = nr;

best_c = c;

end

catch ME

fprintf('failed iteration %d\n', iters);

end

improved on iteration 1 improved on iteration 3

Warning: Failure at t=1.146514e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 4

improved on iteration 6 improved on iteration 7

Warning: Failure at t=2.115100e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.684342e-14) at time t.

failed iteration 8

improved on iteration 10 improved on iteration 11

Warning: Failure at t=6.933508e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 15

Warning: Failure at t=4.658234e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.136868e-13) at time t.

failed iteration 16

Warning: Failure at t=2.178430e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.684342e-14) at time t.

failed iteration 17

Warning: Failure at t=1.178626e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 25

Warning: Failure at t=4.211467e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.136868e-13) at time t.

failed iteration 29

Warning: Failure at t=8.001151e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 30

Warning: Failure at t=8.857406e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 31

Warning: Failure at t=8.085854e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 32

Warning: Failure at t=1.162031e+02. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.273737e-13) at time t.

failed iteration 33

Warning: Failure at t=5.427196e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 36

Warning: Failure at t=2.784805e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 39

Warning: Failure at t=2.469110e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 44

Warning: Failure at t=6.325847e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.136868e-13) at time t.

failed iteration 47

improved on iteration 48

c = best_c

c = 1×2

1.4989 -1.8552

%c = lsqcurvefit(myfunc,c0,x,y)

ytotal = fitting(c,x);

hold on

plot(x,ytotal)

plot(x,y,'o')

hold off

function ytotal = fitting(c,x)

xspan = x;

y0 = zeros(2,1);

[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);

%protect against failure of integration

if t(end) ~= xspan(end)

ytotal = inf(numel(xspan),1);

else

ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));

end

function dy=eq2(x,y,c)

%y(1)=CE,y(2)=LE

dy=zeros(2,1);

dy(1)=0.9728.*(2.5-(1+0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);

dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);

end

Jinglei 2023-1-2

@TorstenThank you for your explanation. Actually I used Excel before to optimize the constants, which, however, changed dramatically under different initial values. I used Matlab just to find a better result in a quicker way since Excel is relatively slow and may easily break down when processing. Here, I just can't understand that since the initial values are already better than many others especially the final chosen ones, why the program will go to a less fitted one. Under this situation, choosing better initial values would be a more tough thing.

@Walter RobersonThe matlab also told a size mismatch and gave a possible reason as the use of matrix operators (*, /, ^) instead of the corresponding elementwise operators (.*, ./, .^), which is obviously incorrect as can be seen in the code. Previously, I used ode to solve the fitting problem and got the function, which was not the optimal one since two of the parameters were not so correct based on experimental data. Therefore, I used this to find those two parameters to get the optimal curve. So I don't think this is caused by ODE. And I would like to know whether there might be any problem with nlinfit.

Jinglei 2023-1-3

编辑：Walter Roberson 2023-1-3

在 MATLAB Online 中打开

I used the following code finally and changed two constants into seven. I found this is much more effective than my former one. I would like to know how to make all constants larger than 0 since the result showed one constant negative. In addition, the result of running seven constants is much less fitted compared with two and I would like to know whether this is caused by some errors in the code and how to correct them.

x = [0;5;10;25;30;35;40;45;50;55;70;80;90;120];
xs = 0:0.1:120;
y = [0;1.34288;1.43759;1.59058;1.59786;1.60515;1.60515;1.55415;1.54687;1.58329;1.64886;1.72171;1.72171;1.72171];
myfunc=@fitting;
rng(123456);
best_r = inf;
best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf];
for iters = 1 : 50
    c0 = [0.9 0.4 1.5 1 0.2 1.5 0.7] + randn(1,7)/5;
    ytotal0 = fitting(c0,xs);
    try
        [c, r] =nlinfit(x,y,myfunc,c0);
        nr = norm(r);
        if nr < best_r
            fprintf('improved on iteration %d\n', iters);
            best_r = nr;
            best_c = c;
        end
    catch ME
        fprintf('failed iteration %d\n', iters);
    end
end
c = best_c
%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
hold on
plot(x,y,'o')
plot(xs,ytotal0)
plot(xs,ytotal)
hold off
function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);
%protect against failure of integration
if t(end) ~= xspan(end)
    ytotal = inf(numel(xspan),1);
else
    ytotal = ye(:,1).*(1+0.39976) + ye(:,2);
end
end 
function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=c(1) .* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(3) - y(1) .* (1 + c(2))) - c(4) .* y(1);
dy(2)=c(5).* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(6) - y(2) - c(4) .* y(2) .* (c(6) - y(2))) - c(7).* y(2);
end

Walter Roberson 2023-1-3

编辑：Walter Roberson 2023-1-3

在 MATLAB Online 中打开

With a repair on plotting, and adding information about change from the original parameters, and tweaking the random divisor to have fewer rejected locations.

I tweaked the code so that it will not offer negative initial coordinates. The nonlinear fitting routine does not offer constraints, so you would need to switch to something like fmincon() to prevent values from going negative; I work around it here by rejecting negative results.

I did not bother to add back in the supression of the warning messages. I figure that you must have a reason for wanting to see them, since you adopted my try/catch but not my warning off code.

x = [0;5;10;25;30;35;40;45;50;55;70;80;90;120];

xs = 0:0.1:120;

y = [0;1.34288;1.43759;1.59058;1.59786;1.60515;1.60515;1.55415;1.54687;1.58329;1.64886;1.72171;1.72171;1.72171];

myfunc=@fitting;

rng(123456);

best_r = inf;

best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf];

c0_orig = [0.9 0.4 1.5 1 0.2 1.5 0.7];

for iters = 1 : 100

c0 = max(0, c0_orig + randn(1,7)/20);

ytotal0 = fitting(c0,xs);

try

[c, r] =nlinfit(x,y,myfunc,c0);

if any(c < 0)

fprintf('rejected negative output on iteration %d\n', iters);

continue;

end

nr = norm(r);

if nr < best_r

fprintf('improved on iteration %d\n', iters);

best_r = nr;

best_c = c;

end

catch ME

fprintf('failed iteration %d\n', iters);

end

improved on iteration 1

Warning: Failure at t=5.214845e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 3

Warning: Failure at t=1.138046e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 4

Warning: Failure at t=3.027034e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 8

rejected negative output on iteration 9

Warning: Failure at t=5.172382e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 10

Warning: Failure at t=3.196325e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 11

rejected negative output on iteration 12 rejected negative output on iteration 15 rejected negative output on iteration 16

Warning: Failure at t=5.034794e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 22

Warning: Failure at t=1.404348e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 24

Warning: Failure at t=1.382413e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 25

Warning: Failure at t=1.858841e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 27

Warning: Failure at t=5.805013e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 28

rejected negative output on iteration 29

Warning: Failure at t=3.031519e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 30

Warning: Failure at t=2.133830e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 31

Warning: Failure at t=1.104143e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 32

Warning: Failure at t=4.857090e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 33

Warning: Failure at t=4.029846e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 34

Warning: Failure at t=5.848573e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 35

Warning: Failure at t=1.051578e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 37

rejected negative output on iteration 38

Warning: Failure at t=2.364404e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 39

Warning: Failure at t=4.498947e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 40

rejected negative output on iteration 41

Warning: Failure at t=2.689470e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 42

Warning: Failure at t=2.677579e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 44

Warning: Failure at t=6.993425e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 46

Warning: Failure at t=2.940197e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 48

rejected negative output on iteration 49

Warning: Failure at t=8.152841e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 50

Warning: Failure at t=2.034104e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 51

Warning: Failure at t=1.137900e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 52

rejected negative output on iteration 53

Warning: Failure at t=1.011267e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 54

Warning: Failure at t=2.787820e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 55

Warning: Failure at t=3.488307e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 56

rejected negative output on iteration 57

Warning: Failure at t=9.999214e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 58

Warning: Failure at t=1.551725e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 61

Warning: Failure at t=3.421249e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 63

rejected negative output on iteration 64 rejected negative output on iteration 65

improved on iteration 66

rejected negative output on iteration 67

Warning: Failure at t=2.873194e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 70

rejected negative output on iteration 71

Warning: The Jacobian at the solution is ill-conditioned, and some model parameters may not be estimated well (they are not identifiable). Use caution in making predictions.

Warning: Failure at t=5.456363e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 74

Warning: Failure at t=6.623573e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 75

Warning: Failure at t=3.539704e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 77

Warning: Failure at t=3.977617e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 78

Warning: Failure at t=8.294844e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 79

Warning: Failure at t=6.023717e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 81

Warning: Failure at t=1.648890e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 82

Warning: Failure at t=2.914583e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 83

rejected negative output on iteration 84 rejected negative output on iteration 85

Warning: Failure at t=4.582848e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 87

Warning: Failure at t=4.580019e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 89

rejected negative output on iteration 92

Warning: Failure at t=2.528501e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 93

rejected negative output on iteration 95

Warning: Failure at t=2.098721e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 97

Warning: Failure at t=1.579961e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 98

Warning: Failure at t=1.076871e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 100

c = best_c

c = 1×7

0.7129 0.3339 1.9853 0.8968 0.0325 0.9857 0.0033

change_from_original = c - c0_orig

change_from_original = 1×7

-0.1871 -0.0661 0.4853 -0.1032 -0.1675 -0.5143 -0.6967

%c = lsqcurvefit(myfunc,c0,x,y)

ytotal = fitting(c,x);

hold on

plot(x,y,'o', 'displayname', 'original')

plot(xs, ytotal0, 'displayname', 'ytotal0')

plot(x, ytotal, 'displayname', 'ytotal')

hold off

xlim auto

ylim auto

legend show

function ytotal = fitting(c,x)

xspan = x;

y0 = zeros(2,1);

[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);

%protect against failure of integration

if t(end) ~= xspan(end)

ytotal = inf(numel(xspan),1);

else

ytotal = ye(:,1).*(1+0.39976) + ye(:,2);

end

function dy=eq2(x,y,c)

%y(1)=CE,y(2)=LE

dy=zeros(2,1);

dy(1)=c(1) .* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(3) - y(1) .* (1 + c(2))) - c(4) .* y(1);

dy(2)=c(5).* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(6) - y(2) - c(4) .* y(2) .* (c(6) - y(2))) - c(7).* y(2);

end

Jinglei 2023-1-4

编辑：Jinglei 2023-1-4

在 MATLAB Online 中打开

@TorstenIt works well this time. Thank you for all your help.

I have got two other questions. Firstly, how could I calculate the R2 of fitting in this case?In addition, is it possible to add several ys in this code to get five curves directly?I maen, making y a matrix with five rows. I wrote it as the following way, which, however, only showed the points without any curve.

myfunc=@fitting;
rng(123456);
best_r = inf;
best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf];
c0_orig = [0.9 0.4 1.5 1 0.2 1.5 0.7];
for iters = 1 : 100
    c0 = max(0, c0_orig + randn(1,7)/20);
    ytotal0 = fitting(c0,xs);
    try
        [c, r] =nlinfit(x,y(:,1),y(:,2),y(:,3),y(:,4),y(:,5),myfunc,c0);
        if any(c < 0)
            fprintf('rejected negative output on iteration %d\n', iters);
            continue;
        end
        nr = norm(r);
        if nr < best_r
            fprintf('improved on iteration %d\n', iters);
            best_r = nr;
            best_c = c;
        end
    catch ME
        fprintf('failed iteration %d\n', iters);
    end
end
Unrecognized function or variable 'xs'.
c = best_c
change_from_original = c - c0_orig
%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
%for adding points
hold on
plot(x,y,'o', 'displayname', 'original')
hold off
xlim auto
ylim auto
legend show
%for adding points
hold on
xs = 0:0.1:150;
ytotal = fitting(c,xs);
plot(xs,ytotal)
hold off

Jinglei 2023-1-6

编辑：Walter Roberson 2023-1-6

在 MATLAB Online 中打开

I editted the code again to make it work for matrix, which, however, still didn't result in five curves successfully. Does anyone know what is wrong with my code?

x = [0 0 0 0 0;5 5 5 5 5; 10 10 10 10 10;
    15 15 15 15 15; 20 20 20 20 20;
    25 25 25 25 25; 30 30 30 30 30; 
    35 35 35 35 35; 40 40 40 40 40;
    45 45 45 45 45; 50 50 50 50 50;
    55 55 55 55 55; 60 60 60 60 60;
    70 70 70 70 70; 80 80 80 80 80;
    90 90 90 90 90;120 120 120 120];
Error using vertcat
Dimensions of arrays being concatenated are not consistent.
xs = 0:0.01:120;
y = [0	0	0	0	0;
    1.342848859	1.466695386	1.495835746	1.524976105	1.189861972;
    1.437555027	1.371989218	0.956739097	1.510405925	1.007734726;
    1.175291792	0.993164546	0.701760952	1.561401554	1.415699757;
    1.699818261	1.510405925	1.590541914	1.481265566	1.437555027;
    1.590541914	1.554116464	1.539546285	1.561401554	1.459410296;
    1.597827003	1.590541914	1.539546285	1.561401554	1.444840117;
    1.605112093	1.590541914	1.437555027	1.510405925	1.473980476;
    1.605112093	1.663392812	1.605112093	1.575971734	1.473980476;
    1.554116464	1.554116464	1.539546285	1.612397183	1.401129577;
    1.546831374	1.590541914	1.561401554	1.561401554	1.481265566;
    1.583256824	1.568686644	1.626967363	0.053387955	1.408414667;
    1.437555027	1.605112093	1.626967363	1.648822632	1.444840117;
    1.648822632	1.648822632	1.707103351	1.561401554	1.546831374;
    1.721673531	1.707103351	1.619682273	1.634252453	1.648822632;
    1.721673531	1.707103351	1.707103351	1.692533171	1.532261195;
    1.721673531	1.707103351	1.677962992	1.546831374	1.619682273
    ];
myfunc=@fitting;
rng(123456);
best_r = inf;
best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf];
c0_orig = [0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7];
for iters = 1 : 100
    c0 = max(0, c0_orig + randn(5,7)/20);
    ytotal0 = fitting(c0,xs);
    try
        [c, r] =nlinfit(x,y,myfunc,c0);
        if any(c < 0)
            fprintf('rejected negative output on iteration %d\n', iters);
            continue;
        end
        nr = norm(r);
        if nr < best_r
            fprintf('improved on iteration %d\n', iters);
            best_r = nr;
            best_c = c;
        end
    catch ME
        fprintf('failed iteration %d\n', iters);
    end
end
c = best_c
change_from_original = c - c0_orig
%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
%for adding points
hold on
plot(x,y,'o', 'displayname', 'original')
hold off
xlim auto
ylim auto
legend show
%for adding points
hold on
xs = 0:0.1:150;
ytotal = fitting(c,xs);
plot(xs,ytotal)
hold off
function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);
%protect against failure of integration
if t(end) ~= xspan(end)
    ytotal = inf(numel(xspan),1);
else
    ytotal = ye(:,1).*(1+0.39976) + ye(:,2);
end
end 
function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=c(1) .* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(3) - y(1) .* (1 + c(2))) - c(4) .* y(1);
dy(2)=c(5).* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(6) - y(2) - c(4) .* y(2) .* (c(6) - y(2))) - c(7).* y(2);
end

Walter Roberson 2023-1-7

在 MATLAB Online 中打开

You cannot do that.

nlinfit() requires that the x and y values and initial model parameters must be vectors. You can call

nlinfit(x(:),y(:),myfunc,c0(:))

and then have myfunc make appropriate adjustments.

But inside myfunc you take the input and use it as tspan values for the ode. The ode*() functions require that the tspan consist of strictly monotonic values -- strictly increasing or strictly decreasing. Which your x(:) would not have, as your x has a lot of rows with duplicate columns.

If what you are trying to do is run a number of different independent fittings simultaneously, hoping that will be more efficient than running them independently, then you are going about that the wrong way. Yes, it is possible to get ode45() or ode15s() to effectively run multiple ODEs simulatenously, but (A) you need to build the code the right way; and (B) it is not at all certain to be more efficient; and (C ) it is questionable whether it would be of sufficient accuracy. @Jan in a recent post made the explicit claim that the numeric accuracy gets worse when you are running multiple not-coupled ODE simulatenously.

If you must make the attempt to run several simultaneous not-coupled ODE, then set up your equations using the Symbolic Toolbox, and then follow the work flow in the first example of odeFunction to build the anonymous function that would be passed to ode15s() or ode45(). (Make sure to build the ytotal equation too, so that you get the proper simultaneous totals.)

But really... just run them independently, the extra work to run them simultaneously is not worth the effort.

Jinglei 2023-1-8

I just run the program with one group of my data, which, however, stopped in the middle. Those warnings and things were just like what were shown when I run successfully so I am confused of what happened for this especially considring the better data compared than others.

failed iteration 63

Warning: Failure at t=1.385053e+00. Unable to meet integration tolerances without reducing the step size below

the smallest value allowed (3.552714e-15) at time t.

> In ode15s (line 730)

In xian2mu1rawdatafor5000rat50du>fitting (line 89)

In nlinfit>@(b,x)w.*model(b,x) (line 283)

In nlinfit>LMfit (line 592)

In nlinfit (line 284)

In xian2mu1rawdatafor5000rat50du (line 47)

failed iteration 68

rejected negative output on iteration 70

Warning: Failure at t=1.187736e+01. Unable to meet integration tolerances without reducing the step size below

the smallest value allowed (2.842171e-14) at time t.

> In ode15s (line 730)

In xian2mu1rawdatafor5000rat50du>fitting (line 89)

In nlinfit>@(b,x)w.*model(b,x) (line 283)

In nlinfit>LMfit (line 592)

In nlinfit (line 284)

In xian2mu1rawdatafor5000rat50du (line 47)

failed iteration 73

Those above are part of the shown things in my matlab.

Walter Roberson 2023-1-18

Then in that case, there are a couple of different ways you could improve the code:

use ga or particleswarm or other similar evolutionary algorithm that tries a wide variety of different positions, instead of looping trying different random positions
use MultiStart instead of looping trying different random positions
do a more detailed mathematical analysis of your equations in order to figure out boundary conditions so that you can generate random starting positions that will be valid. This might be difficult as it would require analyzing the ways in which your ode can fail.