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How to limit the domain for the evaluation of NonlinearModelFit?

Alton Dugas

Posted 10 years ago

I am trying to fit some data using the NonlinearModelFit. z is the data from excel and s is the data after deleting the first row (containing headers). The data follows a log function. I want to use the form for f[t_]. When I run the code, I get the plots and the function that fits the data, but the data causes some issues because there are non-real numbers in the solution. Is there a way to limit the domain for the evaluation in which i know the values will all be real and positive? z = Import[indirname, {"Data", 1}]; s = Delete[z, 1]; f[t_] = a* Log[b + c*t]; nlm = NonlinearModelFit[s, f[t], {a, b, c}, t]; NonlinearModelFit::nrlnum: "The function value {35.2876 +22.2457\ I,33.9223 +22.2457\ I,32.5461 +22.2457\ I,32.0216 +22.2457\ I,<<16>>,29.8249 +22.2457\ I,29.7437 +22.2457\ I,29.8713 +22.2457\ I} is not a list of real numbers with dimensions {23} at {a,b,c} = {7.08102,-248.452,8.99277}."

POSTED BY: Alton Dugas

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Jim Baldwin

Jim Baldwin, Retired

Posted 10 years ago

POSTED BY: Jim Baldwin

Alton Dugas

Posted 10 years ago

Thanks Bill. That does help. I've run into problems before with fitting data and I am trying to learn why some things work and some don't. I am new to Mathematica and would like to learn all the pitfalls, so when I present data, I can explain why certain procedures were used. Again, thanks for the help.

POSTED BY: Alton Dugas

Bill Simpson

Posted 10 years ago

Let's get some initial estimates for the parameters by exponentiating the vertical axis and doing a linear fit. s = {{10.7000, 0.2973}, {10.7333, 1.6486}, {10.7833, 3.0039}, {10.8167, 3.5144}, {10.8500, 3.8395}, {10.9000, 4.1202}, {10.9333, 4.2455}, {10.9667, 4.4361}, {11.0167, 4.5758}, {11.0500, 4.6816}, {11.0833, 4.7129}, {11.1333, 4.8043}, {11.1667, 4.9506}, {11.2000, 5.0406}, {11.2500, 5.0837}, {11.2833, 5.2404}, {11.3167, 5.2456}, {11.3667, 5. {11.4000, 5.3436}, {11.4333, 5.4154}, {11.4833, 5.4245}, {11.5167, 5.4911}, {11.5500, 5.3488}}; exp = Map[{#[[1]], Exp[#[[2]]]} &, s]; line = Fit[exp, {1, t}, t] Show[ListPlot[exp], Plot[line, {t, 10.7, 11.55}]] which gives us -3100.3 + 290.013 t and Now, using those estimates for the parameters as starting values, f[t_] := Log[b + c*t]; nlm = NonlinearModelFit[s, f[t], {{b, -3100}, {c, 290}}, t]; log = Normal[nlm] Show[ListPlot[s], Plot[log, {t, 10.7, 11.55}]] which gives us Log[-3031.56 + 283.439 t] without any error messages and

Let's get some initial estimates for the parameters by exponentiating the vertical axis and doing a linear fit.

s = {{10.7000, 0.2973}, {10.7333, 1.6486}, {10.7833, 3.0039}, {10.8167, 3.5144}, {10.8500, 3.8395},
     {10.9000, 4.1202}, {10.9333, 4.2455}, {10.9667, 4.4361}, {11.0167, 4.5758}, {11.0500, 4.6816},
     {11.0833, 4.7129}, {11.1333, 4.8043}, {11.1667, 4.9506}, {11.2000, 5.0406}, {11.2500, 5.0837},
     {11.2833, 5.2404}, {11.3167, 5.2456}, {11.3667, 5. {11.4000, 5.3436}, {11.4333, 5.4154},
     {11.4833, 5.4245}, {11.5167, 5.4911}, {11.5500, 5.3488}};
exp = Map[{#[[1]], Exp[#[[2]]]} &, s];
line = Fit[exp, {1, t}, t]

Show[ListPlot[exp], Plot[line, {t, 10.7, 11.55}]]

which gives us -3100.3 + 290.013 t

and

enter image description here

Now, using those estimates for the parameters as starting values,

f[t_] := Log[b + c*t];
nlm = NonlinearModelFit[s, f[t], {{b, -3100}, {c, 290}}, t];
log = Normal[nlm]

Show[ListPlot[s], Plot[log, {t, 10.7, 11.55}]]

which gives us Log[-3031.56 + 283.439 t] without any error messages

and

enter image description here

POSTED BY: Bill Simpson

Alton Dugas

Posted 10 years ago

thanks for looking.

POSTED BY: Alton Dugas

Jim Baldwin

Jim Baldwin, Retired

Posted 10 years ago

Sorry, I steered you wrong. The is almost certainly that some of the parameter guesses result in `b + c t` being less than zero (as that is inside a `Log` function). The parameters need to be restricted so that that doesn't happen.

POSTED BY: Jim Baldwin

Alton Dugas

Posted 10 years ago

{10.7000, 0.2973}, {10.7333, 1.6486}, {10.7833, 3.0039}, {10.8167, 3.5144}, {10.8500, 3.8395}, {10.9000, 4.1202}, {10.9333, 4.2455}, {10.9667, 4.4361}, {11.0167, 4.5758}, {11.0500, 4.6816}, {11.0833, 4.7129}, {11.1333, 4.8043}, {11.1667, 4.9506}, {11.2000, 5.0406}, {11.2500, 5.0837}, {11.2833, 5.2404}, {11.3167, 5.2456}, {11.3667, 5.3305}, {11.4000, 5.3436}, {11.4333, 5.4154}, {11.4833, 5.4245}, {11.5167, 5.4911}, {11.5500, 5.3488}

{10.7000,   0.2973}, {10.7333, 1.6486}, {10.7833, 3.0039}, {10.8167, 3.5144},
{10.8500, 3.8395}, {10.9000,    4.1202}, {10.9333, 4.2455}, {10.9667,  4.4361}, 
{11.0167,   4.5758}, {11.0500, 4.6816}, {11.0833, 4.7129}, {11.1333,  4.8043}, 
{11.1667,   4.9506}, {11.2000,    5.0406}, {11.2500, 5.0837}, {11.2833,  5.2404},
{11.3167,   5.2456}, {11.3667,    5.3305}, {11.4000, 5.3436}, {11.4333,  5.4154}, 
{11.4833,   5.4245}, {11.5167,    5.4911}, {11.5500, 5.3488}

POSTED BY: Alton Dugas

Jim Baldwin

Jim Baldwin, Retired

Posted 10 years ago

It looks like the import did not work as desired as it looks like characters ("32.2876 +22.2457\I") rather than a numeric value. Can you show the first few values of z?

POSTED BY: Jim Baldwin

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