how to find the best curve fit for a set of data

i have a set of data which i want to curve fit. how i can determine the best curve fit of it using MATLAB and cftool ? i have to write an script or what?

[John D'Errico]

The best curve fit is an interpolation. The error will be zero. There are an infinite number of such exact interpolatory models. So the phrase "the best curve fit" is meaningless.

[Maryam Mapar]

i have some data which comes from an equation , i dont know the equation and using Matlab, i want to determine the the equaton or at least the general form of it. How i can use Matlab cftool to estimate the best curve fit and the equation finally

[Maryam Mapar]

Let me say some details in fact i have 2 set of data which are outputs of an equation now i want to guess the original equation or function or the best guess using Matlab. i want to use CFTOOL to estimate the best curve fit with the less SSE and RMSE. thank you in advance

[prabha verma]

% type your set of data in this way and call cftool function 
 X=[1 2 3 5 6 8 9];
 Y= [4 6 7 9 12 15 20];
 cftool
% It would leads you to a gui window 
% Select the X values at the "X data" section, and Y values at "Y data" section. Choose the type of fit options you want to selct and click on "fit".
% It will give you the goodness of fit parameters like SSE, R-square value , RMSE etc. On the basis of which you could decide which fitting is best for your data set

Answer 1

[John D'Errico]

I tried not to wade in here, because every time someone asks a question like this, it means they know virtually nothing at all about modeling.

You cannot use the curve fitting toolbox, or ANY such toolbox to know the best fitting curve, IF you are not willing to provide a model form. The curve fitting toolbox is not a magic tool that can look at your data, and somehow know what the underlying model should have been. There is no such tool, although I have heard of tools that try to do so. They cannot.

The best possible fitting model is an exact interpolant. It has ZERO residuals, so it will be the best. And there are infinitely many such exact interpolants. Start with polynomials - infinitely many of them that will fit exactly. (Although a sufficiently high order polynomial to interpolate the data will produce complete garbage here.) Or start with splines, again, infinitely many of them. But they too will produce interpolatory garbage.

So, now lets look at your data.

plot(Var1,Var2,'-o')

Then, just for kicks, I added an interpolating spline.

As you can see, it fits perfectly. But I seriously doubt it is what would be the best fit. Nor would be a Fourier interpolant anything of value here either. You asked about that, and rational models. Don't waste your time.

Anyway, you are kidding me, right? I doubt your data is good enough to be worth anything more than a cubic polynomial fit. Pretty noisy, not much signal there beyond a straight line fit. Maybe you can convince me there is some curvature in there, so I'd accept a cubic polynomial.

If you managed to convince me that the cubic fit is not adequate, because it is not monotonic over the support of the data, I'd tell you to use my SLM toolbox. You can find it at that link on the File Exchange. (It will use the optimization toolbox though.)

slm = slmengine(Var1,Var2,'knots',5,'increasing','on','reg','cr','plot','on')

A careful choice of the parameters for SLM gives me what I'd call a pleasing, monotonic curve fit. Is this the best possible? Arguably it is about as good as you can do. I used cross validation to choose the extent of regularization, over the set of monotonic splines on that support.

[Maryam Mapar]

this is what i got from Matlab help file for an example set of data , now i want to use the same process for my data, is it possible to do so with your SLM ?

Determining the Best Fit

To determine the best fit, you should examine both the graphical and numerical fit results.

Examine the Graphical Fit Results

Determine the best fit by examining the graphs of the fits and residuals. The graphical fit results indicate that:

• The fits and residuals for the polynomial equations are all similar, making it difficult to choose the best one.

• The fit and residuals for the single-term exponential equation indicate it is a poor fit overall. Therefore, it is a poor choice and you can remove the exponential fit from the candidates for best fit.

Evaluate the Numerical Fit Results

When you can no longer eliminate fits by examining them graphically, you should examine the numerical fit results. The Curve Fitting app displays two types of numerical fit results:

• Goodness-of-fit statistics

• Confidence bounds on the fitted coefficients

The goodness-of-fit statistics help you determine how well the curve fits the data. The confidence bounds on the coefficients determine their accuracy.

Examine the sum of squares due to error (SSE) and the adjusted R-square statistics to help determine the best fit. The SSE statistic is the least-squares error of the fit, with a value closer to zero indicating a better fit. The adjusted R-square statistic is generally the best indicator of the fit quality when you add additional coefficients to your model. The largest SSE for exp1 indicates it is a poor fit, which you already determined by examining the fit and residuals. The lowest SSE value is associated with poly6. However, the behavior of this fit beyond the data range makes it a poor choice for extrapolation, so you already rejected this fit by examining the plots with new axis limits. The next best SSE value is associated with the fifth-degree polynomial fit, poly5, suggesting it might be the best fit. However, the SSE and adjusted R-square values for the remaining polynomial fits are all very close to each other. Which one should you choose?

Resolve the best fit issue by examining the confidence bounds for the remaining fits in the Results pane. Double-click a fit in the Table of Fits to open (or focus if already open) the fit figure and view the Results pane. A fit figure displays the fit settings, results pane and plots for a single fit. Display the fifth-degree polynomial and the poly2 fit figures side by side. Examining results side by side can help you assess fits.

Compare the coefficients and bounds (p1, p2, and so on) in the Results pane for both fits, poly5 and poly2. The toolbox calculates 95% confidence bounds on coefficients. The confidence bounds on the coefficients determine their accuracy. Check the equations in the Results pane (f(x)=p1*x+p2*x...) to see the model terms for each coefficient. Note that p2 refers to the p2*x term in Poly2 and the p2*x^4 term in Poly5. Do not compare normalized coefficients directly with non-normalized coefficients.

The bounds cross zero on the p1, p2, and p3 coefficients for the fifth-degree polynomial. This means you cannot be sure that these coefficients differ from zero. If the higher order model terms may have coefficients of zero, they are not helping with the fit, which suggests that this model overfits the census data.

However, the small confidence bounds do not cross zero on p1, p2, and p3 for the quadratic fit, poly2 indicate that the fitted coefficients are known fairly accurately.

Therefore, after examining both the graphical and numerical fit results, you should select poly2 as the best fit to extrapolate the census data.

[John D'Errico]

So what? Sorry, but this type of analysis is not that productive here. Deciding to use a given degree polynomial is not going to give you the "best" fit, because you already know that a least square quadratic or cubic is not the underlying model for that curve. It is surely something nonlinear. The point is, a polynomial fit here, with an order based on statistical heuristics about whether a coefficient may or may not be zero is essentially never a good choice.

And this is also why in the beginning of my response I said that I really did not want to touch this post with a 10 foot pole. Sorry, but true, because you appear to know just enough about curve fitting to be able to parrot back what you read, but not understand what was done there, AND why. That means this can turn into a lengthy class taught by me to you in the comments.

In general, the best fit for something like this is arguably that fit that yields what you expect to see. This is because you, as the creator of the data (i.e., scientist, engineer, technician, analyst, whatever) usually have knowledge about the system. You may know things like having an expectation of monotonicity. You may have some understanding to know how much of those bumps in the curve are probably real, and how much is just noise.

None of that information is available to simple modeling tools. Only you have that knowledge, and it is all in your head. Heuristics that try to determine a polynomial order are meaningless here, again because they lack that information, and because the true model was never a low order polynomial at all.

I will say only one thing more here, that the entire purpose of SLM is to allow the user to translate the information they hold in their brain into a description of the fundamental shape of the curve, and then allow them to produce a model that fits the data.

The best fit is the one that satisfies your eyes, the fit that makes sense to your brain, that is consistent with your expectations about the process.

[Image Analyst]

Maryam, is this homework? Because it looks like it has a fairly specific set of steps it wants you to follow. If so, I'd advise you to do that, because it's what your instructor/grader will expect and what your grade will depend on.

Answer 2

[Amin Oroji]

One of the most interesting problems in supervised machine learning is to find the best fit for the existing data which called regression. Suppose you have three points. Using a polynomial with degree two you can fit the curve to your point. Generally it is not a proper way. It may cause over-fitting. It means that the generalization is not considered here. In other words the error of the chosen model will be high when the model is applied for a new data. Therefore, validation techniques such as cross validation are used to solve this problem. In cross validation, the data is split into k folds (e.g. 10 folds). Then, the first fold is used to validate the model. After that, the next fold is used for validation and so on. Finally the mean of errors is calculated. The model with lower mean error is chosen. For more details you can see

1. https://stanford.edu/class/ee103/lectures/ls-fitting_slides.pdf
2. https://towardsdatascience.com/overfitting-vs-underfitting-a-complete-example-d05dd7e19765?gi=7dbe4692ec9b
3. http://scikit-learn.org/stable/modules/learning_curve.html

Keywords: validation techniques, cross validation, overfitting

Answer 3

[Chad Greene]

Often the term "best fit" refers to linear regression, so perhaps you want polyfit. Here's an example of fitting a line to some scattered data:

x = 100*rand(20,1); 
y = 3*x + 5 + 15*randn(size(x)); 

plot(x,y,'bo') 

P = polyfit(x,y,1); 

xfit = 0:100; 
yfit = polyval(P,xfit); 

hold on
plot(xfit,yfit,'-')

[Maryam Mapar]

hello.I dont know. please guide me. i have a set of data which i want to curve fit. how i can determine the best curve fit of it using MATLAB and cftool ? i have to write an script or what?

[Maryam Mapar]

i have some data which comes from an equation , i dont know the equation and using Matlab, i want to determine the the equaton or at least the general form of it. How i can use Matlab cftool to estimate the best curve fit and the equation finally

[Maryam Mapar]

Let me say some details in fact i have 2 set of data which are outputs of an equation now i want to guess the original equation or function or the best guess using Matlab. i want to use CFTOOL to estimate the best curve fit with the less SSE and RMSE. thank you in advance

Answer 4

[Image Analyst]

I agree 100% with John D'Errico. If you're willing to have a polynomial, you can get an EXACT fit with zero error if you use a Lagange interpolating polynomial: https://en.wikipedia.org/wiki/Lagrange_polynomial It will go through all your points exactly, though for points not at those training point locations, you'll get wildly oscillating numbers. So a zero error "fit" is pretty much useless in the vast majority of situations.

I'm not familiar with the cftool function in the curve fitting toolbox so I can't help you with that.

[Image Analyst]

You have to assume at least some general form of the equation, like it's a polynomial or exponential or Gaussian or whatever. Then you can get parameters for an equation that is of that model type. For example, see my attached demo to fit an exponential decay model, Y = a + exp(-b*x), to some (x,y) data.

[Maryam Mapar]

does it mean that i have to try every models with Matlab ? is there any other short way to estimate the general form or to guess the exact equation ?

[Image Analyst]

Like John said, there are an infinite number of models that could go through the data with various goodnesses of fit. Like I said, if you want an exact fit with no error at all, then use a Lagrange Polynomial, but in general that's no good for smoothing. For smoothing you need to pick a form for the model and then the fitting process will tell you the parameters/coefficients for the various terms in the model. If you have no model whatsoever and can't even guess at one, then you can smooth the data with things like a moving mean filter (movmean), moving median (movmedian I think), or moving polynomial (sgolayfilt), though those methods give discrete numerical curves (arrays) rather than analytical models (formulas) like you'd get if you specified the form/equation of a model.

Do you have any idea at all what simple model might fit your data? Can you at least show us a plot of the raw data? If it's super complicated, like with lots of peaks and valleys, then probably no model will be good and you'll have to use a numerical smoothing like you'd get from sgolayfilt().

[Maryam Mapar]

Here are my numbers : Var1 : 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5

Var2 : 84.766716 85.775747 89.227803 94.347621 98.11531 94.175022 98.449363 108.917848 114.173113 133.533924 140.114545 145.876143 148.447757 165.888946 169.258454 173.680863 i want to guess the original form of the equation with the minimum SSE using CFTOOL

[Image Analyst]

Sorry, I don't have the Curve Fitting Toolbox, and it looks like no one that does have it is answering you. All I can offer is polyfit(). It looks like a polynomial of order 2 or 3 should be good. If you really want a model with residuals of zero, see this code:

Var1 = [0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5]
Var2 = [84.766716 85.775747 89.227803 94.347621 98.11531 94.175022 98.449363 108.917848 114.173113 133.533924 140.114545 145.876143 148.447757 165.888946 169.258454 173.680863]
plot(Var1, Var2, 'b*-', 'LineWidth', 2);
grid on;
[coeffs, s, mu] = polyfit(Var1, Var2, length(Var1)-1);
xFit = linspace(min(Var1), max(Var1), 100);
yFit = polyval(coeffs, xFit);
hold on;
plot(xFit, yFit, 'r.-', 'LineWidth', 2, 'MarkerSize', 13);

However you should not think that it's a good model just because the residuals are zero. Try passing in 2 or 3 instead of "length(Var1)-1" to polyfit and see what you get.

[Maryam Mapar]

thank you for your answer

what if i want to use fourier model or rational model ? is it possible to use the same way to do that ? i have another function calling chevron which i want to use and estimate the best fit using the output numbers ?

Answer 5

[marwan mokbil]

Mr. John D'Errico, I have read your answer regarding the curve best fit. So what I understood that we will never get the exact equation for these known points of (x,y). That mean no one can get the exact equation for known points of (x,y) even if these points coming from real equation. Is that true?

[John D'Errico]

Yes. That is generally true. You can find an infinite number of "equations" that represent any set of data, even if the data is "exact". You can never infer the true equation that generated a set of data.

Answer 6

[marwan mokbil]

Mr. John i am still looking for way can give me the exact equation of these known point, put in confederation that these known point of (x,y) are not random I mean they come from an equation and i want to find this exact equation. Please inform me if you know or any one else know the way to get this equation.

[Image Analyst]

If you've read the above answers you know that there is an unlimited number of equations that can fit a finite set of points. One guaranteed way to get an equation is to use the Lagrange interpolating polynomial, like I gave in my answer https://ww2.mathworks.cn/matlabcentral/answers/349776-how-to-find-the-best-curve-fit-for-a-set-of-data#answer_275220 though I don't think you'd want that method. Otherwise you should make some best guess at the equation using your knowledge of the physics of the situation.

Answer 7

[marwan mokbil]

Mr. John i am still looking for way can give me the exact equation of these known point, put in confederation that these known point of (x,y) are not random I mean they come from an equation and i want to find this exact equation. Please inform me if you know or any one else know the way to get this equation.

[John D'Errico]

Please stop adding answer after answer. It does not matter how many times you ask. You CANNOT know the true equation that generated a set of data, even if the data seems exact.

For example, I'll give you three pairs of points, in the form of (x,y) pairs. Then I'll ask you what the function was that generated those values of y. Deal?

(1,1), (2,1), (3,1)

So three points, with x=1:3. y just happens to have been identically 1, at all three points.

Did you guess that the function was identically a constant? WRONG!

In fact, it was:

fun = @(x) x.^3 - 6*x.^2 + 11*x - 5
fun =
  function_handle with value:
    @(x)x.^3-6*x.^2+11*x-5

fun(1:3)
ans =
     1     1     1

As you can see, fun evaluates to exactly 1 at every point I tried. So we have the exact values. Did you guess the function I created? Is there some good way that you might have guessed that?

Maybe you did guess. But then I'll tell you that, OOPS! sorry. I lied. Here is the true function:

fun = @(x) 2*x.^3 - 12*x.^2 + 22*x - 11
fun =
  function_handle with value:
    @(x) 2*x.^3 - 12*x.^2 + 22*x - 11

fun(1:3)
ans =
     1     1     1

Do you believe me that this last one was the true function? Maybe you should not. How about this one?

fun = @(x) 2*x.^4-12*x.^3+22*x.^2-12*x+1
fun =
  function_handle with value:
    @(x)2*x.^4-12*x.^3+22*x.^2-12*x+1
fun(1:3)
ans =
     1     1     1

How many more of the infinitely many possible functions should I create? And all of the above were trivial modifications. I could go on forever.

Now, I suppose you might argue that you want the "simplest" form. Of course in the example above, the simplest form is y(x)=1. But suppose that I had started with some other set of numbers that were not constant? I could as easily have constructed infinitely many solutions. Which of those would be the "true" one? So, given some arbitrary set of (x,y) pairs,...

(1,1), (2,5), (3,pi), (4,42), ...

Pick any numbers in there, as long as the x values are distinct, a function exists that passes through them, but there are infinitely many choices, and here you cannot as easily know how to just pick the simplest solution, whatever that means to you.

As Walter points out, you can actually prove (and this is trivial to do) that it is possible to generate an infinite set of solutions. In fact, I used that standard trick to generate the various solutions I posed above.

Just because you have true, real, exact function values, this means nothing. You cannot know which function created your data. There are infinitely many (all equally valid) choices of functions that could generate any set of points that represent some function. Simply having the data points is NOT sufficient.

[Walter Roberson]

If you do not know the form of the equation, then this is not possible. There are provably an infinite number (the infinity of continuous numbers, not just the infinity of integers) of different equations that exactly fit any finite set of points that are stated to finite accuracy.

In the past I have posted a constructive proof of this fact, showing how to construct an indefinite number of equations that exactly fit the finite list of points.

[John D'Errico]

Stop adding answers every time you make a comment. Nothing you have said is an answer. Moved answer to a comment by Marwan:

"Sorry for sending my quistion many time ,I new with wetland answer and i thought you did not receive my quistion. Anyway I agree that many equation will pass throw these 3 point (intersected point), but if we select another 3 points may another many equations could pass throw these new 3 points but at the end only one equation can pass throw the six points. Please correct me if I am wrong."

[John D'Errico]

You are wrong. Flat out wrong. Period. Wrong. Did I mention that you are wrong?

I picked 3 points before. What stopped me from picking sets of 6 arbitrary points? Nothing. I could have picked sets of 127000 points. Still the theory is the same. There are infinitely many curves that pass through any set of points.

You keep wanting to see some magical solution, where the magic wand of mathematics will give you the function you need. Sorry, but it does not exist.

There are infinitely many ways to pass a curve exactly through ANY set of points. All are equally as valid as any other.

If you want more, then you need to invest the effort:

1. Choose a model, a family of functions that are consistent with your process. This choice should usually be driven by physical modeling considerations. So you need to understand your process. Or, if the model family is some well known family, i.e., splines, polynomials from one of the many families, trig functions, bessel functions, etc., then you need to know how to work with that set of functions and how to use them, what properties they have.

2. Learn how to fit the model to your data. Different function families are defined in different ways.

3. Learn to use the mathematical tools to help you to fit the model. It may involve linear or nonlinear regression. It may involve spline modeling. It may involve fourier transforms. Learn sufficient mathematics behind the fitting tools to be able to use those tools intelligently.

But I'm sorry, there is no magic wand you can just wave and find the equation that generated any set of data points.

[Walter Roberson]

I gave a constructive proof that there really are an infinite number of equations that fit at https://www.mathworks.com/matlabcentral/answers/347829-how-to-fit-curve-using-derivatives-as-a-constraint#comment_467765 . I also show there that it is not sufficient to match derivatives, by showing how to construct a new equation whose derivative is the same as the original equation at each of the given points.

The argument I gave does not depend upon the number of points (though I suppose it could be argued that it needs to be improved for the degenerate case of there being only one specified point.)

Answer 8

[Alex Sha]

this function is good enough:

y = p1/(1+p2*exp(-p3*x))+p4;

Root of Mean Square Error (RMSE): 3.46432941573129

Sum of Squared Residual: 192.025252811218

Correlation Coef. (R): 0.993674523512372

R-Square: 0.98738905867754

Adjusted R-Square: 0.9854489138587

Determination Coef. (DC): 0.98738905867754

Chi-Square: 0.80116205907639

F-Statistic: 313.184887131818

Parameter Best Estimate

---------- -------------

p1 99.6836025681754

p2 49.1906740359158

p3 0.802021238782871

p4 84.3099000840935

[Victor Zoratti Ferreira]

Alex Sha, why did you choose exactly this type of equation and how did you find the parameters?

Answer 9

[Bhimrao]

For the linear equation, polyfit function for linear curve fitting. The polyfit function fits a polynomial of a specified degree to the data using the method of least squares.