[matlab] curve fitting

Curve fitting

The difference between interpolation and fitting ：

1. Implementation method ： Interpolation requires the curve to pass through the sample points , The fitting does not need to pass through the sample points , Only the minimum overall error is required .
2. The form of the result ： Interpolation is a piecewise approximation of sample points , There is no same approximation function ; Function fitting uses a function to approximate , There is a complete expression .
3. Focus on ： Interpolation can be used to estimate the function values corresponding to some points in the interval ; Fitting can not only estimate the points in the interval , You can also predict points outside the interval .
4. applications ： Interpolation is often used for precise data sets ; Fitting is often used in statistical data sets .

polyfit Function for Univariate multiple curve fitting （ Also polynomial fitting ）, Form like ：$y=a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex+f$, We know $x$、$y$ Sample data , Use this function to solve the parameter （ coefficient ）$a$、$b$、$c$、$d$、$e$、$f$.

regress Function for Univariate or multivariate linear regression （ fitting ）, Form like ：$y=a{x}_1+b{x}_2+c{x}_3+d{x}_4+e$, We know $x_1$、$x_2$、$x_3$、$x_4$、$y$ Sample data , Use this function to solve the parameter （ coefficient ）$a$、$b$、$c$、$d$、$e$.

polyfit function

Invocation format ：

• [P,S,mu]=polyfit(X,Y,m)

• [P,S]=polyfit(X,Y,m)

• P=polyfit(X,Y,m)

Parameter interpretation ：

According to the sample data X and Y, Produce a m Polynomial coefficient vector P And its error data at the sampling point S,mu It's a binary vector ,mu(1) yes mean(X), and mu(2) yes std(X).

x=0: 0.1: 1;
y=[-0.447, 1.978, 3.11, 5.25, 5.02, 4.66, 4.01, 4.58, 3.45, 5, 35];
p = polyfit(x, y, 3) %  Cubic polynomial fitting 
xx = 0: 0.01 : 1;
yy = polyval(p, xx) ; %  According to the coefficient vector p Calculated at xx The function value at point 
plot(xx, yy, '-b', x, y, 'markersize', 20)

When you still need to Linear regression analysis when , Can be reused corrcoef Function to obtain the correlation coefficient .

regress function

stay matlab Of regress Confidence interval in function bint、rint as well as stats The last three values are all infinite , This shows that the data does not obey the linear relationship , Nonlinear fitting function shall be considered for fitting .

Invocation format ：

• [B,BINT,R,RINT,STATS] = regress(Y,X)
• [B,BINT,R,RINT] = regress(Y,X)
• [B,BINT,R] = regress(Y,X)
• [B,BINT] = regress(Y,X)
• B = regress(Y,X)

Parameter interpretation ：

Call this function to ensure that the form of the object function is known , For example, the objective function is ：$y=a{x}_1+b{x}_2+c{x}_3+d{x}_4+e$, Or is it $y=a{x}_1^2+b{x}_2^2+c{x}_1+d{x}_2+e{x}_1\times x_2+f$ Such multinomial polynomials . Pass in the parameter , Call this function to get the fitting coefficient value .

• x It's a matrix . Every sample of the matrix , That is, the number of rows equals the number of samples , Each column of each sample represents the value of each term of the sample in the polynomial excluding the coefficient .
• y It's a column vector . The vector dimension is consistent with the number of rows of the matrix , Same as the number of samples , Each value represents the function value corresponding to each sample .

This function is based on the... Of multiple samples $x_i$ and $y$ Fit the coefficients of the polynomial .

Return value interpretation ：

• B： Regression coefficient , I.e. unknown parameter ,B(1) Constant term ,B(2~……) In turn X Each column （ From the second column ） The coefficient of the corresponding term ;
• BINT： Confidence intervals for regression coefficients .（ confidence interval ： When an estimate is given 95% The confidence interval is [a, b] when , It can be understood that we have 95% It can be said that the average value of the sample is between a To b Between , The probability of error is 5%）
• R： residual （ Residual refers to the observed value and predicted value （ Fit value ） The difference between , That is, the difference between the actual observed value and the regression estimated value ）
• RINT： The confidence interval of the residuals .
• STATS： Statistics used to test the regression model . Yes 4 A numerical ： Determination factor $R^2$（ Statistics to measure goodness of fit ,R² The closer the value of 1, It shows that the better the fitting degree of the regression line to the observed value ）,$F$ Statistics Observations , Tested $p$ Value （$p<0.05$ The regression model is established ）, Estimation of error variance .

x1=[3.91 6.67 5.33 5.56 6.12 7.92 5.82 5.5 5.59 6.12 6.68 6.93]';
x2=[9.43 14.5 15.8 19.8 17.4 23.8 31.6 37.1 36.4 32.2 36.6 41.3]';
X=[ones(12,1), x1, x2];
Y=[280 338 405 432 452 582 596 602 606 621 629 656]';

[b,bint,r,rint,stats] = regress(Y,X)
rcoplot(r,rint) %  Draw a residual diagram 

Residual diagram ： The circle in the residual graph is the actual residual of each data point , The horizontal interval is the residual confidence interval , The confidence interval crosses the origin, indicating that the equation fits well ; Does not pass through the origin , It can be regarded as an outlier , For example, the figure above shows that the second group of data does not pass through the origin , Therefore, the fitting results of the second group of data are poor .

After that, the second group of data can be removed and fitted again , More accurate results .

---

%%   Objective function ：y=Ax1^2+Bx2^2+Cx1+Dx2+Ex1*x2+F  （ This is a quadratic function , Two variables , Capital letters are constants ）
format long;

y=[7613.51 7850.91 8381.86 9142.81 10813.6 8631.43 8124.94 9429.79 10230.81 10163.61 9737.56 8561.06 7781.82 7110.97]';
x1=[7666 7704 8148 8571 8679 7704 6471 5870 5289 3815 3335 2927 2758 2591]';
x2=[16.22 16.85 17.93 17.28 17.23 17 19 18.22 16.3 13.37 11.62 10.36 9.83 9.25]';
X=[ones(size(y)) x1.^2 x2.^2 x1 x2 x1.*x2]; %  structure X matrix ！！！

[b,bint,r,rint,stats] = regress(y,X)

give the result as follows ：

b =

   1.0e+04 *

  -1.353935450267780
   0.000000089381408
  -0.005811190715467
  -0.000605427789545
   0.479983626458515
  -0.000037869040292


bint =

   1.0e+04 *

  -2.621944842897225  -0.085926057638335
   0.000000034253753   0.000000144509063
  -0.027588831662544   0.015966450231609
  -0.001309493882546   0.000098638303455
   0.119564693553897   0.840402559363132
  -0.000105954336341   0.000030216255756


r =

   1.0e+02 *

  -4.397667358984126
  -2.361417286008764
  -1.434643909138249
  -5.904203974279353
   7.511701773844997
   5.570806699070599
  -2.447861341816779
   0.494622057474844
   6.376995507987613
  -6.789520765534544
   2.744335484633611
   1.578124015815701
  -0.803533566911865
  -0.137737336155596


rint =

   1.0e+03 *

  -1.219619853471144   0.340086381674319
  -1.426253867770768   0.953970410569015
  -0.919089416302223   0.632160634474573
  -1.568776909577359   0.387936114721488
   0.111430783412043   1.390909571356956
  -0.533006860832905   1.647168200647025
  -1.168898277755904   0.679326009392548
  -0.977546779130818   1.076471190625786
  -0.398546999643731   1.673946101241254
  -1.439044988776064   0.081140835669156
  -0.919162439561023   1.468029536487746
  -1.088788822632291   1.404413625795431
  -1.271236485719865   1.110529772337492
  -1.017369947314269   0.989822480083149


stats =

   1.0e+05 *

   0.000008444011951   0.000086828553270   0.000000043344434   3.162249735298930

Parameters X The first column of is all 1 Column vector , Starting from the second column, it is the column vector composed of the sample data corresponding to each coefficient corresponding term , The objective function of this problem is a bivariate quadratic , because regress Functions can only solve linear problems , So we will $x_1^2$ As the first item ,$x_2^2$ As the second item ,$x_1$ As the third item ,$x_2$ As the fourth item ,$x_1{x}_2$ As the fifth item , This is equivalent to a five variable objective function .

b Is the corresponding parameter b(1) by F( The last constant term ) ,b(2) by A（ The first parameter ）,b(3) by B,b(4) by C,b(4) by D,b(5) by E.

bint by b Of 95% confidence interval .

stats The third parameter of is F Tested p value ,p Small values （p<0.05）, It shows that the fitting model is effective .

application

The relationship between the known randomness parameter and the diversity measure and the convergence measure is shown in the following table , Diversity is as important as convergence , Q. choose the equilibrium point of random parameters .

surface 1 The relationship between the randomness parameter and the diversity measure

x	0.03	0.06	0.09	0.12	0.15	0.18	0.21	0.24	0.27	0.3
y1	0.01	0.01	0.02	0.03	0.06	0.07	0.13	0.17	0.25	0.37

surface 2 The relationship between the randomness parameter and the convergence measure

x	0.03	0.06	0.09	0.12	0.15	0.18	0.21	0.24	0.27	0.3
y1	0.85	0.76	0.68	0.62	0.54	0.52	0.5	0.49	0.48	0.47

x=0.03:0.03:0.3;
y1=[0.01,0.01,0.02,0.03,0.06,0.07,0.13,0.17,0.25,0.37];
y2=[0.85,0.76,0.68,0.62,0.56,0.52,0.49,0.46,0.43,0.39];
plot(x,y1,'*',x,y2,'o');
legend(' diversity ',' The convergence ')

Problem analysis ：

The increase of random parameter leads to the increase of diversity , Reduced convergence ; Both are equally important , Then take the equilibrium point ; The best position of the equilibrium point is where the diversity and convergence are equal .

Solution ：

First step ： The diversity and convergence are fitted respectively , Get the fitting curve .

The second step ： Find the intersection of the two curves .

p1=polyfit(x,y1,2);
p2=polyfit(x,y2,2);
p=p1-p2;
xi=roots(p); % -1.14148334023966	0.316172995412073
xj=0:0.03:0.36;
yj1=polyval(p1,xj);
yj2=polyval(p2,xj);
yi=polyval(p1,xi(2))
plot(x,y1,'*',x,y2,'o',xj,yj1,xj,yj2,xi(2),yi,'rp');

lsqlin function

Because this function is used less , There is less relevant information on the Internet , Therefore, the following ideas are subjective .

Linear least squares solver with boundary or linear constraints .

Solve the least squares curve fitting problem in the following form ：

See , In fact, the official ones are not detailed , Poor sample

Here I will give the simplest example ：

%% y=ax^2+bxsinx+cx^3
xdata = [3.6,7.7,9.3,4.1,8.6,2.8,1.3,7.9,10.0,5.4];
ydata = [16,150.5,260.1,22.5,206.5,9.9,2.7,165.5,325.0,54.5];
C =[ xdata'.^2, xdata'.*sin(xdata'),xdata'.^3]; d = ydata';
[x, resnorm, residual] = lsqlin(C, d);

%%  mapping 
xi = 0:0.1:11;
f = @(c, x) c(1).*x.^2 + c(2).*x.*sin(x) + c(3).*x.^3;
plot(xdata, ydata, '*', xi, f(x, xi))

The disadvantages of functions ： When the upper boundary （ub） Or the lower boundary （lb） or Aeq or beq The restriction in is when the non polynomial term is , These restriction matrices are difficult to construct . such as , The code above will $x^2$、$xsinx$、$x^3$ Regard each as the first 、 Two 、 Three items , Therefore, the constraint matrix must also be the constraint on these three terms , But if the title only gives the right $x$ What about the limitations of ？ If the title only gives the right $x{e}^{x}lgx$ What about the limitations of ？ It is difficult to translate into constraints for each term .

I don't know how to deal with ” Some terms have upper and lower bounds , Other terms have no upper and lower bounds “ The situation of .

Little is known about this function , And there is too little relevant content on the Internet .

lsqcurvefit function

The least square method is used to solve the nonlinear fitting problem ！！！

Invocation format ：

• x = lsqcurvefit(fun,x0,xdata,ydata)

• x = lsqcurvefit(fun,x0,xdata,ydata,lb,ub)

• x = lsqcurvefit(fun,x0,xdata,ydata,lb,ub,options)

• [x,resnorm,residual,exitflag] = lsqcurvefit(…)

Parameter interpretation ：

• x0 Is the parameter to be found （ Initial solution vector ）, The selection of initial variables affects the final solution
• xdata,ydata Is the data used for fitting
• lb、ub Is the lower and upper bounds of the solution vector lb≤x≤ub, If no boundary is specified , be lb=[ ],ub=[ ]
• fun Is the function to be fitted , Calculation x Fitting function value at , There are generally two types of parameters , One is the parameter of the objective function , One is the unknown of the objective function

Return value interpretation ：

• x Is the parameter vector to be solved
• resnorm=sum ((fun(x,xdata)-ydata).^2), That is to say x Sum of squares of residuals at
• residual=fun(x,xdata)-ydata, That is to say x Residual at
• exitflag Conditions for terminating iterations

%% y=tcos(kx)e^(wx)
x=[0,0.4,1.2,2,2.8,3.6,4.4,5.2,6,7.2,8,9.2,10.4,11.6,12.4,13.6,14.4,15]';
y=[1,0.85,0.29,-0.27,-0.53,-0.4,-0.12,0.17,0.28,0.15,-0.03,-0.15,-0.07,0.059,0.08,0.032,-0.015,-0.02]';
f= @(c,x) c(1)*cos(c(2)*x).*exp(c(3)*x);
c0= [0 0 0];
[c, fval]= lsqcurvefit(f, c0, x, y);
xx=0:0.1:20;
yy=f(c, xx);
plot(x, y, 'r*', xx, yy, 'b-');
disp(c);

Be careful ： The functions defined here are quite special , The parameters to be fitted should also be transferred to the function as parameters of the function , The purpose of treating the parameter to be fitted as a vector input function is to make the form of the function more concise .

xdata = [3.6,7.7,9.3,4.1,8.6,2.8,1.3,7.9,10.0,5.4];
ydata = [16,150.5,260.1,22.5,206.5,9.9,2.7,165.5,325.0,54.5];
c0=[ 0 0 0];
[email protected](c, x) c(1)*x.^2 + c(2)*x.*sin(x) + c(3)*x.^3;
[c, resnorm, r]=lsqcurvefit(f_h, c0, xdata, ydata);

%%  mapping 
xx=0:0.1:11;
yy=f_h(c, xx);
plot(xdata, ydata, 'r*', xx, yy, 'b-');
disp(c);

fittype Functions and fit function

Self study , Use less , Baidu can be used directly when necessary （ For those who play games ）.