More on Curve Fitting

Transcription

1 PHY 310 DataAnalysis 5 More on Curve Fitting In [1]: # All the standard "import" stuff import scipy as sp from scipy.optimize import curve_fit from scipy.linalg import lstsq import matplotlib.pyplot as plt %matplotlib inline In [2]: # M.L. modifications of matplotlib defaults using syntax of v.2.0 # More info at # Changes can also be put in matplotlibrc file, or effected using mpl.rcparams[] plt.rc('figure', figsize = (6, 4.5)) # Reduces overall size of figures plt.rc('axes', labelsize=16, titlesize=14) plt.rc('xtick', labelsize=14) plt.rc('ytick', labelsize=14) #plt.rc('figure', autolayout = True) # Adjusts supblot parameters for Setting things up for the curve-fitting. The following block is the part that you change to determine what type of function you are fitting to. In its current form, you are fitting to a polynomial. In fact, it's just a linear regression. If you want a higher-order polynomial, then uncomment the X2 line and return X2 for a quadratic function, make a X3 = x**3 line and return X3 for a cubic, etc. On the other hand, if you want to fit to a function f(x) = acos(x) + bsin(x) or whatever, then put in something like "X1 = cos(x)" etc. In [3]: # Basis functions for linear model: func = a0*x0 + a1*x1 + a2*x def basis(x): ''' Returns basis functions for linear model The function to be fit is assumed to be of the form f(x) = a0*x0 + a1*x1 + a2*x2 + a3*x where a0, a1, a2,... are constants, and X0, X1, X2,... are defined below. ''' #Uncomment the next line if you want a quadratic fit # X2 = x**2 X1 = x X0 = 0.*X # Need array of len(x), thus the 0.*X1 # Swap the next two lines if you want quadratic # return sp.array([x0,x1,x2]) return sp.array([x0,x1]) Leave the following two blocks alone. (Although you need to evaluate

2 them again if you change the basis.) In [4]: In [5]: def func(x, a): '''Given basis functions and coefficients, returns value of linear function'' return sp.dot(basis(x).t, a) def LinearModelFit(x, y, u): '''Performs linear least squares fit to a set of 2-d data with uncertainties x = array of x values [x0, x1, x2,...] y = array of values of dependent variable [y0, y1, y2,...] u = array of uncertainties for dependent variable [u0, u1, u2,...] ''' X = basis(x).t # Basis functions evaluated at all x (the X_j(x_i)) of N.R.) W = sp.diag(1/u) # Matrix with uncertainties on diagonal Xw = sp.dot(w,x) # A_ij of Eq. (14.3.4) Yw = sp.dot(y,w) # b_i of Eq. (14.3.5) fit = sp.linalg.lstsq(xw,yw) # lstq returns: best values, chi2,... covariance = sp.linalg.inv(sp.dot(xw.t,xw)) uncertainty = sp.sqrt(sp.diag(covariance)) return(fit[0], uncertainty,fit[1], covariance) Okay, we now have everything set up to do the curve-fitting. Onward to the data... Four data sets for class Change whichever data set you want to use to the name "data" and then take it from there. In [33]: #x-values will be the same for all the data sets xvals = sp.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]) #First data set yvals = sp.array([ , , , , , , errvals = sp.array([0.5, 0.4, 0.6, 0.5, 0.4, 0.6, 0.5, 0.5, 0.6, 0.5]) #Second data set yvals2 = sp.array([ , , , , , , errvals2 = sp.array([0.05, 0.04, 0.06, 0.05, 0.04, 0.06, 0.05, 0.05, 0.06, 0.05]) #Third data set yvals3 = sp.array([ , , , , , , errvals3 = sp.array([0.5, 0.6, 1.2, 1.5, 2.4, 3.6, 3.5, 4.5, 4.6, 5.5]) #Fourth data set yvals4 = sp.array([ , , , , , , errvals4 = sp.array([0.5, 0.4, 0.6, 0.5, 0.4, 0.6, 0.5, 0.5, 0.6, 0.5]) yvals5 = sp.array([ , , , , , , , errvals5 = sp.array([1., 1., 1., 1., 1., 1., 1., 1., 1., 1.])

3 In [34]: #We can plot the values with their correponding errorbars plt.errorbar(xvals,yvals,errvals, fmt = 'k.') plt.xlabel('x') plt.ylabel('y') Out[34]: <matplotlib.text.text at 0x > Okay, now do the fit. Returns a: array of coefficients of basis function unc: array of uncertainties for these coefficients chi2: value of χ 2 (not reduced) cov: covariance matrix In [35]: a,unc,chi2,cov = LinearModelFit(xvals,yvals,errvals) In [36]: print('the fitted coefficients',a) print('uncertainties in those values',unc) The fitted coefficients [ ] Uncertainties in those values [ ] Okay, let's plot again but with the fit so that we can see how it looks.

4 In [37]: xfine = sp.linspace(min(xvals), max(xvals), 201) # "quasi-continuous" set of x's # plt.figure(2) # I'm not sure what this does. # plt.title("data with linear best fit",fontsize=14) # This is here if you want a plt.xlabel('$x$') plt.ylabel('$y$') # plt.axhline(0, color='magenta') # This is here if you want to plot a horizontal # plt.xlim(min(xvals)-0.05*(max(xvals)-min(xvals)), max(xvals)+ 0.05*(max(xvals)-m plt.errorbar(xvals, yvals, yerr=errvals, fmt='o') plt.plot(xfine, func(xfine, a)); Okay, now let's determine residuals

5 In [38]: Out[38]: res=func(xvals,a)-yvals #And let's plot them. plt.xlabel('$x$') plt.ylabel('$residuals$') plt.axhline(0, color='magenta') # Let's plot a horizontal line at zero to make it plt.scatter(xvals,res) <matplotlib.collections.pathcollection at 0x886ec18> Now, for normalized residuals

6 In [39]: Out[39]: normres=res/errvals #And let's plot them. plt.xlabel('$x$') plt.ylabel('$normalized residuals$') plt.axhline(0, color='magenta') # Let's plot a horizontal line at zero to make it plt.axhline(1, color='green') # In fact, let's plot horizontal lines at 1 and -1, plt.axhline(-1, color='green') plt.scatter(xvals,normres) <matplotlib.collections.pathcollection at 0xb6fa8d0> χ 2 χ 2 χ 2 Okay, now we will determine and reduced. We are going to calculate this manually, even though the curve fitting program returns.

7 In [40]: chisq = sp.sum(normres**2) print("chi-squared is",chisq) # And let's check to see if the linear model fit came up with the same chi-squared print("chi-squared returned by the Linear Model Fit function:", chi2) DegreesOfFreedom=len(xvals)-len(a) #Number of points in fit minus number of parame RedChisq = chisq/degreesoffreedom print("degrees of freedom is", DegreesOfFreedom) print("reduced Chi-squared is", RedChisq) Chi-squared is Chi-squared returned by the Linear Model Fit function: Degrees of freedom is 8 Reduced Chi-squared is Okay, we're going to get fancy now. We are going to look at the parameter space and plot contour maps of reduced. χ 2 In [41]: In [42]: def RedChiFunc(slope, intercept): # res=func(xvals,sp.array([intercept,slope]))-yvals res=(intercept+slope*xvals-yvals) normres = res/errvals chisq = sp.sum(normres**2) RedChisq = chisq/degreesoffreedom return RedChisq# # Let's test it out. Presumably, if we put the best values of slope and intercep # we should get the reduced chi-squared that we calculated above. print(a[0],a[1]) print(redchifunc(a[1],a[0])) Okay, we got the same result for the reduced chi-squared, so we probably have the function defined correctly. Now, do a contour plot of the reduced chi-squared.

8 In [43]: numvals=50 minslope=-4.0 maxslope=0.0 minint=0.0 maxint=10.0 slopevals=sp.linspace(minslope,maxslope,numvals) intvals=sp.linspace(minint,maxint,numvals) SLOPE, INT = sp.meshgrid(slopevals,intvals) ZVALS = sp.zeros((len(slope),len(int))) for i in range(len(slope)): for j in range(len(int)): ZVALS[i,j]=RedChiFunc(SLOPE[i,j],INT[i,j])# #ZVALS2 = ZVALS - RedChiFunc(a[1],a[0]) # to subtract off the minimum value of ch #cs=plt.contour(slope,int,zvals2,levels=[1,5,10,20,50],colors="k") cs=plt.contour(slope,int,zvals,levels=[1,5,10,20,50],colors="k") plt.xlabel("slope") plt.ylabel("intercept") plt.clabel(cs) Out[43]: <a list of 4 text.text objects> You want to do a few things here. First, zoom in around the minimum. Second, uncomment the lines that calculate zvals2 which is the reduced χ 2 χbest 2 and then replot the contours. The uncertainties are determined by looking at the contour of χr 2 χbest 2 = 1. If I uncommented the line that does the contour plot of χ 2 χ (i.e., subtract off the minimum value of ), I end up with an oval that extends from about to or so, giving an 2.28 ( 2.55) uncertainty α m or so. And the uncertainty in intercept is roughly 2

9 α b or so. The fitting program came up with uncertainties of around 0.3 for the intercept and 0.05 for slope. The reason for the difference is explained in Taylor and Hase: what I did in the previous paragraph isn't quite right. You shouldn't take the min and max values for the χr 2 1 contour. Instead, you should look at max and mins when cutting horizontally and vertically across the oval through the 2.32 ( 2.48) center. When we do that, we find an uncertainty in slope of 0.08, and an uncertainty in intercept of 0.3 or 0.4, both of which are closer to what the fitting routine 2 returned. For data set #5. Presumably, when you have looked at this, you'll see that a linear fit sucks. Of course, you can add more terms in the polynomial, but another approach is to try a log-log fit. Note that to plot uncertainties in the log of the data, we can use (to first approximation) the calculus f approach: if we define a function f(y) = ln(y), then the uncertainty α f = α 1 =. y y y α y In [26]: In [27]: Out[27]: xvalslog = sp.log(xvals) yvalslog = sp.log(yvals) errvalslog = 1.0/yvals # (I'm putting in 1.0 for alpha_y) #We can plot the values with their correponding errorbars plt.errorbar(xvalslog,yvalslog,errvalslog, fmt = 'k.') plt.xlabel('logx') plt.ylabel('logy') <matplotlib.text.text at 0x8775be0>

10 Note that the error bars get smaller with larger values. (By the way, the error bars really would be asymmetrical here.) Now, do a fit. In [32]: a,unc,chi2,cov = LinearModelFit(xvalslog,yvalslog,errvalslog) print('the fitted coefficients',a) print('uncertainties in those values',unc) print('chi-squared is ', chi2) print('reduced chi-squared is ', chi2/8) #Divide by 8 because 10 data points and The fitted coefficients [ ] Uncertainties in those values [ ] Chi-squared is Reduced chi-squared is Plot with the fit. In [31]: xfine = sp.linspace(min(xvalslog), max(xvalslog), 201) # "quasi-continuous" set o plt.xlabel('$logx$') plt.ylabel('$logy$') plt.errorbar(xvalslog, yvalslog, yerr=errvalslog, fmt='o') plt.plot(xfine, func(xfine, a)); In [ ]:

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