Dont know what Polyval is here doing :-/

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Daniel Wurster 2019-5-31

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评论： dpb 2019-6-1

采纳的回答： dpb

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Hello Together, i dont know what polyval is doing here with the blue lines:

What i want is a line through all points (like the black one).

Code for it is at end.

As you can see, the black line with polyval is plotted correct.
The points yd1 for xd1 and yd2 for xd2 are plotted correct
What is not plotted correct, is the polyval for xd1 and xd2 -> What have i done wrong here?

Thanks a lot!

%Black line
xml1 = [ -0.3,  0.4];
yml1 = [-0.15, -0.15];
xml2 = 0.4 + 0.7 * cos(0:pi/50:pi/2) ;
yml2 = -0.85 +0.7 * sin(0:pi/50:pi/2);
xml3 = [1.1, 1.1];
yml3 = [-0.85, -1.15];
load ('xdata310519.mat')
%xd1 =  0    0.3906    0.7430    1.0227    1.2022    1.2641    1.2022
%yd1 =  0   -0.0619   -0.2414   -0.5211   -0.8735   -1.2641   -1.6547
%xd2 =  0    0.1943    0.3696    0.5087    0.5980    0.6288    0.5980
%yd2 =  0   -0.0308   -0.1201   -0.2592   -0.4345   -0.6288   -0.8231
grad = 5;
p1 = polyfit( xd1, yd1, grad);
p2 = polyfit( xd1, yd2, grad);
f1 = polyval( p1, xd1);
f2 = polyval( p2, xd2);
hold on
%Black line
plot(xml1, yml1, '--k');
plot(xml2, yml2,'--k');
plot(xml3, yml3, '--k');
%points
plot(xd1, yd1,'s', xd2, yd2,'s');
plot(xd1,  f1,'b', xd2, f2,'b');

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Answer 1

dpb 2019-5-31

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/465009-dont-know-what-polyval-is-here-doing#answer_377411

编辑：dpb 2019-5-31

在 MATLAB Online 中打开

You fit yd2 vs xd1 but then plotted against xd2. One presumes the fit should be against xd2 as well.

BUT: A perfect illustration why higher order polynomials are rarely the right fitting function--what your fit really is is shown below--

where this was generated from

p2 = polyfit( xd2, yd2, grad);               % fit against intended variable
figure
plot(xd2,yd2,'bx')                           % plot the data points
hold on
plot(xd2,f2,'b-')                            % and the fit at only those points (linear between)
x2=linspace(min(xd2),max(xd2));              % fill in 100 points between first, last
plot(x2,polyval(p2,x2),'r-')                 % and plot those as well...
legend('Data','5th Order at X','5th Order at 100X')

A polynomial is an exceedingly bad choice for such data; use an interpolating spline instead or something similar.

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Daniel Wurster 2019-6-1

编辑：Daniel Wurster 2019-6-1

在 MATLAB Online 中打开

You fit yd2 vs xd1 but then plotted against xd2.  One presumes the fit should be against xd2 as well.

Yes thats right. My bad. I fixed that and it looks now like:

I would add only that the data is highly non-polynomial, although it looks like the relation where x is viewed as a function of y DOES seem to be polynomial in nature.

The data xd1, yd1 and xd2, yd2 is based on sinus and cosinus. So its polynomial.

My overall plan is:

Create datapoints to draw a function through them
Calculate the arc length of the function to compare it.

For 1. i did the following code:

box on; grid on; hold on;
axis ([ -0.5 ,2 , -2 ,2]);
n1 = -pi/2:pi/180:pi/8;
%To get the data points i wrote
%n1 = -pi/2:pi/10:pi/8;
%but to draw it nice i wrote pi/180
i=1:9;
xr1 = 1.2641 * cos(-n1);
yr1 = 1.2641 * sin(-n1) -1.2641;
xr2 = 0.6288 * cos(-n1);
yr2 = 0.6288 * sin(-n1) -0.6288;
h1 = plot(xr1, yr1);
h2 = plot(xr2, yr2);
xd1 = (get(h1, 'XData'))
yd1 = (get(h1, 'YData'))
%save('xdata1.mat','xdata1', 'ydata1');
xd2 = (get(h2, 'XData'))
yd2 = (get(h2, 'YData'))
save('xdata3105.mat','xd1', 'yd1','xd2', 'yd2');
clear all

Which gives me:

For 2. i need to have a polynomial function through the data points to calculte the arc length.

dpb 2019-6-1

If it is the length, then John's FEX submission should be just the ticket <fileexchange/34871-arclength>

dpb 2019-6-1

在 MATLAB Online 中打开

One last fling with respect to modelling the given shape as polynomial -- after your code above then:

xlim([-0.1 1]); ylim([-1 0.5]) % blow up the region of interest to see more detail
b2=polyfit(xd2,yd2,2);         % fit a quadratic in xd2 and add to plot
yhat2=polyval(b2,xd2);
hLY2=plot(xd2,yhat2,'g-');
b2H=polyfit(yd2,xd2,2);          % As J d E observed far more nearly polynomic in Y
xhat2=polyval(b2H,yd2);          % now we're predicting x instead from y...
hLX2=plot(xhat2,yd2,'k-');       % add this possible approximation as well...
legend('YD1','YD2','Quadratic in XD2','Quadratic in YD2')

As can be seen, then open to left parabola comes at least with the right shape and can produce the symmetry axis in the proper direction but doesn't fit the actual data all that closely.

You could, of course, continue to add terms to try to improve, but still the spline will fit directly and is, overall, the far better option here.

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