Fit a spline function to a set of data points

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Deepa Maheshvare
Deepa Maheshvare 2022 年 7 月 3 日
コメント済み: Deepa Maheshvare 2022 年 7 月 4 日
Hi All,
I have a set of coordinate points
x= [0,4,6,10,15,20]
y = [18,17.5,13,12,8,10]
I want to fit a spline function.
Could someone please suggest the the curve fit function that can be used?
  7 件のコメント
Deepa Maheshvare
Deepa Maheshvare 2022 年 7 月 3 日
x = [0,4,6,10,15,20]
y = [18,17.5,13,12,8,10]
vq = interp1(x,y,'linear')
returns NaN
Deepa Maheshvare
Deepa Maheshvare 2022 年 7 月 3 日
I tried
x = [0,4,6,10,15,20]
v = [18,17.5,13,12,8,10]
vq = interp1(x,v, 'linear', 'pp')
xq = x
figure
vq1 = interp1(x,v,xq);
plot(x,v,'o',xq,vq1,':.');
title('(Default) Linear Interpolation');
returns
vq =
struct with fields:
form: 'pp'
breaks: [0 4 6 10 15 20]
coefs: [5×2 double]
pieces: 5
order: 2
dim: 1
orient: 'first'
Here I am not sure why the order (which I infer as the polynomial degree) is 2 for linear interpolation. Could you please explain this a bit?

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採用された回答

Sam Chak
Sam Chak 2022 年 7 月 3 日
Thought you want the 'spline'? Is there a reason to use the 'linear' method?
x = [0, 4, 6, 10, 15, 20];
y = [18, 17.5, 13, 12, 8, 10];
xq = 0:0.01:20;
subplot(2,1,1)
yq1 = interp1(x, y, xq, 'linear');
plot(x, y, 'o', xq, yq1, '--');
ylim([5 22]), grid on, title('Linear Interpolation');
subplot(2,1,2)
yq2 = interp1(x, y, xq, 'spline');
plot(x, y, 'o', xq, yq2, '--');
ylim([5 22]), grid on, title('Spline Interpolation');
  3 件のコメント
Torsten
Torsten 2022 年 7 月 3 日
Seems that order = 2 means that two coefficients are needed to describe the (linear) spline function on each subinterval and is set to 1 + degree of the interpolating polynomial. A bit confusing in my opinion.
Deepa Maheshvare
Deepa Maheshvare 2022 年 7 月 4 日
Thanks a lot for the clarification.

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その他の回答 (1 件)

Simon Chan
Simon Chan 2022 年 7 月 3 日
Use function spline

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