FInd maximum in cubic spline interpolant

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YJ
YJ 2014-3-26
编辑： John D'Errico 2017-12-12
I got a set of data, x and y
I was trying to find an eqaution that bu basic fitting and it comes out that spline interpolant is the best one to fit in. (quad and cubic has a peak lower than actual value, so i can not use them)
Now, I want an actual equation that represents a line between two point.
Also I want to find a maximum y-value in a line of cubic spline interpolant, not from the actual data.
I tried findpeak, but it display the y-vale from my data.
So is there a way to do it??
x =
0.7000
0.7500
0.8000
0.8500
0.9000
0.9500
1.0000
1.0500
1.1000
1.1500
1.2000
1.2500
1.3000
y =
1.0e+003 *
1.8367
1.9180
1.9953
2.0677
2.1333
2.1883
2.2251
2.2318
2.2097
2.1746
2.1356
2.0958
2.0562
plot (x,y)
using basic fit tool, clicked on a spline interpolant
and got stuck from here
Cheers,

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2014-3-26
编辑：Andrei Bobrov 2014-3-27
EDIT
new data
x = [0.70000
0.75000
0.80000
0.85000
0.90000
0.95000
1.00000
1.05000
1.10000
1.15000
1.20000
1.25000
1.30000];
y = [1836.7
1918.0
1995.3
2067.7
2133.3
2188.3
2225.1
2231.8
2209.7
2174.6
2135.6
2095.8
2056.2];
for MATLAB R2013a and later
[~,ii] = findpeaks(y);
F = griddedInterpolant(x,y,'spline');
xmaxs = arrayfun(@(xx)fminsearch(@(x)-F(x),xx),x(ii));
x1 = linspace(x(1),x(end),100);
plot(x,y,'go',x1,F(x1),'b-',xmaxs,F(xmaxs),'r+');
for older releases MATLAB
[~,ii] = findpeaks(y);
F = interp1(x,y,'spline','pp');
xmaxs = arrayfun(@(xx)fminsearch(@(x)-ppval(F,x),xx),x(ii));
x1 = linspace(x(1),x(end),100);
plot(x,y,'go',x1,ppval(F,x1),'b-',xmaxs,ppval(F,xmaxs),'r+');

更多回答（2 个）

Mark Kittisopikul
Mark Kittisopikul 2017-12-12
You can calculate the maximum by using calculus by finding a zero of the first derivative where the second derivative is negative.
x = [0.70000
0.75000
0.80000
0.85000
0.90000
0.95000
1.00000
1.05000
1.10000
1.15000
1.20000
1.25000
1.30000];
y = [1836.7
1918.0
1995.3
2067.7
2133.3
2188.3
2225.1
2231.8
2209.7
2174.6
2135.6
2095.8
2056.2];
Find the spline interpolant in piecewise polynomial form.
>> sp = spline(x,y)
sp =
struct with fields:
form: 'pp'
breaks: [1×13 double]
coefs: [12×4 double]
pieces: 12
order: 4
dim: 1
Take the derivative of the spline:
>> fnder(sp)
ans =
struct with fields:
form: 'pp'
breaks: [1×13 double]
coefs: [12×3 double]
pieces: 12
order: 3
dim: 1
Find the zeros of the 1st derivative:
>> fnzeros(fnder(sp))
ans =
1.0352
1.0352
Confirm the second derivative is negative:
>> fnval(fnder(fnder(sp)),fnzeros(fnder(sp)))
ans =
1.0e+04 *
-1.2996
-1.2996
Check the zero of the first derivative:
>> fnval(fnder(sp),fnzeros(fnder(sp)))
ans =
1.0e-11 *
0.2558
0.2558
Plot a nice figure for visual verification.
>> figure; fnplt(sp);
hold on; plot(fnzeros(fnder(sp)),fnval(sp,fnzeros(fnder(sp))),'o');
grid on
hold on; plot(fnzeros(fnder(sp)).',ylim.','-');
hold on; plot(x,y,'ko');
John D'Errico
John D'Errico 2017-12-12
编辑：John D'Errico 2017-12-12
Too late possibly, but you can always use my SLM toolbox , which has tools that can operate on splines created by tools like spline or pchip.
Based on the vectors x and y as you posed...
pp = spline(x,y);
[maxf,maxloc] = slmpar(pp,'maxfun')
maxf =
2233.2
maxloc =
1.0352
This is the global maximum value of the spline created in pp.
Find SLM for free download from the file exchange .

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