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Interpolating through spline and then get corresponding values

조회 수: 2 (최근 30일)
Hello, I have a certain set of points which create a curve. I want to evaluate what is the y value at other x locations, not given in the problem.
Unfortunately, there is no polynomial which can passes though all these points correctly.
I am sure other people might have asked this question, but I could not find anything.
The X values are:
if true
% code
x=[0
0.00420000000000000
0.00800000000000000
0.0125000000000000
0.0177000000000000
0.0236000000000000
0.0305000000000000
0.0385000000000000
0.0476000000000000
0.0581000000000000
0.0701000000000000
0.0838000000000000
0.0993000000000000
0.116700000000000
0.136300000000000
0.158200000000000
0.182500000000000
0.209200000000000
0.238400000000000
0.269800000000000
0.303600000000000
0.339300000000000
0.376800000000000
0.415600000000000
0.455300000000000
0.495400000000000
0.535600000000000
0.575100000000000
0.613700000000000
0.650900000000000
0.686200000000000
0.719500000000000
0.750500000000000
0.779200000000000
0.805400000000000
0.829200000000000
0.850600000000000
0.869800000000000
0.886900000000000
0.902000000000000
0.915300000000000
0.927000000000000
0.937200000000000
0.946100000000000
0.953900000000000
0.960600000000000
0.965400000000000
0.970400000000000
0.975400000000000
0.980200000000000
0.985200000000000
0.990200000000000
0.995100000000000
1];
end
The y values are:
if true
% code
y=[0.127489040000000
0.111766040000000
0.0167797700000000
0.0337462690000000
0.0266396250000000
0.111117580000000
0.130855270000000
0.138929860000000
0.163963910000000
0.134142850000000
0.150978780000000
0.151461170000000
0.142033560000000
0.143054720000000
0.142033560000000
0.139969710000000
0.139452300000000
0.138929860000000
0.137879260000000
0.141520320000000
0.143562680000000
0.146571800000000
0.159900150000000
0.167058910000000
0.178109250000000
0.196599630000000
0.216707350000000
0.225149480000000
0.251330640000000
0.286469960000000
0.304243410000000
0.330819300000000
0.359129760000000
0.389805860000000
0.429068680000000
0.448364490000000
0.485802150000000
0.531624250000000
0.547334130000000
0.565165850000000
0.582247920000000
0.598792800000000
0.619664680000000
0.639333560000000
0.638537430000000
0.668565600000000
0.713292490000000
0.687499690000000
0.660858750000000
0.633757700000000
0.628645890000000
0.672958870000000
0.660858750000000
0.623358350000000];
end
And the values where I want to evaluate are:
if true
% code
xnew=[0.0250000000000000
0.0500000000000000
0.0750000000000000
0.100000000000000
0.150000000000000
0.200000000000000
0.300000000000000
0.400000000000000
0.500000000000000
0.600000000000000
0.700000000000000
0.800000000000000
0.900000000000000
0.950000000000000];
end
The way I would do is by linear interpolation through the different points and then get for each interval the coefficients of the polynomial and input my data and check if they are within certain limits and use poleval. I think this way is a bit long, and it might be possible to fit a spline or other manners.
Thanks a lot for any suggestions.
Antonio

채택된 답변

Azzi Abdelmalek
Azzi Abdelmalek 2012년 9월 12일
편집: Azzi Abdelmalek 2012년 9월 12일
method='linear'
ynew=interp1(x,y,xnew,method)
%or
method='smoothingspline'
f=fit(x,y,method);
ynew=f(xnew)
  댓글 수: 1
mortain
mortain 2012년 9월 12일
편집: mortain 2012년 9월 12일
Thanks a lot, I was sure there was an easy manner....two lines :-)! Apparently the linear matches better.

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