Linear regression on training set

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katara
katara 2020 年 9 月 10 日
回答済み: Johannes Hougaard 2020 年 9 月 11 日
I have some data that I want to divide into a training set and a validation set in order to do linear regression on the training set to find y0 and r. The training set should contain at least 50% of the data. My code so far is that below:
A=[130, 300, 400, 500, 650, 1075, 2222, 2550, 3300]';
t = [1930, 1943, 1966, 1976, 1991, 1994, 2000, 2005, 2008];
idx=randperm(numel(A))
subSet1=A(idx(1:5)) %Trainingset
subSet2=A(idx(6:end)) %Validationset
If I can assume the function is exponential and is y(t)= y0*e^rt how do I continue to plot the training set to find y0 and r?
Thankful for all help!
  9 件のコメント
J. Alex Lee
J. Alex Lee 2020 年 9 月 10 日
i would take linear least squares anywhere i can get it, including this situation. linear fitting doesn't require initial guesses and guaranteed to give a "result", and is faster. now you could use the result of the polyfit to do a nonlinear fit, if you want to define the least squares differently. But you're still left with a choice on how to define your residual anyway, so you have a lot more things to worrry about if you care to that level with nonlinear fitting.

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回答 (1 件)

Johannes Hougaard
Johannes Hougaard 2020 年 9 月 11 日
the five t values that will correspond to the randomly chosen values are used by using the idx vector similarly to what you do for A.
A=[130, 300, 400, 500, 650, 1075, 2222, 2550, 3300]';
t = [1930, 1943, 1966, 1976, 1991, 1994, 2000, 2005, 2008];
idx=randperm(numel(A));
subSet1=A(idx(1:5)); %Trainingset
subSet2=A(idx(6:end)); %Validationset
t1 = t(idx(1:5)); %t values for Trainingset
y=log(subSet1);
c=polyfit(t1,y, 1)
r=c(1);
lny0=c(2);
y0=exp(c(2));
y2 = y0*exp(r*t);
plot(t,y2,'*')
And to apply your polyfit result you could just use polyval.
% Or you could use
y2 = exp(polyval(c,t));
plot(t,y2);

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