Linear regression on training set

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!

[J. Alex Lee]

you already identified that your regression can be made into linear form, so that's already a big hint for you...

[katara]

Yeah so, I tried rewriting the function as log(y)=log(y0) + rt and then using polyfit(t, log(y),1) but since y0 is unknown that doesn't work.

[katara]

I just realized I could just name a new variable y = log(y) and use polyfit from there. So my code is:

A=[130, 300, 400, 500, 650, 1075, 2222, 2550, 3300]';

t = [1930, 1943, 1966, 1976, 1991, 1994, 2000, 2005, 2008];

t1=[1930, 1943, 1966, 1976, 1991];

idx=randperm(numel(A));
subSet1=A(idx(1:5)); %Trainingset
subSet2=A(idx(6:end)); %Validationset

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,'*')

But now I have chosen that t1 is the first five years of t, which won't correspond correctly to the randomly chosen values of the training set. Is there a way of choosing five t values that will correspond to the randomly chosen values?

[Johannes Hougaard]

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);

[Adam Danz]

Johannes has the right approach (maybe it can be written as an answer). It can be generalized to any size dataset using

idx = randperm(numel(A)); 
nTrain = ceil(numel(A)/2); 
% nTest = numel(A)-nTrain; % if needed
trainIdx = 1:nTrain; 
testIdx = nTrain+1 : numel(A); 

trainSet = [A(trainIdx); t(trainIdx)]; % assuming A and t are row vectors
testSet = [A(testIdx); t(testIdx)];    % same assumptionx

% Then proceed with fitting on the trainSet and measuring 
% error on the testSet

Also note that if you're planning on using a more rigorous cross validation, use cvpartition to partition your data.

[katara]

Thank you!

One question to Johannes, how can I plot the polyfit using polyval. In other problems I have used for example:

c=polyfit(t, temp, 2)
x=polyval(c,t)
plot(t,temp,'*', t, x)

However, for this problem I tried:

y=log(subSet1);
c=polyfit(t1,y, 1)
p=polyval(c,t);

r=c(1);
lny0=(c(2));
y0=exp(c(2));

y2 = y0*exp(r*t);
plot(t,y2,'*',t,p)

And it didn't work. The code You wrote with polyval didn't work either.

The whole code is now:

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)
p=polyval(c,t);

r=c(1);
lny0=(c(2));
y0=exp(c(2));

y2 = y0*exp(r*t);
plot(t,y2,'*',t,p)

[J. Alex Lee]

you just need to exponentiate the result of polyval (remember you took the log), and I would wager the plot you really want is

plot(t,A,'*',t,exp(polyval(c,t)))

Or if I may:

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
t2=t(idx(6:end)); %t values for Trainingset

y=log(subSet1);
c=polyfit(t1,y, 1)
p=polyval(c,t);

r=c(1);
y0=exp(c(2));

yMdlFn = @(t)(y0*exp(r*t));

% to evaluate on test set
yMdlTest = yMdlFn(t2)

% more comprehensive plot
figure(1); cla; hold on
plot(t1,subSet1,'*')
plot(t2,subSet2,'o')
fplot(yMdlFn,[1929,2009])

But also recommend implement Adam's generalization to arbitrarily large data sets partitioned into arbitrarily sized training and test sets (although i think the code posted doesn't work)

[Image Analyst]

If you want a log fit, use fitnlm() rather than polyfit().

[J. Alex Lee]

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.

Answer 1

[Johannes Hougaard]

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);