## Using polyfit and polyval functions with data

### jacob Mitch (view profile)

on 17 Nov 2019 at 18:18
Latest activity Commented on by jacob Mitch

### jacob Mitch (view profile)

on 17 Nov 2019 at 22:09

### Matt J (view profile)

I've been using the polyfit and polyval functions recently and I would like to approximate data between 2 vectors but I'm wondering if Ive made a mistake somewhere.
If
a=[1;2;3;4;5;nan;6;nan;nan;10];%prices with data missing
b=[1;2;3;4;5;6;6;7;8;10]; %original prices
days=[1;2;3;4;5;6;7;8;9;10]%number of days
NotMissing=~isnan(a(:,1));
p=a(NotMissing,1); %prices with data not missing
days=a(NotMissing,1);
pf=polyfit(days,p,1);
pv=polyval(pf,days); %polyval approximation of length 7 compared to b which has length 10
%I've want to calculate the mean square error between the approximate prices pv and the original prices using
%E = sqrt( sum( (b-pv).^2) / numel(b) );
Have I made a mistake using the polyval function, in this case I cannot calcuate E because the vectors are different lengths so I'm wondering if Im using the polyfit and polyval functions incorrectly because my length is being reduced by the number of nan values. Or am I 'calculating the error ' between the 2 vectors incorrectly

### Matt J (view profile)

on 17 Nov 2019 at 21:02
Edited by Matt J

### Matt J (view profile)

on 17 Nov 2019 at 21:04

To do what you were after in your post, you need to fit with the NotMissing data, but then apply the fit at all days,
a=[1;2;3;4;5;nan;6;nan;nan;10];%prices with data missing
b=[1;2;3;4;5;6;6;7;8;10]; %original prices
days=[1;2;3;4;5;6;7;8;9;10];%number of days
NotMissing=~isnan(a);
pf=polyfit(days(NotMissing),a(NotMissing),1);
pv=polyval(pf,days)

jacob Mitch

### jacob Mitch (view profile)

on 17 Nov 2019 at 22:09
Hi thanksMatt, I think this is it, I was trying to do a linear approximation which you have done above the pv you achieved is the same vector length as b allowing for E to be calculated. I think I was fitting my data wrong as you suggested. If you'd be so kind I just wanted to check Am I on the right track for interpt1 with something like
a=[1;2;3;4;5;nan;6;nan;nan;10];%prices with data missing
days=[1;2;3;4;5;6;7;8;9;10];
b=[1;2;3;4;5;6;6;7;8;10]; %number of days
NotMissing=~isnan(a);
p=a(NotMissing,1);
days2=days(NotMissing,1);
x=[days2(1):1:days2(end)]
y=interp1(days2,p,x,'spline');

### Matt J (view profile)

on 17 Nov 2019 at 21:05
Edited by Matt J

### Matt J (view profile)

on 17 Nov 2019 at 21:08

I wonder if what you are really trying to do is a fillmissing operation, instead of a linear fit to the data,
a=[1;2;3;4;5;nan;6;nan;nan;10];%prices with data missing
days=[1;2;3;4;5;6;7;8;9;10]%number of days
pv=fillmissing(a,'linear','SamplePoints',days)