[R-SIG-Finance] Error in lm prediction

Kevin Dhingra kevin.dhingra at appliedacademics.com
Fri Jul 28 15:58:57 CEST 2017

Hi Amol,

The lm function is not intended to be used in the way you are calling it.
Even though you can actually pass y and x as actual data in the formula
argument (y~x), its better to pass the data set in the data argument and
use column names in the formula argument especially when you want to use
the predict function on the fitted object as predict.lm looks for variables
in the function environment. In your example, newdata and those variables
would not have similar length that results in length of y_hat equal to 300.

Now there might be some clever way to get around this with the same
function call that you used (you can try playing with the variable name of
new data to be same as column names in x) but I would rather suggest using
this -

data_fit = data.frame(x = matrix(rnorm(1700*5), ncol = 5), y =
data_fit_is = data_fit[1:a,] #In Sample
data_fit_os = data_fit[(a+1):nrow(data_fit), ] #Out of Sample
m1 = lm(y~., data = data_fit_is)
length(predict(m1, data_fit_os[, 1:5])) #Should be equal to 1400 now and
300 now

Kshitij Dhingra

On Sun, Jul 16, 2017 at 4:31 PM, amol gupta <amolgupta87 at gmail.com> wrote:

> Hi
> I am most likely committing an error in trying to predict  using linear
> regression lm model. please help me figure out what am I doing wrong. I am
> trying to regress a index and its constituents. Here is the code
> #split ts inttwo parts
> a<-300;
> x1<-x[1:a,];
> y1<-y[1:a,];
> x2<-x[(a+1):nrow(x),];
> y2<-y[(a+1):nrow(y),];
> #regression
> m1<-lm( y1~x1)
> r1<-residuals(m1)
> coef(m1)
> ##out of sample
> y_hat<-predict.lm(m1,x2);
> r2<-y_hat-y2;
> x,y are xts. X contains multiple time series. The y_ hat turns out to be of
> 300 samples only, whereas x2 contains 1400 samples.
> Please help me figure out how to predict using model that I have found
> using regression.
> --
> Regards
> Amol
> +91-9897860992
> +91-8889676918
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Kshitij Dhingra
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