Hi
I would like to understand how to extend the function (FUN) I am using in
rollapply below.
######################################
With the following simplified data, test1 yields parameters for a rolling
regression
data = data.frame(Xvar=c(70.67,70.54,69.87,69.51,70.69,72.66,72.65,73.36),
Yvar =c(78.01,77.07,77.35,76.72,77.49,78.70,77.78,79.58))
data.z = zoo(d)
test1 = rollapply(data.z, width=3,
FUN = function(z) coef(lm(z[,1]~z[,2],
data=as.data.frame(z))), by.column = FALSE, align = "right")
print(test1)
######################################
Rewriting this to call myfn1 gives test2 (and is consistent with test1
above)
myfn1 = function(mydata){
dd = as.data.frame(mydata)
l = lm(dd[,1]~dd[,2], data=dd)
c = coef(l)
}
test2 = rollapply(data.z, width=3,
FUN= myfn1, by.column = FALSE, align = "right")
print(test2)
######################################
I would like to be able to use the predict function to obtain a prediction
(and its std error) from the rolling regression I have just calculated.
My effort below issues a warning that 'newdata' had 1 row but variable(s)
found have 3 rows.
(if I run this outside of rollapply I don't get this warning)
Also, I don't see the predicted value or its se with print(fm2[[1]]). Again,
if I run this outside of rollapply I am able to extract the predicted value.
Xpred=c(70.67)
myfn2 = function(mydata){
dd = as.data.frame(mydata)
l = lm(dd[,1]~dd[,2], data=dd)
c = coef(l)
p = predict(l, data.frame(Xvar=Xpred),se=T)
ret=c(l,c,p)
}
fm2 = rollapply(data.z, width=3,
FUN= myfn2, by.column = FALSE, align = "right")
print(fm2[[1]])
Any insights would be gratefully received.
Best regards
Pete
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