I\'m running a rolling regression very similar to the following code:
library(PerformanceAnalytics)
library(quantmod)
data(managers)
FL <- as.formula(Next(HA
The obvious one is to use lm.fit()
instead of lm()
so you don't incur all the overhead in processing the formula etc.
Update: So when I said obvious what I meant to say was blindingly obvious but deceptively difficult to implement!
After a bit of fiddling around, I came up with this
library(PerformanceAnalytics)
library(quantmod)
data(managers)
The first stage is to realise that the model matrix can be prebuilt, so we do that and convert it back to a Zoo object for use with rollapply()
:
mmat2 <- model.frame(Next(HAM1) ~ HAM1 + HAM2 + HAM3 + HAM4, data = managers,
na.action = na.pass)
mmat2 <- cbind.data.frame(mmat2[,1], Intercept = 1, mmat2[,-1])
mmatZ <- as.zoo(mmat2)
Now we need a function that will employ lm.fit()
to do the heavy lifting without having to create design matrices at each iteration:
MyRegression2 <- function(Z) {
## store value we want to predict for
pred <- Z[31, -1, drop = FALSE]
## get rid of any rows with NA in training data
Z <- Z[1:30, ][!rowSums(is.na(Z[1:30,])) > 0, ]
## Next() would lag and leave NA in row 30 for response
## but we precomputed model matrix, so drop last row still in Z
Z <- Z[-nrow(Z),]
## fit the model
fit <- lm.fit(Z[, -1, drop = FALSE], Z[,1])
## get things we need to predict, in case pivoting turned on in lm.fit
p <- fit$rank
p1 <- seq_len(p)
piv <- fit$qr$pivot[p1]
## model coefficients
beta <- fit$coefficients
## this gives the predicted value for row 31 of data passed in
drop(pred[, piv, drop = FALSE] %*% beta[piv])
}
A comparison of timings:
> system.time(Result <- rollapply(managers, 31, FUN="MyRegression",FL,
+ by.column = FALSE, align = "right",
+ na.pad = TRUE))
user system elapsed
0.925 0.002 1.020
>
> system.time(Result2 <- rollapply(mmatZ, 31, FUN = MyRegression2,
+ by.column = FALSE, align = "right",
+ na.pad = TRUE))
user system elapsed
0.048 0.000 0.05
Which affords a pretty reasonable improvement over the original. And now check that the resulting objects are the same:
> all.equal(Result, Result2)
[1] TRUE
Enjoy!