I need to run a bootstrap on a time series with non-standard dependence. So to do this I need to create a function that simulates the time series by making time by time adjustments.
testing<-function(){
sampleData<-as.zoo(data.frame(index=1:1000,vol=(rnorm(1000))^2,x=NA))
sampleData[,"x"]<-sampleData[,"vol"]+rnorm(1000) #treat this is completely exognenous and unknown in connection to vol
sampleData<-cbind(sampleData,mean=rollmean(sampleData[,"vol"],k=3,align="right"))
sampleData<-cbind(sampleData,vol1=lag(sampleData[,"vol"],k=-1),x1=lag(sampleData[,"x"],k=-1),mean1=lag(sampleData[,"mean"],k=-1))
#get estimate
mod<-lm(vol~vol1+x1+mean1,data=sampleData)
res<-mod$residuals
for(i in 5:1000){
#recursively estimate
sampleData[i,"vol"]<-as.numeric(predict(mod,newdata=data.frame(sampleData[i-1,])))+res[i-3]
#now must update other paramaters
#first our rolled average
sampleData[i,"mean"]<-mean(sampleData[(i-3):i,"vol"])
#reupdate our lagged variables
sampleData[i,"vol1"]<-sampleData[i-1,"vol"]
sampleData[i,"mean1"]<-sampleData[i-1,"mean"]
}
lm(vol~vol1+x1+mean1,data=sampleData)
}
When I run this code and measure the run time I get
system.time(testing())
user system elapsed
2.711 0.201 2.915
This is a slight problem for me as a will be integrating this code to construct a bootstrap. This means any time taken here is multiplied by about 100 for each step. And I am updating this at a few thousand times. That means a single run will take hours (to days) to run.
Is there anyway to speed this code up?
Kind regards,
Matthew
Here's how to avoid the overhead of predict.lm. Also note that I used a matrix instead of a zoo object, which would be a tiny bit slower. You can see just how much this slowed down your code. That's the price you pay for convenience.
testing.jmu <- function() {
if(!require(xts)) stop("xts package not installed")
set.seed(21) # for reproducibility
sampleData <- .xts(data.frame(vol=(rnorm(1000))^2,x=NA), 1:1000)
sampleData$x <- sampleData$vol+rnorm(1000)
sampleData$mean <- rollmean(sampleData$vol, k=3, align="right")
sampleData$vol1 <- lag(sampleData$vol,k=1)
sampleData$x1 <- lag(sampleData$x,k=1)
sampleData$mean1 <- lag(sampleData$mean,k=1)
sampleMatrix <- na.omit(cbind(as.matrix(sampleData),constant=1))
mod.fit <- lm.fit(sampleMatrix[,c("constant","vol1","x1","mean1")],
sampleMatrix[,"vol"])
res.fit <- mod.fit$residuals
for(i in 5:nrow(sampleMatrix)){
sampleMatrix[i,"vol"] <-
sum(sampleMatrix[i-1,c("constant","vol1","x1","mean1")] *
mod.fit$coefficients)+res.fit[i-3]
sampleMatrix[i,"mean"] <- mean(sampleMatrix[(i-3):i,"vol"])
sampleMatrix[i,c("vol1","mean1")] <- sampleMatrix[i-1,c("vol","mean")]
}
lm.fit(sampleMatrix[,c("constant","vol1","x1","mean1")], sampleMatrix[,"vol"])
}
system.time(out <- testing.jmu())
# user system elapsed
# 0.05 0.00 0.05
coef(out)
# constant vol1 x1 mean1
# 1.08787779 -0.06487441 0.03416802 -0.02757601
Add the set.seed(21) call to your function and you'll see that my function returns the same coefficients as yours.
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