12
votes

I am running rolling regressions in R, using with the data stored in a data.table.

I have a working version, however it feels like a hack -- I am really using what i know from the zoo package, and none of the magic in data.table ... thus, it feels slower than it ought to be.

Incorporating Joshua's suggestion - below - there is a speedup of ~12x by using lm.fit rather than lm.

(revised) Example code:

require(zoo)
require(data.table)
require(rbenchmark)
set.seed(1)

tt <- seq(as.Date("2011-01-01"), as.Date("2012-01-01"), by="day")
px <- rnorm(366, 95, 1)

DT <- data.table(period=tt, pvec=px)

dtt <- DT[,tnum:=as.numeric(period)][, list(pvec, tnum)]
dtx <- as.matrix(DT[,tnum:=as.numeric(period)][, tnum2:= tnum^2][, int:=1][, list(pvec, int, tnum, tnum2)])

rollreg <- function(dd) coef(lm(pvec ~ tnum + I(tnum^2), data=as.data.frame(dd)))
rollreg.fit <- function(dd) coef(lm.fit(y=dd[,1], x=dd[,-1]))

rr <- function(dd) rollapplyr(dd, width=20, FUN = rollreg, by.column=FALSE)
rr.fit <- function(dd) rollapplyr(dd, width=20, FUN = rollreg.fit, by.column=FALSE)

bmk <- benchmark(rr(dtt), rr.fit(dtx), 
         columns = c('test', 'elapsed', 'relative'),
         replications = 10,
         order = 'elapsed'
       )

     test elapsed relative
2 rr.fit(dtx)    0.48   1.0000
1     rr(dtt)    5.85  12.1875

Trying to apply the knowledge displayed here and here, I cooked up the following simple rolling regression function that I think uses some of the speed of data.table operations.

Note that the problem is a little different (and more realistic): take a vector, add lags, and regress on itself. This class of AR-type problems is pretty broad.

I am sharing it here as it may be of use, and i'm sure that it can be improved (i'll update as I improve):

require(data.table)
set.seed(1)
x  <- rnorm(1000)
DT <- data.table(x)
DTin <- data.table(x)

lagDT <- function(DTin, varname, l=5)
{
    i = 0
    while ( i < l){
        expr <- parse(text = 
                  paste0(varname, '_L', (i+1), 
                     ':= c(rep(NA, (1+i)),', varname, '[-((length(',     varname, ') - i):length(', varname, '))])'
                 )
              )
    DTin[, eval(expr)]
    i <- i + 1
}
return(DTin)
}   

rollRegDT <- function(DTin, varname, k=20, l=5)
{
adj <- k + l - 1
.x <- 1:(nrow(DTin)-adj)
DTin[, int:=1]
dtReg <- function(dd) coef(lm.fit(y=dd[-c(1:l),1], x=dd[-c(1:l),-1]))
eleNum <- nrow(DTin)*(l+1)
outMatx <- matrix(rep(NA, eleNum), ncol = (l+1))
colnames(outMatx) <- c('intercept', 'L1', 'L2', 'L3', 'L4', 'L5')
for (i in .x){
    dt_m <- as.matrix(lagDT(DTin[i:(i+adj), ], varname, l))
    outMatx[(i+(adj)),] <- dtReg(dt_m)
}
return(outMatx)
}

rollCoef <- rollRegDT(DT, varname='x')
2
Is this question similar to stackoverflow.com/questions/11873123/… ?Sameer
Use lm.fit directly and avoid the overhead of the lm function.Joshua Ulrich
Thanks Joshua. @Sameer, just had a look at your direct method, and will try it. I think there is similarity to the extent that your direct approach may suit both - differences being that i have the data in a data.table, and that my real worl problem is likely to be long rather than wide.ricardo
thanks Josh, speedup was ~11.5x (using the actual data and rbenchmark).ricardo
the colnames(outMatx) is hard coded (whereas your function allows for a custom 'l') ..user555265

2 Answers

5
votes

Not as far as I know; data.table doesn't have any special features for rolling windows. Other packages already implement rolling functionality on vectors, so they can be used in the j of data.table. If they are not efficient enough, and no package has faster versions (?), then it's a case of writing faster versions yourself and (of course) contributing them: either to an existing package or creating your own.

Related questions (follow links in links) :

Using data.table to speed up rollapply
R data.table sliding window
Rolling regression over multiple columns in R

2
votes

You can do 14585 / 766 ~ 19 times faster with the roll_regres function from the rollRegres package

require(zoo)
require(data.table)
require(microbenchmark)
set.seed(1)

tt <- seq(as.Date("2011-01-01"), as.Date("2012-01-01"), by="day")
px <- rnorm(366, 95, 1)

DT <- data.table(period=tt, pvec=px)

dtt <- DT[,tnum:=as.numeric(period)][, list(pvec, tnum)]

# this is a quite bad problem as tnum and the square has a high cor
cor(dtt$tnum, dtt$tnum^2)
#R [1] 0.9999951

# so we center it to avoid numerical issues in the comparisons
dtt$tnum <- dtt$tnum - mean(dtt$tnum)
cor(dtt$tnum, dtt$tnum^2)
#R [1] -2.355659e-21

dtx <- as.matrix(DT[,tnum:=as.numeric(period)][, tnum2:= tnum^2][, int:=1][, list(pvec, int, tnum, tnum2)])

rollreg <- function(dd)
  coef(lm(pvec ~ tnum + I(tnum^2), data = as.data.frame(dd)))
rollreg.fit <- function(dd) coef(lm.fit(y=dd[,1], x=dd[,-1]))

rr <- function(dd) rollapplyr(
  dd, width=20, FUN = rollreg, by.column = FALSE, align = "right")
rr.fit <- function(dd) rollapplyr(
  dd, width=20, FUN = rollreg.fit, by.column = FALSE, align = "right")

#####
# use rollRegres
library(rollRegres)
rollreg_out    <- rr(dtt)
rollRegres_out <- roll_regres(pvec ~ tnum + I(tnum^2), dtt, width = 20L)

# show that they give the same
all.equal(rollRegres_out$coefs[-(1:19), ], rollreg_out,
          check.attributes = FALSE)
#R [1] "Mean relative difference: 4.985435e-08"

#####
# benchmark
microbenchmark(
  rr = rr(dtt),
  rr.fit = rr.fit(dtx),
  roll_regres = roll_regres(pvec ~ tnum + I(tnum^2), dtt ,width = 20L),
  times = 5)
#R Unit: microseconds
#R expr        min         lq        mean     median         uq       max neval
#R          rr 279404.357 279456.901 282071.3414 279989.840 282201.396 289304.21     5
#R      rr.fit  13744.598  14017.981  14585.2106  14147.166  14887.117  16129.19     5
#R roll_regres    621.037    660.939    766.7364    721.383    843.853    986.47     5