Optimizing search in time series data frame - r

Question

I have a data frame of 50 columns by 2.5 million rows in R, representing a time series. The time column is of class POSIXct. For analysis, I repeatedly need to find the state of the system for a given class at a particular time.
My current approach is the following (simplified and reproducible):
set.seed(1)
N <- 10000
.time <- sort(sample(1:(100*N),N))
class(.time) <- c("POSIXct", "POSIXt")
df <- data.frame(
time=.time,
distance1=sort(sample(1:(100*N),N)),
distance2=sort(sample(1:(100*N),N)),
letter=sample(letters,N,replace=TRUE)
)
# state search function
time.state <- function(df,searchtime,searchclass){
# find all rows in between the searchtime and a while (here 10k seconds)
# before that
rows <- which(findInterval(df$time,c(searchtime-10000,searchtime))==1)
# find the latest state of the given class within the search interval
return(rev(rows)[match(T,rev(df[rows,"letter"]==searchclass))])
}
# evaluate the function to retrieve the latest known state of the system
# at time 500,000.
df[time.state(df,500000,"a"),]
However, the call to which is very costly. Alternatively, I could first filter by class and then find the time, but that doesn't change the evaluation time much. According to Rprof, it's which and == that cost the majority of the time.
Is there a more efficient solution? The time points are sorted weakly increasing.

Answer 1

Because which, == and [ are all linear with the size of the data frame, the solution is to generate subset data frames for bulk operations, as follows:
# function that applies time.state to a series of time/class cominations
time.states <- function(df,times,classes,day.length=24){
result <- vector("list",length(times))
day.end <- 0
for(i in 1:length(times)){
if(times[i] > day.end){
# create subset interval from 1h before to 24h after
day.begin <- times[i]-60*60
day.end <- times[i]+day.length*60*60
df.subset <- df[findInterval(df$time,c(day.begin,day.end))==1,]
}
# save the resulting row from data frame
result[[i]] <- df.subset[time.state(df.subset,times[i],classes[i]),]
}
return(do.call("rbind",result))
}
With dT=diff(range(df$times)) and dT/day.length large, this reduces the evaluation time with a factor of dT/(day.length+1).