Vfrdtyky

I have a data frame (da) where each row has a timestamp in ascending order (intervals between each timestamp is random).

I wanted to keep rows of da based on whether its time fell within the times found between those in two other vectors (first.times and second.times). So I'd go down the vectors of first.time and second.time iteratively and see if da has times within those intervals (min = first times and max = second.times), with which I keep, and the rest I don't.

The only way I've figured out how to do it is with a for loop, but it can take a while. Here's the code with some example data:

#Set start and end dates

date1 <- as.POSIXct(strptime('1970-01-01 00:00', format = '%Y-%m-%d %H:%M'))

date2 <- as.POSIXct(strptime('1970-01-05 23:00', format = '%Y-%m-%d %H:%M'))



#Interpolate 250000 dates in between (dates are set to random intervals)

dates <- c(date1 + cumsum(c(0, round(runif(250000, 20, 200)))), date2)



#Set up dataframe

da <- data.frame(dates = dates,

                 a = round(runif(1, 1, 10)),

                 b = rep(c('Hi', 'There', 'Everyone'), length.out = length(dates)))

head(da); dim(da)



#Set up vectors of time

first.times <- seq(date1,      #First time in sequence is date1

                   date2,      #Last time in sequence is date2

                   by = 13*60) #Interval of 13 minutes between each time (13 min * 60 sec)



second.times <- first.times + 5*60 #Second time is 5 min * 60 seconds later

head(first.times); length(first.times)

head(second.times); length(second.times)



#Loop to obtain rows

subsetted.dates <- da[0,]

system.time(for(i in 1:length(first.times)){

  subsetted.dates <- rbind(subsetted.dates, da[da$dates >= first.times[i] & da$dates < second.times[i],])

})

 user  system elapsed 

2.590   0.825   3.520 

I was wondering if there is a more efficient and faster way of doing what I did in the for loop. It goes pretty fast with this example dataset, but my actual dataset can take 45 seconds for each iteration, and with 1000 iterations to make, this can take a while!

Any help will go a long way!

Thanks!

Never use rbind or cbind within a loop! This leads to excessive copying in memory. See Patrick Burns' R Interno: Circle 2 - Growing Objects. Instead, build a list of data frames to rbind once outside the loop:

Since you iterate element wise between equal length vectors, consider mapply or its list wrapper, Map:

df_list <- Map(function(f, s) da[da$dates >= f & da$dates < s,],

               first.times, second.times)



# EQUIVALENT CALL

df_list <- mapply(function(f, s) da[da$dates >= f & da$dates < s,],

                  first.times, second.times, SIMPLIFY=FALSE)

Even consider adding first and second times into data frame with transform to add columns:

df_list <- Map(function(f, s) transform(da[da$dates >= f & da$dates < s,], 

                                        first_time = f, second_time = s),

               first.times, second.times)

From there, use a host of solutions to row bind list of data frames:

# BASE

final_df <- do.call(rbind, df_list)



# PLYR

final_df <- rbind.fill(df_list)



# DPLYR

final_df <- bind_rows(df_list)



# DATA TABLE

final_df <- rbindlist(df_list)

Check benchmark examples here: Convert a list of data frames into one data frame

Comparing to the original setup ...

> subsetted.dates <- da[0,]

> system.time(for(i in 1:length(first.times)){

+   subsetted.dates <- rbind(subsetted.dates, da[da$dates >= first.times[i] & da$dates < second.times[i],])

+ })

   user  system elapsed 

   3.97    0.35    4.33 

... it is possible to get a slight performance improvement using lapply:

> system.time({

+   subsetted.dates <- lapply(1:length(first.times),function(i) da[da$dates >= first.times[i] & da$dates < second.times[i],])

+   subsetted.dates <- do.call(rbind,subsetted.dates)

+ })

   user  system elapsed 

   3.37    0.26    3.75 

Changing a bit the algorithm, if you first create index of dates with a bit smaller set of data and then apply it, that leads to even a better performance:

> system.time({

+   da_dates <- da$dates

+   da_inds <- lapply(1:length(first.times),function(i) which(da_dates >= first.times[i] & da_dates < second.times[i]))

+   subsetted.dates <- da[unlist(da_inds),]

+ })

   user  system elapsed 

   2.60    0.31    2.94 

Suggesting that that the time intervals can be ordered in time order (in this case they were already in time order) and that they are not overlapping, the problem becomes even faster:

system.time({ 

  da_date_order <- order(da$dates)

  da_date_back_order <- order(da$dates)

  da_sorted_dates <- sort(da$dates)

  da_selected_dates <- rep(FALSE,length(da_sorted_dates))

  j = 1

  for (i in 1:length(da_dates)) {

    if (da_sorted_dates[i] >= first.times[j] & da_sorted_dates[i] < second.times[j]) {

      da_selected_dates[i] <- TRUE

    } else if (da_sorted_dates[i] >= second.times[j]) {

      j = j + 1

      if (j > length(second.times)) {

        break

      }

    }

  }

  subsetted.dates <- da[da_date_back_order[da_selected_dates],]

})



user  system elapsed 

0.98    0.00    1.01 

And if you allow sorting the original da dataset, then the solution is even faster:

system.time({

  da <- da[order(da$dates),]

  da_sorted_dates <- da$dates

  da_selected_dates <- rep(FALSE,length(da_sorted_dates))

  j = 1

  for (i in 1:length(da_dates)) {

    if (da_sorted_dates[i] >= first.times[j] & da_sorted_dates[i] < second.times[j]) {

      da_selected_dates[i] <- TRUE

    } else if (da_sorted_dates[i] >= second.times[j]) {

      j = j + 1

      if (j > length(second.times)) {

        break

      }

    }

  }

  subsetted.dates <- da[da_selected_dates,]

})



user  system elapsed 

0.63    0.00    0.63