Performance of rbind.data.frame

问题:

I have a list of dataframes for which I am certain that they all contain at least one row (in fact, some contain only one row, and others contain a given number of rows), and that they all have the same columns (names and types). In case it matters, I am also certain that there are no NA's anywhere in the rows.

The situation can be simulated like this:

#create one row
onerowdfr<-do.call(data.frame, c(list(), rnorm(100) , lapply(sample(letters[1:2], 100, replace=TRUE), function(x){factor(x, levels=letters[1:2])})))
colnames(onerowdfr)<-c(paste("cnt", 1:100, sep=""), paste("cat", 1:100, sep=""))
#reuse it in a list
someParts<-lapply(rbinom(200, 1, 14/200)*6+1, function(reps){onerowdfr[rep(1, reps),]})

I've set the parameters (of the randomization) so that they approximate my true situation.

Now, I want to unite all these dataframes in one dataframe. I thought using rbind would do the trick, like this:

system.time(
result<-do.call(rbind, someParts)
)

Now, on my system (which is not particularly slow), and with the settings above, this takes is the output of the system.time:

   user  system elapsed 
   5.61    0.00    5.62

Nearly 6 seconds for rbind-ing 254 (in my case) rows of 200 variables? Surely there has to be a way to improve the performance here? In my code, I have to do similar things very often (it is a from of multiple imputation), so I need this to be as fast as possible.

回答1:

Can you build your matrices with numeric variables only and convert to a factor at the end? rbind is a lot faster on numeric matrices.

On my system, using data frames:

> system.time(result<-do.call(rbind, someParts))
   user  system elapsed 
  2.628   0.000   2.636 

Building the list with all numeric matrices instead:

onerowdfr2 <- matrix(as.numeric(onerowdfr), nrow=1)
someParts2<-lapply(rbinom(200, 1, 14/200)*6+1, 
                   function(reps){onerowdfr2[rep(1, reps),]})

results in a lot faster rbind.

> system.time(result2<-do.call(rbind, someParts2))
   user  system elapsed 
  0.001   0.000   0.001

EDIT: Here's another possibility; it just combines each column in turn.

> system.time({
+   n <- 1:ncol(someParts[[1]])
+   names(n) <- names(someParts[[1]])
+   result <- as.data.frame(lapply(n, function(i) 
+                           unlist(lapply(someParts, `[[`, i))))
+ })
   user  system elapsed 
  0.810   0.000   0.813  

Still not nearly as fast as using matrices though.

EDIT 2:

If you only have numerics and factors, it's not that hard to convert everything to numeric, rbind them, and convert the necessary columns back to factors. This assumes all factors have exactly the same levels. Converting to a factor from an integer is also faster than from a numeric so I force to integer first.

someParts2 <- lapply(someParts, function(x)
                     matrix(unlist(x), ncol=ncol(x)))
result<-as.data.frame(do.call(rbind, someParts2))
a <- someParts[[1]]
f <- which(sapply(a, class)=="factor")
for(i in f) {
  lev <- levels(a[[i]])
  result[[i]] <- factor(as.integer(result[[i]]), levels=seq_along(lev), labels=lev)
}

The timing on my system is:

   user  system elapsed 
   0.090    0.00    0.091 

回答2:

Not a huge boost, but swapping rbind for rbind.fill from the plyr package knocks about 10% off the running time (with the sample dataset, on my machine).

回答3:

If you really want to manipulate your data.frames faster, I would suggest to use the package data.table and the function rbindlist(). I did not perform extensive tests but for my dataset (3000 dataframes, 1000 rows x 40 columns each) rbindlist() takes only 20 seconds.

回答4:

This is ~25% faster, but there has to be a better way...

system.time({
  N <- do.call(sum, lapply(someParts, nrow))
  SP <- as.data.frame(lapply(someParts[[1]], function(x) rep(x,N)))
  k <- 0
  for(i in 1:length(someParts)) {
    j <- k+1
    k <- k + nrow(someParts[[i]])
    SP[j:k,] <- someParts[[i]]
  }
})

回答5:

Make sure you're binding dataframe to dataframe. Ran into huge perf degradation when binding list to dataframe.