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I have a list of data frames with different sets of columns. I would like to combine them by rows into one data frame. I use plyr::rbind.fill to do that. I am looking for something that would do this more efficiently, but is similar to the answer given here

require(plyr)

set.seed(45)
sample.fun <- function() {
   nam <- sample(LETTERS, sample(5:15))
   val <- data.frame(matrix(sample(letters, length(nam)*10,replace=TRUE),nrow=10))
   setNames(val, nam)  
}
ll <- replicate(1e4, sample.fun())
rbind.fill(ll)
Michael Szczepaniak
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mrkilinc
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4 Answers4

43

UPDATE: See this updated answer instead.

UPDATE (eddi): This has now been implemented in version 1.8.11 as a fill argument to rbind. For example:

DT1 = data.table(a = 1:2, b = 1:2)
DT2 = data.table(a = 3:4, c = 1:2)

rbind(DT1, DT2, fill = TRUE)
#   a  b  c
#1: 1  1 NA
#2: 2  2 NA
#3: 3 NA  1
#4: 4 NA  2

FR #4790 added now - rbind.fill (from plyr) like functionality to merge list of data.frames/data.tables

Note 1:

This solution uses data.table's rbindlist function to "rbind" list of data.tables and for this, be sure to use version 1.8.9 because of this bug in versions < 1.8.9.

Note 2:

rbindlist when binding lists of data.frames/data.tables, as of now, will retain the data type of the first column. That is, if a column in first data.frame is character and the same column in the 2nd data.frame is "factor", then, rbindlist will result in this column being a character. So, if your data.frame consisted of all character columns, then, your solution with this method will be identical to the plyr method. If not, the values will still be the same, but some columns will be character instead of factor. You'll have to convert to "factor" yourself after. Hopefully this behaviour will change in the future.

And now here's using data.table (and benchmarking comparison with rbind.fill from plyr):

require(data.table)
rbind.fill.DT <- function(ll) {
    # changed sapply to lapply to return a list always
    all.names <- lapply(ll, names)
    unq.names <- unique(unlist(all.names))
    ll.m <- rbindlist(lapply(seq_along(ll), function(x) {
        tt <- ll[[x]]
        setattr(tt, 'class', c('data.table', 'data.frame'))
        data.table:::settruelength(tt, 0L)
        invisible(alloc.col(tt))
        tt[, c(unq.names[!unq.names %chin% all.names[[x]]]) := NA_character_]
        setcolorder(tt, unq.names)
    }))
}

rbind.fill.PLYR <- function(ll) {
    rbind.fill(ll)
}

require(microbenchmark)
microbenchmark(t1 <- rbind.fill.DT(ll), t2 <- rbind.fill.PLYR(ll), times=10)
# Unit: seconds
#                      expr      min        lq    median        uq       max neval
#   t1 <- rbind.fill.DT(ll)  10.8943  11.02312  11.26374  11.34757  11.51488    10
# t2 <- rbind.fill.PLYR(ll) 121.9868 134.52107 136.41375 184.18071 347.74724    10


# for comparison change t2 to data.table
setattr(t2, 'class', c('data.table', 'data.frame'))
data.table:::settruelength(t2, 0L)
invisible(alloc.col(t2))
setcolorder(t2, unique(unlist(sapply(ll, names))))

identical(t1, t2) # [1] TRUE

It should be noted that plyr's rbind.fill edges past this particular data.table solution until list size of about 500.

Benchmarking plot:

Here's the plot on runs with list length of data.frames with seq(1000, 10000, by=1000). I've used microbenchmark with 10 reps on each of these different list lengths.

enter image description here

Benchmarking gist:

Here's the gist for benchmarking, in case anyone wants to replicate the results.

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Arun
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  • +1 but `rbind.fill.DT` returns invisibly. Was that intentional? – GSee Aug 01 '13 at 21:18
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    very nice - this really should go into `data.table` - maybe you could add a FR – eddi Aug 01 '13 at 21:20
  • Any theories as to why plyr would be faster at 6000 than 5000? The difference from 7000 to 8000 I'm willing to write off as an insignificant difference, but ~35 seconds...? – joran Aug 02 '13 at 00:16
  • @Arun Thank you for your answer. I just want to note that related to [this bug](http://stackoverflow.com/questions/15933846/rbindlist-two-data-tables-where-one-has-factor-and-other-has-character-type-for) I had to download newer version [(1.8.9)](https://r-forge.r-project.org/R/?group_id=240). But still t1 and t2 are not still identical. t1 has factor and char type columns where t2 has only factor columns – mrkilinc Aug 02 '13 at 15:17
  • decent, but lacks parallellization! :) – Xachriel Aug 02 '13 at 19:52
15

Now that rbindlist (and rbind) for data.table has improved functionality and speed with the recent changes/commits in v1.9.3 (development version), and dplyr has a faster version of plyr's rbind.fill, named rbind_all, this answer of mine seems a bit too outdated.

Here's the relevant NEWS entry for rbindlist:

o  'rbindlist' gains 'use.names' and 'fill' arguments and is now implemented entirely in C. Closes #5249    
  -> use.names by default is FALSE for backwards compatibility (doesn't bind by 
     names by default)
  -> rbind(...) now just calls rbindlist() internally, except that 'use.names' 
     is TRUE by default, for compatibility with base (and backwards compatibility).
  -> fill by default is FALSE. If fill is TRUE, use.names has to be TRUE.
  -> At least one item of the input list has to have non-null column names.
  -> Duplicate columns are bound in the order of occurrence, like base.
  -> Attributes that might exist in individual items would be lost in the bound result.
  -> Columns are coerced to the highest SEXPTYPE, if they are different, if/when possible.
  -> And incredibly fast ;).
  -> Documentation updated in much detail. Closes DR #5158.

So, I've benchmarked the newer (and faster versions) on relatively bigger data below.

New Benchmark:

We'll create a total of 10,000 data.tables with columns ranging from 200-300 with the total number of columns after binding to be 500.

Functions to create data:

require(data.table) ## 1.9.3 commit 1267
require(dplyr)      ## commit 1504 devel
set.seed(1L)
names = paste0("V", 1:500)
foo <- function() {
    cols = sample(200:300, 1)
    data = setDT(lapply(1:cols, function(x) sample(10)))
    setnames(data, sample(names)[1:cols])
}
n = 10e3L
ll = vector("list", n)
for (i in 1:n) {
    .Call("Csetlistelt", ll, i, foo())
}

And here are the timings:

## Updated timings on data.table v1.9.5 - three consecutive runs:
system.time(ans1 <- rbindlist(ll, fill=TRUE))
#   user  system elapsed 
#  1.993   0.106   2.107 
system.time(ans1 <- rbindlist(ll, fill=TRUE))
#   user  system elapsed 
#  1.644   0.092   1.744 
system.time(ans1 <- rbindlist(ll, fill=TRUE))
#   user  system elapsed 
#  1.297   0.088   1.389 


## dplyr's rbind_all - Timings for three consecutive runs
system.time(ans2 <- rbind_all(ll))
#   user  system elapsed  
#  9.525   0.121   9.761 

#   user  system elapsed  
#  9.194   0.112   9.370 

#   user  system elapsed  
#  8.665   0.081   8.780 

identical(ans1, setDT(ans2)) # [1] TRUE
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Arun
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    When linking to this answer from a `dplyr`-ish question, I noted in the `dplyr 0.7.0` NEWS that "`rbind_list()` and `rbind_all()` now call `.Deprecated()`, they will be removed in the next CRAN release. Please use `bind_rows()` instead". You may consider updating your answer so it can be used as a canonical also with current `dplyr`. Cheers. – Henrik Jan 14 '18 at 22:27
5

There is still something to be gained if you parallelize both rbind.fill and rbindlist. The results are done with data.table version 1.8.8 as version 1.8.9 got bricked when I tried it with the parallelized function. So the results aren't identical between data.table and plyr, but they are identical within data.table or plyr solution. Meaning parallel plyr matches to unparallel plyr, and vice versa.

Here's the benchmark/scripts. The parallel.rbind.fill.DT looks horrible, but that's the fastest one I could pull.

require(plyr)
require(data.table)
require(ggplot2)
require(rbenchmark)
require(parallel) 

# data.table::rbindlist solutions
rbind.fill.DT <- function(ll) {
  all.names <- lapply(ll, names)
  unq.names <- unique(unlist(all.names))
  rbindlist(lapply(seq_along(ll), function(x) {
    tt <- ll[[x]]
    setattr(tt, 'class', c('data.table', 'data.frame'))
    data.table:::settruelength(tt, 0L)
    invisible(alloc.col(tt))
    tt[, c(unq.names[!unq.names %chin% all.names[[x]]]) := NA_character_]
    setcolorder(tt, unq.names)
  }))
}

 parallel.rbind.fill.DT <- function(ll, cluster=NULL){
   all.names <- lapply(ll, names)
   unq.names <- unique(unlist(all.names)) 
   if(is.null(cluster)){
     ll.m <- rbindlist(lapply(seq_along(ll), function(x) {
       tt <- ll[[x]]
       setattr(tt, 'class', c('data.table', 'data.frame'))
       data.table:::settruelength(tt, 0L)
       invisible(alloc.col(tt))
       tt[, c(unq.names[!unq.names %chin% all.names[[x]]]) := NA_character_]
       setcolorder(tt, unq.names)
     }))
   }else{
     cores <- length(cluster)
     sequ <- as.integer(seq(1, length(ll), length.out = cores+1))
     Call <- paste(paste("list", seq(cores), sep=""), " = ll[", c(1, sequ[2:cores]+1), ":", sequ[2:(cores+1)], "]", sep="", collapse=", ") 
     ll <- eval(parse(text=paste("list(", Call, ")")))
     rbindlist(clusterApply(cluster, ll, function(ll, unq.names){
        rbindlist(lapply(seq_along(ll), function(x, ll, unq.names) {
          tt <- ll[[x]]
          setattr(tt, 'class', c('data.table', 'data.frame'))
          data.table:::settruelength(tt, 0L)
          invisible(alloc.col(tt))
          tt[, c(unq.names[!unq.names %chin% colnames(tt)]) := NA_character_]
          setcolorder(tt, unq.names)
        }, ll=ll, unq.names=unq.names))
      }, unq.names=unq.names))
    }
  }           


# plyr::rbind.fill solutions
rbind.fill.PLYR <- function(ll) {
  rbind.fill(ll)
}

parallel.rbind.fill.PLYR <- function(ll, cluster=NULL, magicConst=400){
  if(is.null(cluster) | ceiling(length(ll)/magicConst) < length(cluster)){
    rbind.fill(ll)
  }else{
    cores <- length(cluster)
    sequ <- as.integer(seq(1, length(ll), length.out = ceiling(length(ll)/magicConst)))
    Call <- paste(paste("list", seq(cores), sep=""), " = ll[", c(1, sequ[2:(length(sequ)-1)]+1), ":", sequ[2:length(sequ)], "]", sep="", collapse=", ") 
    ll <- eval(parse(text=paste("list(", Call, ")")))
    rbind.fill(parLapply(cluster, ll, rbind.fill))
  }
} 

# Function to generate sample data of varying list length
set.seed(45)
sample.fun <- function() {
  nam <- sample(LETTERS, sample(5:15))
  val <- data.frame(matrix(sample(letters, length(nam)*10,replace=TRUE),nrow=10))
  setNames(val, nam)
}

ll <- replicate(10000, sample.fun())
cl <- makeCluster(4, type="SOCK")
clusterEvalQ(cl, library(data.table))
clusterEvalQ(cl, library(plyr))
benchmark(t1 <- rbind.fill.PLYR(ll),
  t2 <- rbind.fill.DT(ll),
  t3 <- parallel.rbind.fill.PLYR(ll, cluster=cl, 400),
  t4 <- parallel.rbind.fill.DT(ll, cluster=cl),
  replications=5)
stopCluster(cl)

# Results for rbinding 10000 dataframes
# done with 4 cores, i5 3570k and 16gb memory
# test                          reps elapsed relative 
# rbind.fill.PLYR                 5  321.80    16.682   
# rbind.fill.DT                   5   26.10    1.353    
# parallel.rbind.fill.PLYR        5   28.00    1.452     
# parallel.rbind.fill.DT          5   19.29    1.000    

# checking are results equal
t1 <- as.matrix(t1)
t2 <- as.matrix(t2)
t3 <- as.matrix(t3)
t4 <- as.matrix(t4)

t1 <- t1[order(t1[, 1], t1[, 2]), ]
t2 <- t2[order(t2[, 1], t2[, 2]), ]
t3 <- t3[order(t3[, 1], t3[, 2]), ]
t4 <- t4[order(t4[, 1], t4[, 2]), ]

identical(t2, t4) # TRUE
identical(t1, t3) # TRUE
identical(t1, t2) # FALSE, mismatch between plyr and data.table

As you can see parallesizing rbind.fill made it comparable to data.table, and you could get marginal increase of speed by parallesizing data.table even with this low of a dataframe count.

Xachriel
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4

simply dplyr::bind_rows will do the job, as

library(dplyr)

merged_list <- bind_rows(ll)

#check it

> nrow(merged_list)
[1] 100000
> ncol(merged_list)
[1] 26

Time taken

> system.time(merged_list <- bind_rows(ll))
   user  system elapsed 
   0.29    0.00    0.28
AnilGoyal
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