R: `split` preserving natural order of factors

r, split

Solution

`split` converts the `f` (second) argument to factors, if it isn't already one. So, if you want the order to be retained, factor the column yourself with the desired level. That is:

df$yearmon <- factor(df$yearmon, levels=unique(df$yearmon))
# now split
split(df, df$yearmon)
# $`4_2013`
#   Date.of.Inclusion Securities.Included Securities.Excluded yearmon
# 1        2013-04-01          INDUSINDBK             SIEMENS  4_2013
# 2        2013-04-01                NMDC               WIPRO  4_2013

# $`9_2012`
#   Date.of.Inclusion Securities.Included Securities.Excluded yearmon
# 3        2012-09-28               LUPIN                SAIL  9_2012
# 4        2012-09-28          ULTRACEMCO                STER  9_2012

# $`4_2012`
#   Date.of.Inclusion Securities.Included Securities.Excluded yearmon
# 5        2012-04-27          ASIANPAINT                RCOM  4_2012
# 6        2012-04-27          BANKBARODA              RPOWER  4_2012

But do not use `split`. Use `data.table` instead:

However normally, `split` tends to be terribly slow as the levels increase. So, I'd suggest using `data.table` to subset to a list. I'd suppose that'd be much faster!

require(data.table)
dt <- data.table(df)
dt[, grp := .GRP, by = yearmon]
setkey(dt, grp)
o2 <- dt[, list(list(.SD)), by = grp]$V1

Benchmarking on huge data:

set.seed(45)
dates <- seq(as.Date("1900-01-01"), as.Date("2013-12-31"), by = "days")
ym <- do.call(paste, c(expand.grid(1:500, 1900:2013), sep="_"))

df <- data.frame(x1 = sample(dates, 1e4, TRUE), 
                 x2 = sample(letters, 1e4, TRUE), 
                 x3 = sample(10, 1e4, TRUE), 
                 yearmon = sample(ym, 1e4, TRUE), 
      stringsAsFactors=FALSE)

require(data.table)
dt <- data.table(df)

f1 <- function(dt) {
    dt[, grp := .GRP, by = yearmon]
    setkey(dt, grp)

    o1 <- dt[, list(list(.SD)), by=grp]$V1
}

f2 <- function(df) {
    df$yearmon <- factor(df$yearmon, levels=unique(df$yearmon))
    o2 <- split(df, df$yearmon)
}

require(microbenchmark)
microbenchmark(o1 <- f1(dt), o2 <- f2(df), times = 10)

# Unit: milliseconds
         expr        min         lq     median        uq      max neval
#  o1 <- f1(dt)   43.72995   43.85035   45.20087  715.1292 1071.976    10
#  o2 <- f2(df) 4485.34205 4916.13633 5210.88376 5763.1667 6912.741    10

Note that the solution from `o1` will be an unnamed list. But you can set the names simply by doing `names(o1) <- unique(dt$yearmon)`

Problem

Original source