Update
The stri_split_fixed
function in more recent versions of "stringi" have a simplify
argument that can be set to TRUE
to return a matrix. Thus, the updated solution would be:
stri_split_fixed(df1$combCol2, ",", 2, simplify = TRUE)
Original answer (with updated benchmarks)
If you are comfortable with the "stringr" syntax and don't want to veer too far from it, but you also want to benefit from a speed boost, try the "stringi" package instead:
library(stringr)
library(stringi)
system.time(temp1 <- str_split_fixed(df1$combCol2, ',', 2))
# user system elapsed
# 3.25 0.00 3.25
system.time(temp2a <- do.call(rbind, stri_split_fixed(df1$combCol2, ",", 2)))
# user system elapsed
# 0.04 0.00 0.05
system.time(temp2b <- stri_split_fixed(df1$combCol2, ",", 2, simplify = TRUE))
# user system elapsed
# 0.01 0.00 0.01
Most of the "stringr" functions have "stringi" parallels, but as can be seen from this example, the "stringi" output required one extra step of binding the data to create the output as a matrix instead of as a list.
Here's how it compares with @RichardScriven's suggestion in the comments:
fun1a <- function() do.call(rbind, stri_split_fixed(df1$combCol2, ",", 2))
fun1b <- function() stri_split_fixed(df1$combCol2, ",", 2, simplify = TRUE)
fun2 <- function() {
do.call(rbind, regmatches(df1$combCol2, regexpr(",", df1$combCol2),
invert = TRUE))
}
library(microbenchmark)
microbenchmark(fun1a(), fun1b(), fun2(), times = 10)
# Unit: milliseconds
# expr min lq mean median uq max neval
# fun1a() 42.72647 46.35848 59.56948 51.94796 69.29920 98.46330 10
# fun1b() 17.55183 18.59337 20.09049 18.84907 22.09419 26.85343 10
# fun2() 370.82055 404.23115 434.62582 439.54923 476.02889 480.97912 10