更新
最新版本的“stringi”中的stri_split_fixed 函数有一个simplify 参数,可以将其设置为TRUE 以返回一个矩阵。因此,更新后的解决方案是:
stri_split_fixed(df1$combCol2, ",", 2, simplify = TRUE)
原始答案(带有更新的基准)
如果您对“stringr”语法感到满意并且不想偏离它太远,但又想从速度提升中受益,请尝试使用“stringi”包:
library(stringr)
library(stringi)
system.time(temp1 <- str_split_fixed(df1$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
大多数“stringr”函数都有“stringi”并行,但从这个例子可以看出,“stringi”输出需要一个额外的步骤来绑定数据以将输出创建为矩阵而不是列表.
这是它与 @RichardScriven 在 cmets 中的建议的比较:
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