Replace multiple strings in one gsub() or chartr() statement in R? - r

Question

I have a string variable containing alphabet[a-z], space[ ], and apostrophe['],eg. x <- "a'b c"
I want to replace apostrophe['] with blank[], and replace space[ ] with underscore[_].
x <- gsub("'", "", x)
x <- gsub(" ", "_", x)
It works absolutely, but when I have a lot of condition, the code becomes ugly. Therefore, I want to use chartr(), but chartr() can't deal with blank, eg.
x <- chartr("' ", "_", x)
#Error in chartr("' ", "_", "a'b c") : 'old' is longer than 'new'
Is there any way to solve this problem? thanks!

Answer 1

You can use gsubfn
library(gsubfn)
gsubfn(".", list("'" = "", " " = "_"), x)
# [1] "ab_c"
Similarly, we can also use mgsub which allows multiple replacement with multiple pattern to search
mgsub::mgsub(x, c("'", " "), c("", "_"))
#[1] "ab_c"

Answer 2

I am a fan of the syntax that the %<>% and %>% opperators from the magrittr package provide.
library(magrittr)
x <- "a'b c"
x %<>%
gsub("'", "", .) %>%
gsub(" ", "_", .)
x
##[1] "ab_c"
gusbfn is wonderful, but I like the chaining %>% allows.

Answer 3

I'd go with the quite fast function stri_replace_all_fixed from library(stringi):
library(stringi)
stri_replace_all_fixed("a'b c", pattern = c("'", " "), replacement = c("", "_"), vectorize_all = FALSE)
Here is a benchmark taking into account most of the other suggested solutions:
library(stringi)
library(microbenchmark)
library(gsubfn)
library(mgsub)
library(magrittr)
library(dplyr)
x_gsubfn <-
x_mgsub <-
x_nested_gsub <-
x_magrittr <-
x_stringi <- "a'b c"
microbenchmark("gsubfn" = { gsubfn(".", list("'" = "", " " = "_"), x_gsubfn) },
"mgsub" = { mgsub::mgsub(x_mgsub, c("'", " "), c("", "_")) },
"nested_gsub" = { gsub("Find", "Replace", gsub("Find","Replace", x_nested_gsub)) },
"magrittr" = { x_magrittr %<>% gsub("'", "", .) %>% gsub(" ", "_", .) },
"stringi" = { stri_replace_all_fixed(x_stringi, pattern = c("'", " "), replacement = c("", "_"), vectorize_all = FALSE) }
)
Unit: microseconds
expr min lq mean median uq max neval
gsubfn 458.217 482.3130 519.12820 513.3215 538.0100 715.371 100
mgsub 180.521 200.8650 221.20423 216.0730 231.6755 460.587 100
nested_gsub 14.615 15.9980 17.92178 17.7760 18.7630 40.687 100
magrittr 113.765 133.7125 148.48202 142.9950 153.0680 296.261 100
stringi 3.950 7.7030 8.41780 8.2960 9.0860 26.071 100

Answer 4

I know it is a bit old but it is hard to pass on an efficient base R solution. Just use the pipe:
test <- "abcegdfk461mnb"
test2 <- gsub("e|4|6","",test)
print(test2)

Answer 5

I think nested gsub will do the job.
gsub("Find","Replace",gsub("Find","Replace",X))

Answer 6

I would opt for a magrittr and/or dplyr solution, as well. However, I prefer not making a new copy of the object, especially if it is in a function and can be returned cheaply.
i.e.
return(
catInTheHat %>% gsub('Thing1', 'Thing2', .) %>% gsub('Red Fish', 'Blue
Fish', .)
)
...and so on.

Answer 7

gsub("\\s", "", chartr("' ", " _", x)) # Use whitespace and then remove it

Answer 8

Try this replace multi text character in column:
df$TYPE <- str_replace_all(df$TYPE, c("test" = "new_test", "G" = "N", "T" = "W"))

Answer 9

I use this function, which also allows omitting the argument for the replacement if the replacement is empty:
s=function(x,...,ignore.case=F,perl=F,fixed=F,useBytes=F){
a=match.call(expand.dots=F)$...
l=length(a)
for(i in seq(1,l,2))x=gsub(a[[i]],if(i==l)""else a[[i+1]],x,ignore.case=ignore.case,perl=perl,fixed=fixed,useBytes=useBytes)
x
}
> s("aa bb cc","aa","dd","bb")
[1] "dd cc"