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

问题

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!

回答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"

回答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.

回答3:

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.

回答4:

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 soultions:

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

回答5:

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

回答6:

I think nested gsub will do the job.

gsub("Find","Replace",gsub("Find","Replace",X))

来源：https://stackoverflow.com/questions/33949945/replace-multiple-strings-in-one-gsub-or-chartr-statement-in-r

标签

string

gsub