Replace multiple letters with accents with gsub

问题

of course I could replace specific arguments like this:

    mydata=c(\"á\",\"é\",\"ó\")
    mydata=gsub(\"á\",\"a\",mydata)
    mydata=gsub(\"é\",\"e\",mydata)
    mydata=gsub(\"ó\",\"o\",mydata)
    mydata

but surely there is a easier way to do this all in onle line, right? I dont find the gsub help to be very comprehensive on this.

回答1:

Use the character translation function

chartr("áéó", "aeo", mydata)

回答2:

An interesting question! I think the simplest option is to devise a special function, something like a "multi" gsub():

mgsub <- function(pattern, replacement, x, ...) {
  if (length(pattern)!=length(replacement)) {
    stop("pattern and replacement do not have the same length.")
  }
  result <- x
  for (i in 1:length(pattern)) {
    result <- gsub(pattern[i], replacement[i], result, ...)
  }
  result
}

Which gives me:

> mydata <- c("á","é","ó")
> mgsub(c("á","é","ó"), c("a","e","o"), mydata)
[1] "a" "e" "o"

回答3:

Maybe this can be usefull:

iconv('áéóÁÉÓçã', to="ASCII//TRANSLIT")
[1] "aeoAEOca"

回答4:

You can use stringi package to replace these characters.

> stri_trans_general(c("á","é","ó"), "latin-ascii")

[1] "a" "e" "o"

回答5:

Another mgsub implementation using Reduce

mystring = 'This is good'
myrepl = list(c('o', 'a'), c('i', 'n'))

mgsub2 <- function(myrepl, mystring){
  gsub2 <- function(l, x){
   do.call('gsub', list(x = x, pattern = l[1], replacement = l[2]))
  }
  Reduce(gsub2, myrepl, init = mystring, right = T) 
}

回答6:

This is very similar to @kith, but in function form, and with the most common diacritcs cases:

removeDiscritics <- function(string) {
  chartr(
     "ŠŽšžŸÀÁÂÃÄÅÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖÙÚÛÜÝàáâãäåçèéêëìíîïðñòóôõöùúûüýÿ"
    ,"SZszYAAAAAACEEEEIIIIDNOOOOOUUUUYaaaaaaceeeeiiiidnooooouuuuyy"
    , string
  )
}


removeDiscritics("test áéíóú")

"test aeiou"

回答7:

A problem with some of the implementations above (e.g., Theodore Lytras's) is that if the patterns are multiple characters, they may conflict in the case that one pattern is a substring of another. A way to solve this is to create a copy of the object and perform the pattern replacement in that copy. This is implemented in my package bayesbio, available on CRAN.

mgsub <- function(pattern, replacement, x, ...) {
  n = length(pattern)
  if (n != length(replacement)) {
    stop("pattern and replacement do not have the same length.")
  }
  result = x
  for (i in 1:n) {
    result[grep(pattern[i], x, ...)] = replacement[i]
  }
  return(result)
}

Here is a test case:

  asdf = c(4, 0, 1, 1, 3, 0, 2, 0, 1, 1)

  res = mgsub(c("0", "1", "2"), c("10", "11", "12"), asdf)

回答8:

Not so elegant, but it works and does what you want

> diag(sapply(1:length(mydata), function(i, x, y) {
+   gsub(x[i],y[i], x=x)
+ }, x=mydata, y=c('a', 'b', 'c')))
[1] "a" "b" "c"

回答9:

You can use the match function. Here match(x, y) returns the index of y where the element of x is matched. Then you can use the returned indices, to subset another vector (say z) that contains the replacements for the values of x, appropriately matched with y. In your case:

mydata <- c("á","é","ó")
desired <- c('a', 'e', 'o')

desired[match(mydata, mydata)]

In a simpler example, consider the situation below, where I was trying to substitute a for 'alpha', 'b' for 'beta' and so forth.

x <- c('a', 'a', 'b', 'c', 'b', 'c', 'e', 'e', 'd')

y <- c('a', 'b', 'c', 'd', 'e')
z <- c('alpha', 'beta', 'gamma', 'delta', 'epsilon')

z[match(x, y)]

回答10:

Related to Justin's answer:

> m <- c("á"="a", "é"="e", "ó"="o")
> m[mydata]
  á   é   ó 
"a" "e" "o"

And you can get rid of the names with names(*) <- NULL if you want.

回答11:

In this case, doesn't have so much sense, but if they are just two, you can also combine them with gsub:

mydata <- gsub("á","a", gsub("é","e",mydata))

来源：https://stackoverflow.com/questions/15253954/replace-multiple-letters-with-accents-with-gsub

标签

regex

gsub