6

我想用对应的 ascii 替换非 ascii 字符（目前只有西班牙语）。如果我有“á”，我想用“a”替换它，依此类推。

我构建了这个函数（工作正常），但我不想使用循环（包括像 sapply 这样的内部循环）。

latin2ascii<-function(x) {
if(!is.character(x)) stop ("input must be a character object")
require(stringr)
mapL<-c("á","é","í","ó","ú","Á","É","Í","Ó","Ú","ñ","Ñ","ü","Ü")
mapA<-c("a","e","i","o","u","A","E","I","O","U","n","N","u","U")
for(y in 1:length(mapL)) {
  x<-str_replace_all(x,mapL[y],mapA[y])
  }
x
}

有没有优雅的方法来解决它？任何帮助、建议或修改表示赞赏

4

2 回答 2

7

gsubfn()在同名的包中对于这种事情真的很好：

library(gsubfn)

# Create a named list, in which:
#   - the names are the strings to be looked up
#   - the values are the replacement strings
mapL <- c("á","é","í","ó","ú","Á","É","Í","Ó","Ú","ñ","Ñ","ü","Ü")
mapA <- c("a","e","i","o","u","A","E","I","O","U","n","N","u","U")

# ll <- setNames(as.list(mapA), mapL) # An alternative to the 2 lines below
ll <- as.list(mapA)
names(ll) <- mapL


# Try it out
string <- "ÍÓáÚ"
gsubfn("[áéíóúÁÉÍÓÚñÑüÜ]", ll, string)
# [1] "IOaU"

编辑：

G. Grothendieck 指出，基数 R 也有一个功能：

A <- paste(mapA, collapse="")
L <- paste(mapL, collapse="")
chartr(L, A, "ÍÓáÚ")
# [1] "IOaU"
于 2012-05-22T15:47:22.307 回答
2

我喜欢 Josh 的版本，但我想我可能会添加另一个“矢量化”解决方案。它返回一个无重音字符串的向量。它也只依赖于base函数。

x=c('íÁuÚ','uíÚÁ')

mapL<-c("á","é","í","ó","ú","Á","É","Í","Ó","Ú","ñ","Ñ","ü","Ü")
mapA<-c("a","e","i","o","u","A","E","I","O","U","n","N","u","U")
split=strsplit(x,split='')
m=lapply(split,match,mapL)
mapply(function(split,m) paste(ifelse(is.na(m),split,mapA[m]),collapse='') , split, m)
# "iAuU" "uiUA"
于 2012-05-22T16:10:45.017 回答