用 R 中没有循环的已定义字符串列表替换非 ascii 字符答案

【问题标题】：Replace non-ascii chars with a defined string list without a loop in R用 R 中没有循环的已定义字符串列表替换非 ascii 字符
【发布时间】：2012-05-29 02:22:59
【问题描述】：

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

【问题讨论】：

标签： r replace special-characters non-ascii-characters

【解决方案1】：

gsubfn() 同名包里的这种东西真的很不错：

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，base R 也有这样的功能：

A <- paste(mapA, collapse="")
L <- paste(mapL, collapse="")
chartr(L, A, "ÍÓáÚ")
# [1] "IOaU"

【讨论】：

谢谢！完美运行。只有一个问题（只是想知道）；你知道 gsubfn 函数是否使用任何类型的内部循环吗？应该比 sapply 快吗？
@Álvaro -- 我不认为gsubfn() 特别快 -- '只是' 方便和优雅。
另见 chartr 在 R 的基础上，这似乎可以解决上述问题，尽管如果实际问题存在变化，例如替换两个字符序列，那么 gsubfn 仍然可以处理它，但是不是chartr。
@G.Grothendieck -- 感谢您指出这一点。我已将其附加到答案中。

【解决方案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"

【讨论】：