r - 如何在 R 中删除包含任何非字母字符（连字符和撇号除外）的单词

Question

我需要删除 R 中中间有非字母字符（连字符和撇号除外）的字符串中的所有单词（或用空格替换它们）。有人可以帮忙吗？谢谢。

例如
str = "he@llo wor*ld i'm using state-of-the-art technologies it's i4u"
预期输出 " i'm using state-of-the-art technologies it's "

我尝试了以下正则表达式。

lines <- c("i'm",
           'gas-lighting',
           "i'm gas-lighting",
           "i-love-you",
           "i@u",
           "b2b",
           "i'm gas-lighting u i@u b2b")

gsub("\\w+[^a-z'-]+\\w+", " ", lines) 
[1] "i'm"          "gas-lighting" "i' -lighting" "i-love-you"   " "            
" "            "i' -     "

问题是单词之间的空格？试图跳过空间。

gsub("\\w+[^a-z\\s'-]+\\w+", " ", lines)**  
[1] "i'm"          "gas-lighting" "i' -lighting" "i-love-you"   " "            
" "            "i' -     "

它不会跳过空格吗？预期以下字符串。

[1] "i'm"          "gas-lighting" "i'm gas-lighting" "i-love-you"   " "            
" "            "i'm gas-lighting u    "

更新 2：好的，到目前为止效果很好。

> lines <- c("i'm",
+            'gas-lighting',
+            "i'm gas-lighting",
+            "i-love-you",
+            "i@u",
+            "b2b",
+            "i'm gas-lighting u and you and you i@u b2b",
+            " he@llo wor$ld how*are&you ")
>
> # split a string at spaces then remove the words 
> # that contain any non-alphabetic characters (excpet "-", "'")
> # then paste them together (separate them with spaces)
> unlist(lapply(lines, function(line){
+   words <- unlist(strsplit(line, "\\s+"))
+   words <- words[!grepl("[^a-z'-]", words, perl=TRUE)]
+   paste(words, collapse=" ")}))
[1] "i'm"                                "gas-lighting"                      
[3] "i'm gas-lighting"                   "i-love-you"                        
[5] ""                                   ""                                  
[7] "i'm gas-lighting u and you and you" "" 

更新 1：到目前为止，我正在使用以下正则表达式。

> # replace word at the beginning of a string
> lines <- gsub("^\\s*\\w*[^a-z'-]+\\w*", " ", lines); lines
[1] "i'm"                     "gas-lighting"            "i'm gas-lighting"        "i-love-you"             
[5] " "                       " "                       "i'm gas-lighting u i@u "
> # replace word at the end of a string
> lines <- gsub("\\s[a-z]+[^a-z'-]+\\w*$", " ", lines); lines 
[1] "i'm"                     "gas-lighting"            "i'm gas-lighting"        "i-love-you"             
[5] " "                       " "                       "i'm gas-lighting u i@u "
> # replace words between spaces
> gsub("\\s\\w*[^a-z'-]+\\w*\\s", " ", lines)
[1] "i'm"                 "gas-lighting"        "i'm gas-lighting"    "i-love-you"          " "                  
[6] " "                   "i'm gas-lighting u "

标签： rregex

Answer 1

作为带有 grepl 的 Harro Cyranka 的变体

paste0(sapply(break_1, function(x) x[!grepl("[^Aa-zZ|'|-]", x)]), collapse = " ")