您可以更简单地使用stringr包的str_extract_all首先提取所有关键字，然后使用str_replace_all替换它们：

ll <- lapply(str_extract_all(df$title, regex("keyword1+|keyword2+|kword3+|kword4+", ignore_case = TRUE)),
             paste, collapse = "; ")
df$title_keyword <- unlist(lapply(ll, str_replace_all, regex(c("keyword1" = "k1",
                                                     "keyword2" = "k2",
                                                     "kword3" = "k3",
                                                     "kword4" = "k4"),
                                                   ignore_case = TRUE)))

正则表达式中的+查找模式的一个或多个示例。
输出：

# A tibble: 6 × 6
        id treatment_modalities no_patients  year title                                                 title_keyword
     <dbl> <chr>                      <dbl> <dbl> <chr>                                                 <chr>        
1 36933814 HIFU                          32  2023 Title with keyword1 and keyword2 inside of it.        "k1; k2"     
2 36921141 UAE                           NA  2022 A second title with kword3 and keyword1 inside of it. "k3; k1"     
3 39601489 UAE; RFA; HIFU               152  2023 Here we have kword4 and nothing else to see.          "k4"         
4 36898335 UAE; RFA                      NA  2023 A title with kword4 and kword3 inside of it.          "k4; k3"     
5 36432859 UAE; HIFU                     15  2023 And one with keyword1, keyword2 and kword3 in it.     "k1; k2; k3" 
6 33447951 UAE                          428  2023 This title does not contain a keyword.                ""

一种方法：

paste2 <- function(x,...){
  newx <- x[nzchar(x)]
  out <- paste(newx,...)
  if_else(nzchar(out)==0,NA_character_,out)
}

df2 <- df%>%
  rowwise()%>%
  mutate(key1=ifelse(str_detect(title, regex("keyword1", ignore_case = T)),"k1",""),
         key2=ifelse(str_detect(title, regex("keyword2", ignore_case = T)),"k2",""),
         key3=ifelse(str_detect(title, regex("kword3", ignore_case = T)),"k3",""),
         key4=ifelse(str_detect(title, regex("kword4", ignore_case = T)),"k4",""),
         new=paste2(c(key1,key2,key3,key4),sep=";",collapse=";")) %>%
  select(-starts_with("key")) %>%
  ungroup()
df2

# A tibble: 6 x 6
        id treatment_modalities no_patients  year title                                                 new     
     <dbl> <chr>                      <dbl> <dbl> <chr>                                                 <chr>   
1 36933814 HIFU                          32  2023 Title with keyword1 and keyword2 inside of it.        k1;k2   
2 36921141 UAE                           NA  2022 A second title with kword3 and keyword1 inside of it. k1;k3   
3 39601489 UAE; RFA; HIFU               152  2023 Here we have kword4 and nothing else to see.          k4      
4 36898335 UAE; RFA                      NA  2023 A title with kword4 and kword3 inside of it.          k3;k4   
5 36432859 UAE; HIFU                     15  2023 And one with keyword1, keyword2 and kword3 in it.     k1;k2;k3
6 33447951 UAE                          428  2023 This title does not contain a keyword.                NA

您可以分离标记关键字的列，然后将它们合并：

df %>% 
       mutate(
         k1 = case_when(str_detect(title, regex("keyword1", ignore_case = T)) ~ "k1", TRUE ~ NA_character_),
         k2 = case_when(str_detect(title, regex("keyword2", ignore_case = T)) ~ "k2", TRUE ~ NA_character_),
         k3 = case_when(str_detect(title, regex("kword3", ignore_case = T)) ~   "k3", TRUE ~ NA_character_),
         k4 = case_when(str_detect(title, regex("kword4", ignore_case = T)) ~   "k4", TRUE ~ NA_character_)) %>%
       rowwise() %>%
       mutate(
         title_keyword=paste(c(na.omit(k1), na.omit(k2), na.omit(k3), na.omit(k4)), collapse = ";"),
         title_keyword = ifelse(title_keyword=="", NA, title_keyword)
  )

更新

根据评论，这里有一个不同的通用方法，可以避免写出这么多条件：

key <- c("keyword1", "keyword2", "kword3", "kword4")
value <- c("k1", "k2", "k3", "k4")

df |>
  mutate(title_keyword = map_chr(title, ~ str_flatten(value[str_detect(.x, key)], ";")),
         title_keyword = na_if(title_keyword, ""))

如何运作

1.首先为键值对创建一个向量。key是你想要匹配的，value是你想要在key匹配时返回的元素。
1.迭代title，对于每个标题，我们使用str_detect测试所有关键字。这将返回一个逻辑值向量，我们使用它来访问这些值。
1.使用str_flatten，我们将返回值折叠成字符串（"k1;k2"）。

1. str_flatten将返回""，因此我们使用na_if将其转换为NA。

输出

id treatment_modalities no_patients  year title                  title~1
     <dbl> <chr>                      <dbl> <dbl> <chr>                  <chr>  
1 36933814 HIFU                          32  2023 Title with keyword1 a~ k1;k2  
2 36921141 UAE                           NA  2022 A second title with k~ k1;k3  
3 39601489 UAE; RFA; HIFU               152  2023 Here we have kword4 a~ k4     
4 36898335 UAE; RFA                      NA  2023 A title with kword4 a~ k3;k4  
5 36432859 UAE; HIFU                     15  2023 And one with keyword1~ k1;k2;~
6 33447951 UAE                          428  2023 This title does not c~ NA

老

与其有条件地设置title_keyword的值，为什么不直接提取您要查找的值：

df |> 
  mutate(title_keyword = map_chr(str_match_all(title, "(k)e?y?word(\\d)"), ~ str_c(str_c(.x[,2], .x[,3]), collapse = ";")),
         title_keyword = na_if(title_keyword, ""))

1.在这里，我们匹配您的kword和keyword，并提取k和末尾的数字（\\d）。
1.我们将这些捕获粘贴在一起，然后折叠它们，并以;分隔。