如何将向量字符串与R中的数据框列匹配

Question

我在 r 中有以下数据框

Id    titles
1     emami paper mills slips 10% on dismal q4 numbers
2     jsw steel q4fy17 standalone net profit rises 173.33%
3     fmcg major hul q4fy17 standalone net profit rises 6.2
4     chennai petroleum, allsec tech slip 6-7% on poor q4

而且，我在向量中有名字

names <- c("emami ltd","jsw steel ltd","abc","hul india ltd","tcs","chennai petroleum corp ltd")

我想将数据框列标题与向量字符串匹配，并在新列中打印相应的字符串。我想要的数据框是

 Id    titles                                                    names
1     emami paper mills slips 10% on dismal q4 numbers           emami ltd
2     jsw steel q4fy17 standalone net profit rises 173.33%       jsw steel ltd
3     fmcg major hul q4fy17 standalone net profit rises 6.2      hul india ltd
4     chennai petroleum, allsec tech slip 6-7% on poor q4        chennai petroleum corp ltd

我正在使用以下代码进行操作，但它并没有给我想要的。

df[grepl(paste(names, collapse="|"), df$titles),]

如何在 R 中做到这一点？

Answer 1

如果我理解正确的话，您可以使用 BaseR 的 gregexpr 以及 regematches 和 gsub 来完成您的任务。

数据：在 OP 更改问题后编辑

options(stringsAsFactors = F)
df <- data.frame(titles = c("emami paper mills slips 10% on dismal q4 numbers",
                            "jsw steel q4fy17 standalone net profit rises 173.33%",
                            "fmcg major hul q4fy17 standalone net profit rises 6.2",
                            "chennai petroleum, allsec tech slip 6-7% on poor q4"),stringsAsFactors = F)

names <- c("emami ltd","jsw steel ltd","abc","hul india ltd","tcs","chennai petroleum corp ltd")

正则表达式：

library(dplyr)
library(stringr)

newnames <- gsub("^(\\w+).*","\\1",names)
regmat <- regmatches(df$titles,gregexpr(paste0(newnames,collapse="|"),df$titles))
regmat[lapply(regmat,length) == 0] <- NA
df <- data.frame(cbind(df,newnames =do.call("rbind",regmat)),stringsAsFactors = F)
df1 <- data.frame(names=names,newnames=newnames,stringsAsFactors = F)
left_join(df,df1,by="newnames")

您也可以使用stringr 库，如下所示：

library(stringr)
newnames <- str_replace(names,"^(\\w+).*","\\1")
df$newnames <- str_extract(df$titles,paste0(newnames,collapse="|"))
df1 <- data.frame(names=names,newnames=newnames,stringsAsFactors = F)
left_join(df,df1,by="newnames")

输出：

    > left_join(df,df1,by="newnames")
                                                 titles newnames                      names
1      emami paper mills slips 10% on dismal q4 numbers    emami                  emami ltd
2  jsw steel q4fy17 standalone net profit rises 173.33%      jsw              jsw steel ltd
3 fmcg major hul q4fy17 standalone net profit rises 6.2      hul              hul india ltd
4   chennai petroleum, allsec tech slip 6-7% on poor q4  chennai chennai petroleum corp ltd

Answer 2

从你的名字中删除 ltd ：

names <- gsub(" ltd","",names)

Answer 3

也可以使用sqldf 进行这种“模糊”合并。

构造查找：

names <- data.frame(name = c("emami ltd","jsw steel ltd","abc","hul india ltd","tcs"))
names$lookup <- gsub("(\\w+).*", "\\1", names$name)

执行合并：

library(sqldf)
res <- sqldf("SELECT l.*, r.name
       FROM df as l
       LEFT JOIN names as r
       ON l.titles LIKE '%'||r.lookup||'%'")

一些注意事项：我从查找中提取了第一个单词，因为您说您只想要"hul"，而不是"hul india"。同样在sql 中，|| 表示连接，% 表示通配符（它将匹配任何内容），因此如果任何查找出现在文本中的任何位置，无论它之前或之后是什么，这将匹配。

---

使用Reduce 然后合并的另一个选项是：

df$lookup <- Reduce( function(x, y) {x[grepl(y,x)] <- y; x}, c(list(df$titles), names$lookup))
merge(df, names)

Answer 4

为了添加到之前的答案，我制作了一个函数，包括之前的一些 cmets：

df <-  data.frame(title=c("emami paper mills slips 10% on dismal q4 numbers",
                            "jsw steel q4fy17 standalone net profit rises 173.33%",
                            "fmcg major hul q4fy17 standalone net profit rises 6.2"))


names <- c("emami ltd","jsw steel ltd","abc","hul india ltd","tcs")

find_string <- function(data,names){

    ### Clean the names 
    newnames <- gsub("^(\\w+).*","\\1",names)

    ### Loop over the names to find which sentence contain it
    for(i in 1:length(newnames)){

        if(length(grep(newnames[i],df$title)) != 0){
            df$names[grep(newnames[i],df$title)] <- newnames[i]

        }else{
            print(paste(names[i],"not found in the data!"))
        }
    }
    return(df)
}

### Run the function

find_string(df,names)

希望这会有所帮助！