如果两个向量之间存在grep匹配，则返回数据帧的不同列

Question

我有一个文件名向量和一个数据框，其中包含每个文件名的“组”名称。

files <- c("data/backup/LATEST/20181514.X1235",
           "data/backup/LATEST/X1255+20181514",
           "data/backup/LATEST/20181514-X1237",
           "data/backup/LATEST/20181514-E1235",
           "data/backup/LATEST/20181514F1235",
           "data/backup/LATEST/M32_-X6635__20181514",
           "data/backup/LATEST/20181514-X1205",
           "data/backup/LATEST/l-A1230.20181514-XX")

groups <- data.frame(
                    ID = c("X1235","X1255","A1230","K93430",
                           "LOP0343","J3490","X1205","X6635",
                           "F1235","E1235","X1237"), 
                    Group = c("A","A","A",
                              "B","A","A",
                              "B","B","B",
                              "B","A")
)

作为最终结果，我希望有一个数据框，其中包含来自files的完整文件路径，第二列显示其group。

我怎样才能做到这一点？

结果

                           filepath         group
1 data/backup/LATEST/20181514.X1235         A
2 data/backup/LATEST/X1255+20181514         A
3 data/backup/LATEST/20181514-X1237         A
4 data/backup/LATEST/20181514-E1235         B
5 data/backup/LATEST/20181514F1235          B
6 data/backup/LATEST/M32_-X6635__20181514   B
7 data/backup/LATEST/20181514-X1205         B
8 data/backup/LATEST/l-A1230.20181514-XX    A

Answer 1

这是使用stringr::str_detect的一种方式

library(stringr)
strdet <- function(x){
      #browser()
      groups[str_detect(x,groups$ID),'Group']
      }

apply(df, 1, strdet)

[1] "A" "A" "A" "B" "B" "B" "B" "A"

PS：

我将文件更改为数据框和

我假设你在文件和组之间有一对一的关系

我用stringAsFactor=FALSE读了df 数据 df <- data.frame(files, stringsAsFactors = FALSE)

Answer 2

使用基数R，您可以创建group向量：

group_list <- lapply(groups$ID,
       function(patt) groups$Group[which(grepl(patt, files))])
data.frame(files=files, group=unlist(group_list))
    files                                    group
    data/backup/LATEST/20181514.X1235        A
    data/backup/LATEST/X1255+20181514        A
    data/backup/LATEST/20181514-X1237        B
    data/backup/LATEST/20181514-E1235        B
    data/backup/LATEST/20181514F1235         A
    data/backup/LATEST/M32_-X6635__20181514  A
    data/backup/LATEST/20181514-X1205        B
    data/backup/LATEST/l-A1230.20181514-XX   A

那是你在找什么？

Answer 3

如果你可以假设ID字符串的构建方式（一个字母，四个数字），使用tidverse：

data.frame(file=files) %>%
  mutate(ID=str_extract(file,"[A-Z]\\d{4}")) %>%
  left_join(groups,by="ID")

我在创建组时添加了stringsAsFactors=FALSE以避免警告。

如果你不能：

library(fuzzyjoin)
data.frame(file=files,stringsAsFactors=FALSE) %>%
  fuzzy_left_join(groups, by=list(x="file",y="ID"), match_fun=str_detect)