Return 如果两个向量之间存在 grep 匹配，则数据框的不同列

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 "group" 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，我想要一个数据框，其中一列包含来自 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

的方法

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，然后

我假设文件和组之间存在一对一关系

我使用 stringAsFactor=FALSE

阅读了 df

数据

df <- data.frame(files, stringsAsFactors = FALSE)

Answer 2

使用 base 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

这就是你要找的吗？

Answer 3

如果您可以假设 ID 字符串的构建方式（一个字母，四位数字），使用 tidverse：

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

我在创建群组时添加了 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)

Return 如果两个向量之间存在 grep 匹配，则数据框的不同列

Return a different column of the dataframe if there is a grep match between two vectors

regex

r

dataframe

grepl

数据