字符串拆分、合并和堆叠多列

Question

我有以下原始数据，如下所示：

rawData <- data.frame(ID = c(1,2,3), 
          Name = c("Company B; Company A; Company C", "Company A; Company D", "Company E"),
          Name_location = c("Company A (USA (Primary)); Company B (Japan(Primary)); Company C (Korea,South (Primary))", "Company A (USA (Primary)); Company D (USA (Primary))", "European (Primary)" ))


 ID    Name                              Name_location
 1     Company B; Company A;Company C    Company A (USA (Primary)); Company B (Japan(Primary)); Company C (Korea,South (Primary))
 2     Company A; Company D              Company A (USA (Primary)); Company D (USA (Primary))
 3     Company E                         European (Primary)

我需要将数据转换为如下所示：

Name_location 字段在“名称”字段中包含每个公司的位置数据，但它可能是乱序的。此外，如果名称字段中只有 1 家公司，Name_location 字段将只有位置，而如果名称字段中有多家公司，Name_location 字段将遵循语法“公司（位置）（主要））；公司（位置（主要））”

我需要一种方法将公司及其位置隔离为单独的行，可通过 ID 识别。

 IdealData <- data.frame(ID = c(1,1,1,2,2,3), 
                  Name = c("Company B", "Company A", "Company C", "Company A","Company D", "Company E"),
                  Location = c("Japan","USA", "Korea,South","USA","USA","European"))


     ID      Name            Location
     1       Company B       Japan
     1       Company A       USA
     1       Company C       Korea,South
     2       Company A       USA
     2       Company D       USA
     3       Company E       European

希望在 R 中实现这一点

Answer 1

使用separate_rows后，我们可以用str_extract

提取具体成分

library(stringr)
library(dplyr)
library(tidyr)
rawData %>% 
   separate_rows(c(Name, Name_location), sep=";\s*") %>%
   separate(Name_location, into = c('Name1', 'Location'), sep= "\s+(?=\()",
         extra = "merge") %>%
   mutate(Location = case_when(Name1 == 'European' ~ Name1,
         TRUE ~ trimws(str_extract(Location,
               "(?<=\()[^(]+"))[match(Name, Name1)])) %>%
   select(-Name1)
# A tibble: 6 x 3
#     ID Name      Location   
#  <dbl> <chr>     <chr>      
#1     1 Company B Japan      
#2     1 Company A USA        
#3     1 Company C Korea,South
#4     2 Company A USA        
#5     2 Company D USA        
#6     3 Company E European

Answer 2

如果你想在没有包和库的情况下做到这一点，你可以循环遍历条目并创建一个新的 data.frame:

rawData <- data.frame("ID" = c(1,2,3), 
                      "Name" = c("Company B; Company A; Company C", "Company A; Company D", "Company E"),
                      "Name_location" = c("Company A (USA (Primary)); Company B (Japan(Primary)); Company C (Korea,South (Primary))", "Company A (USA (Primary)); Company D (USA (Primary))", "European (Primary)" ))
rawData$Name = as.character(rawData$Name)
rawData$Name_location = as.character(rawData$Name_location)

idealData = list("ID"=c(),"Company"=c(),"Location"=c())
for(i in 1:length(rawData$ID)){
  print(strsplit(rawData$Name[i],";"))
  ncomp = length(strsplit(rawData$Name[i],";")[[1]])
  print(ncomp)
  if(ncomp==1){
    idealData[["ID"]]=c(idealData[["ID"]],rawData$ID[i])
    idealData[["Company"]]=c(idealData[["Company"]],rawData$Name[i])
    idealData[["Location"]]=c(idealData[["Location"]],strsplit(rawData$Name_location[i]," \(")[[1]][1])
  }else{
    vcomp = strsplit(rawData$Name[i],"; ")[[1]]
    for(compi in 1:ncomp){
      idealData[["ID"]]=c(idealData[["ID"]],rawData$ID[i])
      idealData[["Company"]]=c(idealData[["Company"]],vcomp[compi])
      loc = strsplit(rawData$Name_location[i],";")[[1]]
      print(loc)
      loc = loc[grep(vcomp[compi],loc)][1]
      idealData[["Location"]]=c(idealData[["Location"]],strsplit(loc,"\(")[[1]][2])
    } 
  }
}

idealData = as.data.frame(idealData)

给出输出：

> idealData
  ID   Company     Location
1  1 Company B        Japan
2  1 Company A         USA 
3  1 Company C Korea,South 
4  2 Company A         USA 
5  2 Company D         USA 
6  3 Company E     European

字符串拆分、合并和堆叠多列

String Split, Merge, and Stack multiple columns

r

transform

strsplit

stringr

tidyverse