如何从R中带有模式的字符串中提取特定单词

Question

我有一个数据框，其中包含教师中学生论文的导师和顾问的姓名，例如：

 DF<-data.frame(Names=c("Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3",
  "Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3",
  "Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3"))

我会将主管和顾问分成两个不同的列（如我所愿），如下所示：

DF1<-data.frame(Supervisor=c("Ali Ahmadi","Ali Ahmadi","Ali Ahmadi"),Advisors=c("Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi","Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi","Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi"))

DF1
  Supervisor                                             Advisors
1 Ali Ahmadi Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi
2 Ali Ahmadi Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi
3 Ali Ahmadi Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi

我尝试了以下代码：

DF1<-strsplit(DF$Names, "Name :")

stopwords = c(":","Type","Family","Name","1","2", "3", "Advisor", "Family")

DF2 <- lapply(DF1,function(x) unlist(strsplit(x," ")) )

DF3 <- lapply(DF2,function(x)  x[!x %in% stopwords] )

DF4<-lapply(DF3,function(x)  paste(x, collapse = " "))

但最终结果如下所示不是我的预期，显然需要进一步的工作才能转换为数据框！：

DF4
[[1]]
[1] " Ali , Ahmadi , First supervisor  Aram , Rezaeei ,  Omid , Saeedi ,  Nima , Shaki ,  Sohrab , Karimi ,"

[[2]]
[1] " Ali , Ahmadi , First supervisor  Aram , Rezaeei ,  Omid , Saeedi ,  Nima , Shaki ,  Sohrab , Karimi ,"

[[3]]
[1] " Ali , Ahmadi , First supervisor  Aram , Rezaeei ,  Omid , Saeedi ,  Nima , Shaki ,  Sohrab , Karimi ,"

有没有简化的方法来解决这个问题？ 我发现 regexp 可能会有所帮助，但至少在我的示例中我不知道如何使用它。 提前感谢您的任何回答...

Answer 1

这是一个使用extract的尝试：

library(tidyr)
DF %>%
  # clean strings:
  mutate(Names = gsub("\\s?(Name|Family|First supervisor|Advisor|Type|\\d|\\s[,:])", "", Names, perl = TRUE)) %>%
  # extract data into columns:
  extract(Names,
          into = c("Supervisor", "Advisor"),
          regex = "(\\w+\\s\\w+)\\s(.*)") %>%
  # insert commas into `Advisor`:
  mutate(Advisor = gsub("(\\w+\\s\\w+\\b)(?!$)", "\\1,", Advisor, perl = TRUE))
  Supervisor                                              Advisor
1 Ali Ahmadi Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi
2 Ali Ahmadi Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi
3 Ali Ahmadi Aram Rezaeei, Omid Saeedi, Nima Shaki, Sohrab Karimi

说明（根据 OP 的要求）：

extract的regex表达式中的正则表达式旨在完成两项任务：

• (i) 它必须从头到尾将字符串描述为一个整体
• (ii) 它必须挑选出那些应该填充新创建的列的元素

任务 (i) 是通过(\\w+\\s\\w+)捕获组成Supvervisor名称的两个词来实现的，而\\s描述（但不捕获）以下空格，而(.*)描述/ 匹配该空格后面的任何内容 - 即在本例中为四个Advisor名称。

任务 (ii) 是通过将Supvervisor名称和Advisor名称包装在括号中给出的捕获组中来实现的； 这些括号是函数extract “意识到”它们的内容应该进入新列的“语法”。

最后使用捕获组再次在Advisor名称之间插入逗号，可以使用反向引用（ \\1 ）在gsub的替换参数中重新收集该逗号。 (?!$)表达式是一个否定的前瞻，它断言只有当单词边界锚\\b后面的内容不是（因此前瞻中的! ）字符串的结尾（以$表示）时才插入逗号）。 希望这可以帮助！

数据：

DF<-data.frame(Names=c("Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3",
                       "Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3",
                       "Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3"))

Answer 2

这是一个基本的 R 解决方案。

DF <- data.frame(Names=c("Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3",
                       "Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3",
                       "Name : Ali , Family : Ahmadi , Type : First supervisor Name : Aram , Family : Rezaeei , Type : Advisor Name : Omid , Family : Saeedi , Type : Advisor 1 Name : Nima , Family : Shaki , Type : Advisor 2 Name : Sohrab , Family : Karimi , Type : Advisor 3"))

stopwords <- c(":","Type","Family","Name","1","2", "3", "Advisor", "Family")
stoppattern <- paste(stopwords, collapse = "|")

DF1 <- strsplit(DF$Names, "Name :")
DF1 <- lapply(DF1, \(x) trimws(x[sapply(x, nchar) > 0L]))

DF2 <- lapply(DF1, \(x) {
  gsub(stoppattern, "", x)
})

DF3 <- lapply(DF2, \(x) {
  y <- gsub(stoppattern, "", x)
  y <- strsplit(x, ",")
  y <- lapply(y, trimws)
  lapply(y, \(.y) {
    .y <- trimws(.y)
    .y[sapply(.y, nchar) > 0L]
  })
})

DF4 <- lapply(DF3, \(x) {
  Supervisor <- x[[1]][1:2]
  Supervisor <- paste(trimws(Supervisor), collapse = " ")
  Advisors <- unlist(x[-1])
  Advisors <- paste(trimws(Advisors), collapse = ", ")
  data.frame(Supervisor, Advisors)
})

Final <- do.call(rbind, DF4)
Final
#>   Supervisor                                                 Advisors
#> 1 Ali Ahmadi Aram, Rezaeei, Omid, Saeedi, Nima, Shaki, Sohrab, Karimi
#> 2 Ali Ahmadi Aram, Rezaeei, Omid, Saeedi, Nima, Shaki, Sohrab, Karimi
#> 3 Ali Ahmadi Aram, Rezaeei, Omid, Saeedi, Nima, Shaki, Sohrab, Karimi

^{由reprex 包(v2.0.1) 创建于 2022-06-05}

Answer 3

凌乱的基地R：

# Store a vector of names: ir_names => character vector
ir_names <- c("Name", "Family", "Type")

# Compute it's lenght: ir_name_len => string scalar
ir_name_len <- length(ir_names)

# Compute the desired result: res => data.frame
res <- do.call(
  rbind, 
  lapply(
    strsplit(
      DF$Names,
      "Name\\s+\\:\\s+"
    ),
    function(x){
      y <- data.frame(tmp = unlist(strsplit(x, " , ")))
      ir1 <- setNames(
        data.frame(
          do.call(
            rbind, 
            lapply(
              split(
                y, 
                ceiling(seq_len(nrow(y))/ir_name_len)
              ), 
              t
            )
          ),
          row.names = NULL,
          stringsAsFactors = FALSE
        ),
        ir_names
      )
      ir2 <- transform(
        ir1,
        Name = trimws(paste(Name, gsub("Family\\s+\\:\\s+", "", Family))),
        Type = trimws(gsub("Type\\s+\\:\\s+", "", Type))
      )[,c("Name", "Type")]
      ir3 <- data.frame(
        Supervisor = ir2$Name[which(grepl("supervisor", ir2$Type))],
        Advisor = toString(ir2$Name[-which(grepl("supervisor", ir2$Type))]),
        stringsAsFactors = FALSE,
        row.names = NULL
      )
    }
  )
)
# Print to console: data.frame => stdout(console)
res