r - 使用带有 grepl 和循环的名称列表从字符串中提取名称，并将它们添加到 R 中的新列

Question

我有一个数据集，其中一列包含姓名，一列指示该人白天做了什么。我正在尝试使用 R 找出那天在我的数据集中谁会见了谁。我创建了一个包含数据集中名称的向量，并在循环中使用 grepl 来确定名称出现在详细说明人们事件的列中的位置在数据集中。

name <- c("Dupont","Dupuy","Smith") 

activity <- c("On that day, he had lunch with Dupuy in London.", 
              "She had lunch with Dupont and then went to Brighton to meet Smith.", 
              "Smith remembers that he was tired on that day.")

met_with <- c("Dupont","Dupuy","Smith")

df<-data.frame(name, activity, met_with=NA)


for (i in 1:length(met_with)) {
df$met_with<-ifelse(grepl(met_with[i], df$activity), met_with[i], df$met_with)
}

然而，由于两个原因，该解决方案并不令人满意。当这个人遇到一个以上的人时，我不能提取一个以上的名字(在我的例子中是 Dupuy)，我不能告诉 R 在我的名字中使用这个名字而不是代词时不要返回这个人的名字事件列(例如史密斯)。
理想情况下，我希望 df 看起来像:

  name         activity                                            met_with                             
  Dupont       On that day, he had lunch with Dupuy in London.     Dupuy
  Dupuy        She had lunch with Dupont and then (...).           Dupont Smith
  Smith        Smith remembers that he was tired on that day.      NA

我正在清理字符串以构建边缘列表和节点列表，以便稍后进行网络分析。
谢谢

Answer 1

与@Gki 相同的逻辑，但使用 stringr函数和 mapply而不是循环。

library(stringr)

pat <- str_c('\\b', df$name, '\\b', collapse = '|')
df$met_with <- mapply(function(x, y) str_c(setdiff(x, y), collapse = ' '), 
       str_extract_all(df$activity, pat), df$name)

df

#    name                                                           activity
#1 Dupont                    On that day, he had lunch with Dupuy in London.
#2  Dupuy She had lunch with Dupont and then went to Brighton to meet Smith.
#3  Smith                     Smith remembers that he was tired on that day.

#      met_with
#1        Dupuy
#2 Dupont Smith
#3