在r中的不同数据框中匹配两个以上单词的单词

Question

我有两个这样的数据框 DF1 和 DF2。

ID = c(1, 2, 3, 4) 
Issues = c('Issue1, Issue4', 'Issue2, Issue5, Issue6', 'Issue3, Issue4', 'Issue1, Issue5')
Location = c('x', 'y', 'z', 'w')
Customer = c('a', 'b', 'c', 'd')
DF1 = data.frame(ID, Issues, Location, Customer)

Root_Cause = c('R1', 'R2', 'R3', 'R4')
List_of_Issues = c('Issue1, Issue3, Issue5', 'Issue2, Issue1, Issue4', 'Issue6, Issue7', 'Issue5, Issue6')  
DF2 = data.frame(Root_Cause, List_of_Issues)

我想比较 DF1 的 "Issues" 和 DF2 的 "List_of_Issues" 的数据帧，如果 "Issues" 列中的两个以上的单词存在于 "List_of_Issues" 中DF2 中的列，然后我想从 DF2 填充后续 "Root_Cause"。我生成的数据框应该看起来像 DF3。

ID = c(1, 2, 3, 4)
Issues = c('Issue1, Issue4', 'Issue2, Issue5, Issue6', 'Issue3, Issue4', 'Issue1, Issue5')
Location = c('x', 'y', 'z', 'w')
Customer = c('a', 'b', 'c', 'd')
Root_Cause = c('R2', 'R4', NA, 'R1')
DF3 = data.frame(ID, Issues, Location, Customer, Root_Cause)

Answer 1

使用data.table：

编辑： 我已经编辑了您的示例数据以解决多根本原因的可能性。在此数据中，ID==1 对应于 R2 和 R3。

数据

ID = c(1, 2, 3, 4) 
Issues = c('Issue1, Issue4, Issue6, Issue7', 'Issue2, Issue5, Issue6', 'Issue3, Issue4', 'Issue1, Issue5')
Location = c('x', 'y', 'z', 'w')
Customer = c('a', 'b', 'c', 'd')
DF1 = data.table(ID, Issues, Location, Customer)

Root_Cause = c('R1', 'R2', 'R3', 'R4')
List_of_Issues = c('Issue1, Issue3, Issue5', 'Issue2, Issue1, Issue4', 'Issue6, Issue7', 'Issue5, Issue6')  
DF2 = data.table(Root_Cause, List_of_Issues)

代码

DF1[, Issues := strsplit(Issues, split = ', ')]
DF2[, List_of_Issues := strsplit(List_of_Issues, split = ', ')]

DF1[, RootCause := lapply(Issues, function(x){

  matchvec = sapply(DF2[, List_of_Issues], function(y) length(unlist(intersect(y, x))))
  ids = which(matchvec > 1)
  str = DF2[, paste(Root_Cause[ids], collapse = ', ')]

  ifelse(str == '', NA, str)

})]

结果

> DF1
   ID                      Issues Location Customer RootCause
1:  1 Issue1,Issue4,Issue6,Issue7        x        a    R2, R3
2:  2        Issue2,Issue5,Issue6        y        b        R4
3:  3               Issue3,Issue4        z        c        NA
4:  4               Issue1,Issue5        w        d        R1

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