如何在R中的字符串向量中找到相似的词？

  • Z时代
  • 2024-01-10
  • 分类：综合

有时字符串向量中的字符串有拼写错误，我们希望提取相似的单词以避免这种拼写错误，因为相似的单词可能表示单词的正确和不正确形式。这可以通过使用 agrep 和 lapply 函数来实现。

在线示例1

x1<-c("India","United Kingdoms","Indiaa","Egyypt","United
Kingdom","Turkey","Egypt","Belaarus","Belarus")
lapply(x1,agrep,x1,value=TRUE)

输出结果

[[1]]
[1] "India" "Indiaa"
[[2]]
[1] "United Kingdoms" "United Kingdom"
[[3]]
[1] "India" "Indiaa"
[[4]]
[1] "Egyypt" "Egypt"
[[5]]
[1] "United Kingdoms" "United Kingdom"
[[6]]
[1] "Turkey"
[[7]]
[1] "Egyypt" "Egypt"
[[8]]
[1] "Belaarus" "Belarus"
[[9]]
[1] "Belaarus" "Belarus"

在线示例2

x2<-c("Alhadi","Umair","Omar","Alhadi","Shanti","Shant","Umaer","Peter","Rahul","Pattrick","P
eeter","Rahuls")
lapply(x2,agrep,x2,value=TRUE)

输出结果

[[1]]
[1] "Al-hadi" "Alhadi"
[[2]]
[1] "Umair" "Umaer"
[[3]]
[1] "Omar"
[[4]]
[1] "Al-hadi" "Alhadi"
[[5]]
[1] "Shanti" "Shant"
[[6]]
[1] "Shanti" "Shant"
[[7]]
[1] "Umair" "Umaer"
[[8]]
[1] "Peter" "Peeter"
[[9]]
[1] "Rahul" "Rahuls"
[[10]]
[1] "Pattrick"
[[11]]
[1] "Peter" "Peeter"
[[12]]
[1] "Rahul" "Rahuls"

在线示例3

x3<-c("Alabamaa","New Yorky","New
Yok","Alabma","Florida","Illinois","Texas","Illinoise")
lapply(x3,agrep,x3,value=TRUE)

输出结果

[[1]]
[1] "Alabamaa"
[[2]]
[1] "New Yorky"
[[3]]
[1] "New Yorky" "New Yok"
[[4]]
[1] "Alabamaa" "Alabma"
[[5]]
[1] "Florida"
[[6]]
[1] "Illinois" "Illinoise"
[[7]]
[1] "Texas"
[[8]]
[1] "Illinois" "Illinoise"

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