如何在R中的字符串向量中找到相似的词？
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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