Question

我有一个看起来像这样的数据框

    Word<-c("bat", "cat", "cab", "some", "ban", "bait", "at", "done", "dot", "ran", "cant")
    S1<-c("b","c","c","s", "b", "b", "a", "d","d", "r", "c")
    S2<-c("a","a","a","o","a","a","t","o","o","a","a")
    S3<-c("t","t","b","m", "n", "i", "", "n","t", "n", "n")
    S4<-c("","","","e", "", "t", "", "e","", "", "t")
    df<-data.frame(Word, S1, S2, S3, S4,  stringsAsFactors=FALSE)

我想计算相似发音的单词的数量和名称。相似的发音词是指通过添加，替换或删除而相差一个声音的单词。简而言之，我想要这样的东西

    Word<-c("bat", "cat", "cab", "some", "ban", "bait", "at", "done", "dot", "ran", "cant")
    S1<-c("b","c","c","s", "b", "b", "a", "d","d", "r", "c")
    S2<-c("a","a","a","o","a","a","t","o","o","a","a")
    S3<-c("t","t","b","m", "n", "i", "", "n","t", "n", "n")
    S4<-c("","","","e", "", "t", "", "e","", "", "t")
    Number<-c(4,4,1,0,2,1,2,0,0,1,2)
    Names<-c("cat, ban, bait, at", "bat, cab, at, cant","cat","","bat, ran","bat","bat, cat","","","ban","can, cat")
    df2<-data.frame(Word, S1, S2, S3, S4, Number, Names,  stringsAsFactors=FALSE)

Answer 1

如果我的理解正确，似乎您正在寻找主题词之间的Levenshtein distance。 utils软件包中的adist函数可以为您计算Levenshtein距离。它返回一个矩阵，其中包含要从第i个单词到第j个单词的替换/插入/删除的数量。

dist <- utils::adist(Word)
dist

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
 [1,]    0    1    2    4    1    1    1    4    2     2     2
 [2,]    1    0    1    4    2    2    1    4    2     2     1
 [3,]    2    1    0    4    2    3    2    4    3     2     2
 [4,]    4    4    4    0    4    4    4    2    3     4     4
 [5,]    1    2    2    4    0    2    2    3    3     1     2
 [6,]    1    2    3    4    2    0    2    4    3     3     2
 [7,]    1    1    2    4    2    2    0    4    2     2     2
 [8,]    4    4    4    2    3    4    4    0    2     3     3
 [9,]    2    2    3    3    3    3    2    2    0     3     3
[10,]    2    2    2    4    1    3    2    3    3     0     2
[11,]    2    1    2    4    2    2    2    3    3     2     0

然后，您可以在行或列上循环并返回距离为1的任何单词。

links <- apply(dist, 1, function(d) {
  paste0(Word[d == 1], collapse = ", ")
})
cbind.data.frame(Word, links)

   Word              links
1   bat cat, ban, bait, at
2   cat bat, cab, at, cant
3   cab                cat
4  some                   
5   ban           bat, ran
6  bait                bat
7    at           bat, cat
8  done                   
9   dot                   
10  ran                ban
11 cant                cat

现在您已经以编程方式派生了df2的第一列和最后一列。对于计数，您可以简单地使用：

counts <- apply(dist, 1, function(d){sum(d == 1)})

计算数据框中类似发音的单词的数量和名称

1 个答案: