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我需要创建一个相似度矩阵，下面的代码就是我目前所拥有的。但是，结果不是我需要的。该代码返回一个有 16 行的矩阵，它是文档术语矩阵中的 8 个唯一术语与工作标题中的 2 个唯一术语的乘积。

我需要的是一个只有 4 行（每个标题一个）的矩阵，每一行表示 workTitle 中每个单词与标题中每个术语之间的编辑距离之和。

require(tm)

workTitle <- c("biomechanical engineer")
titles <- c("train machinist", "operations supervisor", "pharmacy tech", "mechanical engineer")

# create Corpus and a document-term matrix from the titles
titleCorpus <- Corpus(VectorSource(titles))
titleDtm <- DocumentTermMatrix(titleCorpus)

# print out the document-term matrix
inspect(titleDtm)

# calculate edit distance between every word from the test_var and the column names in the document-term matrix
d <- apply(titleDtm, 1, function(x) {
  terms <- unlist(strsplit(as.character(workTitle), " "))
  adist(colnames(titleDtm), terms)
})

这是上面代码的输出：

       Docs
         1  2  3  4
   [1,] 11 11 11 11
   [2,]  8  8  8  8
   [3,]  3  3  3  3
   [4,]  9  9  9  9
   [5,] 11 11 11 11
   [6,] 11 11 11 11
   [7,] 10 10 10 10
   [8,] 11 11 11 11
   [9,]  0  0  0  0
  [10,]  7  7  7  7
  [11,]  8  8  8  8
  [12,]  9  9  9  9
  [13,]  8  8  8  8
  [14,]  8  8  8  8
  [15,]  7  7  7  7
  [16,]  6  6  6  6
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1 回答 1

1

如果我理解正确的话，怎么样：

terms <- as.character(Dictionary(titleDtm))
dat <- data.frame(adist(titles, terms), row.names = titles)
colnames(dat) <- terms
dat

这导致

                       engineer machinist mechanical operations pharmacy supervisor tech train
 train machinist             12         6         11         12       11         14   12    10
 operations supervisor       16        17         18         11       18         11   19    17
 pharmacy tech               12        10         11         11        5         13    9    11
 mechanical engineer         11        13          9         16       16         16   16    16

然后是总和

data.frame(sum = rowSums(dat))

具有以下输出

                      sum
train machinist        88
operations supervisor 127
pharmacy tech          82
mechanical engineer   113
于 2013-09-14T10:46:17.800 回答