循环遍历 tm 语料库而不丢失语料库结构

Question

我有一个 tm 文档语料库和一个单词列表。我想在语料库上运行for 循环，以便循环按顺序从语料库中删除列表中的每个单词。

一些复制数据：

library(tm)
m <- cbind(c("Apple blue two","Pear yellow five","Banana yellow two"),
             c(1, 2, 3))
tm_corpus <- Corpus(VectorSource(m[,1]))
words <- as.list(c("Apple", "yellow", "two"))

tm_corpus 现在是一个包含 3 个文档的语料库对象：

<<SimpleCorpus>>
Metadata:  corpus specific: 1, document level (indexed): 0
Content:  documents: 3

words 是 3 个单词的列表：

[[1]]
[1] "Apple"

[[2]]
[1] "yellow"

[[3]]
[1] "two"

我尝试了三种不同的循环。第一个是：

tm_corpusClean <- tm_corpus
for (i in seq_along(tm_corpusClean)) {
  for (u in seq_along(words)) {
    tm_corpusClean[i] <- tm_map(tm_corpusClean[i], removeWords, words[[u]])
  }
}

返回以下错误 7 次（编号 1-7）：

Error in x$dmeta[i, , drop = FALSE] : incorrect number of dimensions
In addition: Warning messages:
1: In tm_corpusClean[i] <- tm_map(tm_corpusClean[i], removeWords,                 
words[[u]]) :
  number of items to replace is not a multiple of replacement length
2: In tm_corpusClean[i] <- tm_map(tm_corpusClean[i], removeWords,         
words[[u]]) :
  number of items to replace is not a multiple of replacement length
[...]

第二个是：

tm_corpusClean <- tm_corpus
for (i in seq_along(words)) {
  for (u in seq_along(tm_corpusClean)) {
    tm_corpusClean[u] <- tm_map(tm_corpusClean[u], removeWords, words[[i]])
  }
}

返回错误：

Error in x$dmeta[i, , drop = FALSE] : incorrect number of dimensions

最后一个循环是：

tm_corpusClean <- tm_corpus
for (i in seq_along(words)) {
  tm_corpusClean <- tm_map(tm_corpusClean, removeWords, words[[i]])
}

这实际上返回了一个名为tm_corpusClean的对象，但是这个对象只返回了第一个文档，而不是原来的三个：

inspect(tm_corpusClean[[1]])

<<PlainTextDocument>>
Metadata:  7
Content:  chars: 6

 blue 

我哪里错了？

Answer 1

在我们进行顺序删除之前，测试tm_map 是否适用于您的示例：

obj1 <- tm_map(tm_corpus, removeWords, unlist(words))
sapply(obj1, `[`, "content")

$`1.content`
[1] " blue "

$`2.content`
[1] "Pear  five"

$`3.content`
[1] "Banana  "

接下来，使用 lapply 依次删除一个单词，即"Apple", "yellow", "two"：

obj2 <- lapply(words, function(word) tm_map(tm_corpus, removeWords, word))
sapply(obj2, function(x) sapply(x, `[`, "content"))

          [,1]                [,2]             [,3]              
1.content " blue two"         "Apple blue two" "Apple blue "     
2.content "Pear yellow five"  "Pear  five"     "Pear yellow five"
3.content "Banana yellow two" "Banana  two"    "Banana yellow "  

请注意，生成的语料库在一个嵌套列表中（为什么使用两个 sapply 来查看内容）。