3

我的文件如下：

 doc1 = very good, very bad, you are great
 doc2 = very bad, good restaurent, nice place to visit

我想让我的语料库分开，,以便我的最终DocumentTermMatrix变成：

      terms
 docs       very good      very bad        you are great   good restaurent   nice place to visit
  doc1       tf-idf          tf-idf         tf-idf          0                    0
  doc2       0                tf-idf         0                tf-idf             tf-idf

我知道如何计算DocumentTermMatrix单个单词，但不知道如何separated for each phrase在 R中制作语料库。R首选解决方案，但Python也欢迎解决方案。

我尝试过的是：

> library(tm)
> library(RWeka)
> BigramTokenizer <- function(x) NGramTokenizer(x, Weka_control(min = 1, max = 3))
> options(mc.cores=1)
> texts <- c("very good, very bad, you are great","very bad, good restaurent, nice place to visit")
> corpus <- Corpus(VectorSource(texts))
> a <- TermDocumentMatrix(corpus, control = list(tokenize = BigramTokenizer))
> as.matrix(a)

我正进入（状态：

                         Docs
  Terms                   1 2
  bad good restaurent   0 1
  bad you are           1 0
  good restaurent nice  0 1
  good very bad         1 0
  nice place to         0 1
  place to visit        0 1
  restaurent nice place 0 1
  very bad good         0 1
  very bad you          1 0
  very good very        1 0
  you are great         1 0

我想要的不是单词的组合，而是我在矩阵中显示的短语。

4

3 回答 3

1

这是使用qdap+tm包的一种方法：

library(qdap); library(tm); library(qdapTools)

dat <- list2df(list(doc1 = "very good, very bad, you are great",
 doc2 = "very bad, good restaurent, nice place to visit"), "text", "docs")

x <- sub_holder(", ", dat$text)

m <- dtm(wfm(x$unhold(gsub(" ", "~~", x$output)), dat$docs) )
weightTfIdf(m)

inspect(weightTfIdf(m))

## A document-term matrix (2 documents, 5 terms)
## 
## Non-/sparse entries: 4/6
## Sparsity           : 60%
## Maximal term length: 19 
## Weighting          : term frequency - inverse document frequency (normalized) (tf-idf)
## 
##       Terms
## Docs   good restaurent nice place to visit very bad very good you are great
##   doc1       0.0000000           0.0000000        0 0.3333333     0.3333333
##   doc2       0.3333333           0.3333333        0 0.0000000     0.0000000

您也可以一举返回 a DocumentTermMatrix，但这可能更难理解：

x <- sub_holder(", ", dat$text)

apply_as_tm(t(wfm(x$unhold(gsub(" ", "~~", x$output)), dat$docs)), 
    weightTfIdf, to.qdap=FALSE)
于 2014-06-04T14:28:40.483 回答
0

如果您只是使用 strsplit 以逗号分隔，然后通过与某些字符组合将您的短语变成单个“单词”怎么办。例如

library(tm)
docs <- c(D1 = "very good, very bad, you are great", 
    D2 = "very bad, good restaurent, nice place to visit")

dd <- Corpus(VectorSource(docs))
dd <- tm_map(dd, function(x) {
    PlainTextDocument(
       gsub("\\s+","~",strsplit(x,",\\s*")[[1]]), 
       id=ID(x)
     )
})
inspect(dd)

# A corpus with 2 text documents
# 
# The metadata consists of 2 tag-value pairs and a data frame
# Available tags are:
#   create_date creator 
# Available variables in the data frame are:
#   MetaID 

# $D1
# very~good
# very~bad
# you~are~great
# 
# $D2
# very~bad
# good~restaurent
# nice~place~to~visit

dtm <- DocumentTermMatrix(dd, control = list(weighting = weightTfIdf))
as.matrix(dtm)

这将产生

# Docs good~restaurent nice~place~to~visit very~bad very~good you~are~great
#   D1       0.0000000           0.0000000        0 0.3333333     0.3333333
#   D2       0.3333333           0.3333333        0 0.0000000     0.0000000
于 2014-06-04T15:21:00.300 回答
0

对于使用 text2vec 的任何人来说，这是基于自定义词汇的非常方便的解决方案：

library(text2vec)
doc1 <- 'very good, very bad, you are great'
doc2 <- 'very bad, good restaurent, nice place to visit'
docs <- list(doc1, doc2)
docs <- sapply(docs, strsplit, split=', ')
vocab <- vocab_vectorizer(create_vocabulary(unique(unlist(docs))))
dtm <- create_dtm(itoken(docs), vocab)
dtm

这将导致：

2 x 5 sparse Matrix of class "dgCMatrix"
  very good very bad you are great good restaurent nice place to visit
1         1        1             1               .                   .
2         .        1             .               1                   1

这种方法允许在加载文件和准备词汇时进行更多定制。

于 2018-11-26T15:24:02.657 回答