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我的目标是使用 R 进行基于词典的情感分析！

我有两个字符向量。一个用正面词，一个用负面词。例如

pos <- c("good", "accomplished", "won", "happy")
neg <- c("bad", "loss", "damaged", "sued", "disaster")

我现在有一个包含数千篇新闻文章的语料库，我想知道每篇文章，我的向量 pos 和 neg 在文章中有多少元素。

例如（不确定语料库功能如何在这里工作，但你明白了：我的语料库中有两篇文章）

mycorpus <- Corpus("The CEO is happy that they finally won the case.", "The disaster caused a huge loss.")

我想得到这样的东西：

article 1: 2 element of pos and 0 element of neg
article 2: 0 elements of pos, 2 elements of neg

另一件好事是，如果我能为每篇文章获得以下信息：

(pos 词数 - neg 词数)/(文章总词数)

非常感谢你！！

编辑：

@Victorp：这似乎不起作用

我得到的矩阵看起来不错：

mytdm[1:6,1:10]
               Docs
Terms          1 2 3 4 5 6 7 8 9 10
aaron          0 0 0 0 0 1 0 0 0  0
abandon        1 1 0 0 0 0 0 0 0  0
abandoned      0 0 0 3 0 0 0 0 0  0
abbey          0 0 0 0 0 0 0 0 0  0
abbott         0 0 0 0 0 0 0 0 0  0
abbotts        0 0 1 0 0 0 0 0 0  0

但是当我执行您的命令时，每个文档都会得到零！

colSums(mytdm[rownames(mytdm) %in% pos, ])
   1    2    3    4    5    6    7    8    9   10   11   12   13   14   15 
   0    0    0    0    0    0    0    0    0    0    0    0    0    0    0 
  16   17   18   19   20   21   22   23   24   25   26   27   28   29   30 
   0    0    0    0    0    0    0    0    0    0    0    0    0    0    0

这是为什么？？

4

2 回答 2

1

您好，您可以使用 TermDocumentMatrix 来执行此操作：

mycorpus <- Corpus(VectorSource(c("The CEO is happy that they finally won the case.", "The disaster caused a huge loss.")))
mytdm <- TermDocumentMatrix(mycorpus, control=list(removePunctuation=TRUE))
mytdm <- as.matrix(mytdm)

# Positive words
colSums(mytdm[rownames(mytdm) %in% pos, ])
1 2 
2 0 

# Negative words
colSums(mytdm[rownames(mytdm) %in% neg, ])
1 2 
0 2 

# Total number of words per documents
colSums(mytdm)
1 2 
9 5
于 2014-02-25T15:00:17.527 回答
1

这是另一种方法：

## pos <- c("good", "accomplished", "won", "happy")
## neg <- c("bad", "loss", "damaged", "sued", "disaster")
## 
## mycorpus <- Corpus(VectorSource(
##     list("The CEO is happy that they finally won the case.", 
##     "The disaster caused a huge loss.")))

library(qdap)
with(tm_corpus2df(mycorpus), termco(text, docs, list(pos=pos, neg=neg)))

##   docs word.count       pos       neg
## 1    1         10 2(20.00%)         0
## 2    2          6         0 2(33.33%)
于 2014-02-25T15:40:51.380 回答