Question

我的目标是使用R进行基于词典的情感分析！

我有两个角色向量。一个有积极的话，一个有负面的词。例如

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

我现在有成千上万的新闻文章，我想知道每篇文章，如何我的向量pos和neg的许多元素都在文章中。

e.g。（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

如果我能为每篇文章获得以下内容，那么另一件好事就是：

（pos字数 - 字数）/（文章中总字数）

非常感谢你！

编辑：

@Victorp：这似乎不起作用

矩阵我看起来很好：

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

但是当我执行你的命令时，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

为什么会这样？

Answer 1

您好，您可以使用TermDocumentMatrix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

Answer 2

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%)

获取语料库中的字符向量元素的数量

2 个答案:

获取语​​料库中的字符向量元素的数量

2 个答案:

获取语料库中的字符向量元素的数量