我正在尝试使用AFINN字典(get_sentiments(“afinn”)对Tweets数据集的情绪。下面提供了数据集的示例:
A tibble: 10 x 2
Date TweetText
<dttm> <chr>
1 2018-02-10 21:58:19 "RT @RealSirTomJones: Still got the moves! That was a lo~
2 2018-02-10 21:58:19 "Yass Tom \U0001f600 #snakehips still got it #TheVoiceUK"
3 2018-02-10 21:58:19 Yasss tom he’s some chanter #TheVoiceUK #ItsNotUnusual
4 2018-02-10 21:58:20 #TheVoiceUK SIR TOM JONES...HE'S STILL HOT... AMAZING VO~
5 2018-02-10 21:58:21 I wonder how many hips Tom Jones has been through? #TheV~
6 2018-02-10 21:58:21 Tom Jones has still got it!!! #TheVoiceUK
7 2018-02-10 21:58:21 Good grief Tom Jones is amazing #TheVoiceuk
8 2018-02-10 21:58:21 RT @tonysheps: Sir Thomas Jones you’re a bloody legend #~
9 2018-02-10 21:58:22 @ITV Tom Jones what a legend!!! ❤️ #StillGotIt #TheVoice~
10 2018-02-10 21:58:22 "RT @RealSirTomJones: Still got the moves! That was a lo~
我想做的是: 1.将推文分成单个单词。 2.使用AFINN词典对这些单词进行评分。 3.总结每条推文的所有单词的分数 4.将此总和返回到新的第三列,这样我就能看到每条推文的分数。
对于类似的词典,我发现了以下代码:
# Initiate the scoreTopic
scoreTopic <- 0
# Start a loop over the documents
for (i in 1:length (myCorpus)) {
# Store separate words in character vector
terms <- unlist(strsplit(myCorpus[[i]]$content, " "))
# Determine the number of positive matches
pos_matches <- sum(terms %in% positive_words)
# Determine the number of negative matches
neg_matches <- sum(terms %in% negative_words)
# Store the difference in the results vector
scoreTopic [i] <- pos_matches - neg_matches
} # End of the for loop
dsMyTweets$score <- scoreTopic
但是我无法调整此代码以使其与afinn字典一起使用。
答案 0 :(得分:1)
对于整洁的数据原则,这将是一个很好的用例。让我们设置一些示例数据(这些是我的真实推文)。
library(tidytext)
library(tidyverse)
tweets <- tribble(
~tweetID, ~TweetText,
1, "Was Julie helping me because I don't know anything about Python package management? Yes, yes, she was.",
2, "@darinself OMG, this is my favorite.",
3, "@treycausey @ftrain THIS IS AMAZING.",
4, "@nest No, no, not in error. Just the turkey!",
5, "The @nest people should write a blog post about how many smoke alarms went off yesterday. (I know ours did.)")
现在我们有一些示例数据。在下面的代码中,unnest_tokens()
对文本进行了标记,即将其分解为单个单词(tidytext包允许您使用特殊的标记器进行推文),inner_join()
实现情感分析。
tweet_sentiment <- tweets %>%
unnest_tokens(word, TweetText, token = "tweets") %>%
inner_join(get_sentiments("afinn"))
#> Joining, by = "word"
现在我们可以找到每条推文的分数。将推文的原始数据集和left_join()
的每个推文的得分sum()
加到其上。来自tidyr的便捷函数replace_na()
允许您将生成的NA
值替换为零。
tweets %>%
left_join(tweet_sentiment %>%
group_by(tweetID) %>%
summarise(score = sum(score))) %>%
replace_na(list(score = 0))
#> Joining, by = "tweetID"
#> # A tibble: 5 x 3
#> tweetID TweetText score
#> <dbl> <chr> <dbl>
#> 1 1. Was Julie helping me because I don't know anything about … 4.
#> 2 2. @darinself OMG, this is my favorite. 2.
#> 3 3. @treycausey @ftrain THIS IS AMAZING. 4.
#> 4 4. @nest No, no, not in error. Just the turkey! -4.
#> 5 5. The @nest people should write a blog post about how many … 0.
由reprex package(v0.2.0)创建于2018-05-09。
如果您对情绪分析和文本挖掘感兴趣,我邀请您查看extensive documentation and tutorials we have for tidytext。
答案 1 :(得分:0)
供将来参考:
Score_word <- function(x) {
word_bool_vec <- get_sentiments("afinn")$word==x
score <- get_sentiments("afinn")$score[word_bool_vec]
return (score) }
Score_tweet <- function(sentence) {
words <- unlist(strsplit(sentence, " "))
words <- as.vector(words)
scores <- sapply(words, Score_word)
scores <- unlist(scores)
Score_tweet <- sum(scores)
return (Score_tweet)
}
dsMyTweets$score<-apply(df, 1, Score_tweet)
这执行我最初的想法! :)