1

我正在尝试对已经获取并存储在 MongoDb 中的推文进行情感分析。获取数据框格式的推文后，我收到以下错误：

ip.txt=laply(ip.lst,function(t) t$getText())
Error in t$getText : $ operator is invalid for atomic vectors

整个代码如下：

iphone.tweets <- searchTwitter('#iphone', n=15, lang="en")
iphone.text=laply(iphone.tweets,function(t) t$getText())
df_ip <- as.data.frame(iphone.text)

m <- mongo("iphonecollection",db="project")
m$insert(df_ip)
df_ip<-m$find()
ip.lst<-as.list(t(df_ip))
ip.txt=laply(ip.lst,function(t) t$getText())

我想要做的是计算情绪分数如下：

iphone.scores <- score.sentiment(ip.txt, pos.words,neg.words, .progress='text')

score.sentiment 例程如下：

  score.sentiment = function(sentences, pos.words, neg.words, .progress='none')
{
  require(plyr)
  require(stringr)
   # we got a vector of sentences. plyr will handle a list or a vector as an "l" for us
   # we want a simple array of scores back, so we use "l" + "a" + "ply" = laply:
  scores = laply(sentences, function(sentence, pos.words, neg.words) {
    # clean up sentences with R's regex-driven global substitute, gsub():
    sentence = gsub('[[:punct:]]', '', sentence)
    sentence = gsub('[[:cntrl:]]', '', sentence)
    sentence = gsub('\\d+', '', sentence)
    # and convert to lower case:
    sentence = tolower(sentence)
    # split into words. str_split is in the stringr package
    word.list = str_split(sentence, '\\s+')
    # sometimes a list() is one level of hierarchy too much
    words = unlist(word.list)
    # compare our words to the dictionaries of positive & negative terms
    pos.matches = match(words, pos.words)
    neg.matches = match(words, neg.words)
    # match() returns the position of the matched term or NA
    # we just want a TRUE/FALSE:
    pos.matches = !is.na(pos.matches)
    neg.matches = !is.na(neg.matches)
    # and conveniently enough, TRUE/FALSE will be treated as 1/0 by sum():
    score = sum(pos.matches) - sum(neg.matches)
    return(score)
   }, pos.words, neg.words, .progress=.progress )
   scores.df = data.frame(score=scores, text=sentences)
   return(scores.df)
 }
4

1 回答 1

1

我认为您想使用sapply，它会使返回的状态对象列表变平searchTwitter。在任何情况下，这都有效。请注意，您需要安装然后启动MongoDB它才能工作：

library(twitteR)
library(plyr)
library(stringr)
library(mongolite)

# you have to set up a Twitter Application at https://dev.twitter.com/ to get these 
#
ntoget <- 600 # get 600 tweets

iphone.tweets <- searchTwitter('#iphone', n=ntoget, lang="en")
iphone.text <- sapply(iphone.tweets,function(t) t$getText())
df_ip <- as.data.frame(iphone.text)

# MongoDB must be installed and the service started (mongod.exe in Windows)
#
m <- mongo("iphonecollection",db="project")
m$insert(df_ip)
df_ip_out<-m$find()

# Following routine (score.sentiment) was copied from:
# http://stackoverflow.com/questions/32395098/r-sentiment-analysis-with-phrases-in-dictionaries
#
score.sentiment = function(sentences, pos.words, neg.words, .progress='none')
{
  require(plyr)  
  require(stringr)  
  # we got a vector of sentences. plyr will handle a list  
  # or a vector as an "l" for us  
  # we want a simple array ("a") of scores back, so we use  
  # "l" + "a" + "ply" = "laply":  
  scores = laply(sentences, function(sentence, pos.words, neg.words) {
    # clean up sentences with R's regex-driven global substitute, gsub():
    sentence = gsub('[[:punct:]]', '', sentence)
    sentence = gsub('[[:cntrl:]]', '', sentence)
    sentence = gsub('\\d+', '', sentence)    
    # and convert to lower case:    
    sentence = tolower(sentence)    
    # split into words. str_split is in the stringr package    
    word.list = str_split(sentence, '\\s+')    
    # sometimes a list() is one level of hierarchy too much    
    words = unlist(word.list)    
    # compare our words to the dictionaries of positive & negative terms
    pos.matches = match(words, pos)
    neg.matches = match(words, neg)   
    # match() returns the position of the matched term or NA    
    # we just want a TRUE/FALSE:    
    pos.matches = !is.na(pos.matches)   
    neg.matches = !is.na(neg.matches)   
    # and conveniently enough, TRUE/FALSE will be treated as 1/0 by sum():
    score = sum(pos.matches) - sum(neg.matches)    
    return(score)    
  }, pos.words, neg.words, .progress=.progress )  
  scores.df = data.frame(score=scores, text=sentences)  
  return(scores.df)  
}

tweets <- as.character(df_ip_out$iphone.text)
neg = c("bad","prank","inferior","evil","poor","minor")
pos = c("good","great","superior","excellent","positive","super","better")
analysis <- score.sentiment(tweets,pos,neg)
table(analysis$score)

产生以下结果（4 分差，592 分中立，4 分好）：

 -1   0   1 
  4 592   4
于 2015-12-26T13:03:35.983 回答