错误与列数据框有R

Question

我有一个已经阻止了我2天的问题，希望我能在这里找到一个解决方案：

我创建一个包含单词列表的数据帧和其正，负极性为情感分析任务。

word   positive.polarity  negative.polarity 
1 interesting     1     0       
2  boring     0     1       

对于每个字I提取它的上下文，其是一组3分以上的字句。

我一直的助推器单词的列表：

-booster_words <- c("more","enough", "a lot", "as", "so") 
-negative_words <- c("not", "rien", "ni", "aucun", "nul", "jamais", "pas", "non plus", "sans") 

我想创建一个新的列positive.ponderate.polarity其中包含如果有一个助推器和否定词，每3 devided的正极性值上下文，如果在上下文中只有助推词（在上下文中有ni个否定词），则每3个字符相乘。

当我用这句话运行：

"The course was so interesting, but the professor was not boring" 

我得到这个数据帧：

word positive.polarity negative.polarity  positive.ponderate.polarity 
1 interesting     1     0     0.3333333 
2  boring     0     1     0.0000000 

但我已经找到，因为结果这个数据帧：

word positive.polarity negative.polarity positive.ponderate.polarity 
1 interesting     1     0     3 
2  boring     0     1     0.0000000 

这里代码：

calcPolarity <- function(sentiment_DF,sentences){ 
    booster_words <- c("more","enough", "a lot", "as", "so") 
    negative_words <- c("not", "rien", "ni", "aucun", "nul", "jamais", "pas", "non plus", "sans") 
    reduce_words <- c("peu", "presque", "moins", "seulement") 
    # pre-allocate the polarity result vector with size = number of sentences 
    polarity <- rep.int(0,length(sentences)) 

    # loop per sentence 
    for(i in 1:length(polarity)){ 
     sentence <- sentences[i] 

     # separate each sentence in words using regular expression 
     wordsOfASentence <- unlist(regmatches(sentence,gregexpr("[[:word:]]+",sentence,perl=TRUE))) 

     # get the rows of sentiment_DF corresponding to the words in the sentence using match 
     # N.B. if a word occurs twice, there will be two equal rows 
     # (but I think it's correct since in this way you count its polarity twice) 
     subDF <- sentiment_DF[match(wordsOfASentence,sentiment_DF$word,nomatch = 0),] 


     # Find (number) of matching word. 
     wordOfInterest <- wordsOfASentence[which(wordsOfASentence %in% levels(sentiment_DF$word))] # No multigrepl, so working with duplicates instead. eg interesting 
     regexOfInterest <- paste0("([^\\s]+\\s){0,3}", wordOfInterest, "(\\s[^\\s]+){0,3}") 

     # extract a context of 3 words before the word in the dataframe 
     context <- stringr::str_extract(sentence, regexOfInterest) 
     names(context) <- wordOfInterest # Helps in forloop 

     print(context) 
     if(any(unlist(strsplit(context, " ")) %in% booster_words)) 

     { 
      print(booster_words) 
      if(any(unlist(strsplit(context, " ")) %in% negative_words)) 

      { 
       subDF$positive.ponderate.polarity <- subDF$positive.polarity/3 

      } 
      else 
      { 
       subDF$positive.ponderate.polarity <- subDF$positive.polarity * 3 

      } 
     } 



     # Debug option 
     print(subDF) 

     # calculate the total polarity of the sentence and store in the vector 
     polarity[i] <- sum(subDF$positive.ponderate.polarity) - sum(subDF$negative.ponderate.polarity) 

    } 
    return(polarity) 
} 

sentiment_DF <- data.frame(word=c('interesting','boring','pretty'), 
          positive.polarity=c(1,0,1), 
          negative.polarity=c(0,1,0)) 
sentences <- c("The course was so interesting, but the professor was not boring") 
result <- calcPolarity(sentiment_DF,sentences) 

用法：

result <- calcPolarity(sentiment_DF,sentences) 
       interesting      boring 
"course was so interesting" "professor was not boring" 
[1] "more" "enough" "a lot" "as"  "so"  
     word positive.polarity negative.polarity positive.ponderate.polarity 
1 interesting     1     0     0.3333333 
2  boring     0     1     0.0000000 

编辑：

calcPolarity <- function(sentiment_DF,sentences){ 
    booster_words <- c("more","enough", "a lot", "as", "so") 
    negative_words <- c("not", "rien", "ni", "aucun", "nul", "jamais", "pas", "non plus", "sans") 
    reduce_words <- c("peu", "presque", "moins", "seulement") 
    # pre-allocate the polarity result vector with size = number of sentences 
    polarity <- rep.int(0,length(sentences)) 

    # loop per sentence 
    for(i in 1:length(polarity)){ 
     sentence <- sentences[i] 

     # separate each sentence in words using regular expression 
     wordsOfASentence <- unlist(regmatches(sentence,gregexpr("[[:word:]]+",sentence,perl=TRUE))) 

     # get the rows of sentiment_DF corresponding to the words in the sentence using match 
     # N.B. if a word occurs twice, there will be two equal rows 
     # (but I think it's correct since in this way you count its polarity twice) 
     subDF <- sentiment_DF[match(wordsOfASentence,sentiment_DF$word,nomatch = 0),] 


     # Find (number) of matching word. 
     wordOfInterest <- wordsOfASentence[which(wordsOfASentence %in% levels(sentiment_DF$word))] # No multigrepl, so working with duplicates instead. eg interesting 
     regexOfInterest <- paste0("([^\\s]+\\s){0,3}", wordOfInterest, "(\\s[^\\s]+){0,3}") 

     # extract a context of 3 words before the word in the dataframe 
     context <- stringr::str_extract(sentence, regexOfInterest) 
     names(context) <- wordOfInterest # Helps in forloop 

     print(context) 
     for(i in 1:length(context)){ 
      if(any(unlist(strsplit(context[i], " ")) %in% booster_words)) 

      { 
       print(booster_words) 
       if(any(unlist(strsplit(context[i], " ")) %in% negative_words)) 

       { 
        subDF$positive.ponderate.polarity <- subDF$positive.polarity + 4 

       } 
       else 
       { 
        subDF$positive.ponderate.polarity <- subDF$positive.polarity + 9 

       } 
      } 
     } 



     # Debug option 
     print(subDF) 

     # calculate the total polarity of the sentence and store in the vector 
     polarity[i] <- sum(subDF$positive.ponderate.polarity) - sum(subDF$negative.ponderate.polarity) 

    } 
    return(polarity) 
} 

sentiment_DF <- data.frame(word=c('interesting','boring','pretty'), 
          positive.polarity=c(1,0,1), 
          negative.polarity=c(0,1,0)) 
sentences <- c("The course was interesting, but the professor was not so boring") 
result <- calcPolarity(sentiment_DF,sentences) 

我得到这样的结果：

 word positive.polarity negative.polarity positive.ponderate.polarity 
1 interesting     1     0       5 
2  boring     0     1       4 

但它是incorrest，我必须有这样的结果

 word positive.polarity negative.polarity positive.ponderate.polarity 
1 interesting     1     0       1 
2  boring     0     1       4 

有什么想法吗？

Answer 1

问题是，你正在搜索你的函数中的两套程序单词，即正面和负面的有趣和无聊。

我加了一个for循环的calcPolarity功能，您通过运行之前，如果为我工作的语句：

for(i in 1:length(context)){ 
    if(any(unlist(strsplit(context[i], " ")) %in% booster_words)) 

    { 
    print(booster_words) 
    if(any(unlist(strsplit(context[i], " ")) %in% negative_words)) 

    { 
     subDF$positive.ponderate.polarity <- subDF$positive.polarity/3 

    } 
    else 
    { 
     subDF$positive.ponderate.polarity <- subDF$positive.polarity * 3 

    } 
    } 
} 

这给了预期的效果：

 word positive.polarity negative.polarity positive.ponderate.polarity 
    1 interesting     1     0       3 
    2  boring     0     1       0 

编辑：

问题是你没有正确索引你想要替换的值。因为您没有指定正极性极性的行，它正在改变整个列。

下面的编辑应该能够满足您的需求，但几乎可以肯定的是一种更有效的方法。

for(j in 1:length(context)){ 
    if(any(unlist(strsplit(context[j], " ")) %in% booster_words)) { 
    print(booster_words) 
    if(any(unlist(strsplit(context[j], " ")) %in% negative_words)){ 
     subDF$positive.ponderate.polarity[j] <- subDF$positive.polarity[j] + 4 
    } 
    else 
    { 
     subDF$positive.ponderate.polarity[j] <- subDF$positive.polarity[j] + 9 
    } 
    } 
    else { 
    subDF$positive.ponderate.polarity[j] <- subDF$positive.polarity[j] 
    } 

}