Question

道歉，如果这是一个非常具体的问题，可能不会推广到其他人的问题。

背景

我希望做一些情感分析，从词典中单词的基本二进制匹配开始，然后转向更复杂的情感分析形式，利用语法规则等。

问题

要做一些二元匹配 - 这将形成情感分析的第一阶段 - 我提供了两个表，一个包含单词，另一个包含这些单词的词性。

    V1     V2        V3          V4   V5
1    R     is fantastic    language <NA>
2 Java     is       far        from good
3 Data mining        is fascinating <NA>


   V1  V2  V3 V4   V5
1  NN VBZ  JJ NN <NA>
2 NNP VBZ  RB IN   JJ
3 NNP  NN VBZ JJ <NA>

我想执行一些基本的情感分析，如下所示：我想应用一个带有两个参数的函数，一个字（来自第一个数据框）和相应的POS标记（来自第二个）以确定哪个列表用于确定单词的正/负方向的单词。例如，单词“奇妙”将与POS标签“JJ”一起被提取，因此单独的形容词列表将被检查该单词的存在/不存在。

最后，我想得到一个显示匹配结果的数据框：

   V1  V2  V3 V4   V5
1  0   0   1   0   <NA>
2  0   0  -1   0   1
3  0   0   0   1   <NA>

我尝试制定自己的代码，但一直出现错误，之后我觉得这样做无法解决。

#test sentences
sentences<- as.list(c("R is fantastic language", "Java is far from good", "Data mining is fascinating"))

#using the OpenNLP package
require(openNLP)

#perform tagging
taggedSentences<- tagPOS(sentences)

#split to words
individualWords<- unname(sapply(taggedSentences, function(x){strsplit(x,split=" ")}))

#Strip Tags
individualWordsClean<- unname(sapply(individualWords, function(x){gsub("/.+","",x)}))

#Strip words
individualTags<- unname(sapply(individualWords, function(x){gsub(".+/","",x)}))

#create a dataframe for words; courtesy @trinker
numberRow<- length(individualWords)
numberCol<- unname(sapply(individualWords, length))
df1<- as.data.frame(matrix(nrow=numberRow, ncol=max(numberCol)))
for (i in 1:numberRow){
df1[i,1:numberCol[i]]<- individualWordsClean [[i]]
}


#create a dataframe for tags; courtesy @trinker
numberRow<- length(individualWords)
numberCol<- unname(sapply(individualTags, length))
df2<- as.data.frame(matrix(nrow=numberRow, ncol=max(numberCol)))
for (i in 1:numberRow){
df2[i,1:numberCol[i]]<- individualTags [[i]]
}

#Create negative/positive words' lists
posAdj<- c("fantastic","fascinating","good")
negAdj<- c("bad","poor")
posNoun<- "R"
negNoun<- "Java"

#Function to match words and assign sentiment score
checkLexicon<- function(word,tag){
if (grep("JJ|JJR|JJS",tag)){
ifelse(word %in% posAdj, +1, ifelse(word  %in% negAdj, -1, 0))
}
else if(grep("NN|NNP|NNPS|NNS",tag)){
ifelse(word %in% posNoun, +1, ifelse(word %in% negNoun, -1, 0))
}
else if(grep("VBZ",tag)){
ifelse(word %in% "is","ok","none")
}
else if(grep("RB",tag)){
ifelse(word %in% "not",-1,0)
}
else if(grep("IN",tag)){
ifelse(word %in% "far",-1,0)
}
}

#Method to output a single value when used in conjuction with apply
justShow<- function(x){
    x
    }

#Main method that intends to extract word/POS tag pair, and determine sentiment score
mapply(FUN=checkLexicon, word=apply(df1,2,justShow),tag=apply(df2,2,justShow))

不幸的是，我对此方法没有成功，收到的错误是

Error in if (grep("JJ|JJR|JJS", tag)) { : argument is of length zero

我对R来说比较新，但似乎我无法在这里使用apply函数，因为它不会向mapply函数返回任何参数。另外，我不确定mapply是否会实际产生另一个数据帧。

请批评/建议。感谢

PS。 {3}}对于那些感兴趣的人来说，对于Rinker关于R的笔记。{/ 3>

Answer 1

错误是尝试将grep用作grepl。在乔兰指出之后，这一点得到了纠正。工作职能如下。

>df1

    V1     V2        V3          V4   V5
1    R     is fantastic    language <NA>
2 Java     is       far        from good
3 Data mining        is fascinating <NA>

>df2

   V1  V2  V3 V4   V5
1  NN VBZ  JJ NN <NA>
2 NNP VBZ  RB IN   JJ
3 NNP  NN VBZ JJ <NA>

#Function to match words and assign sentiment score
checkLexicon<- function(word,tag){
if (grepl("JJ|JJR|JJS",tag)){
ifelse(word %in% posAdj, +1, ifelse(word  %in% negAdj, -1, 0))
}
else if(grepl("NN|NNP|NNPS|NNS",tag)){
ifelse(word %in% posNoun, +1, ifelse(word %in% negNoun, -1, 0))
}
else if(grepl("VBZ",tag)){
ifelse(word %in% "is","ok","none")
}
else if(grepl("RB",tag)){
ifelse(word %in% "not",-1,0)
}
else if(grepl("IN",tag)){
ifelse(word %in% "far",-1,0)
}
}

#Method to output a single value when used in conjuction with apply
justShow<- function(x){
    x
    }

#Main method that intends to extract word/POS tag pair, and determine sentiment score
myObject<- mapply(FUN=checkLexicon, word=apply(df1,2,justShow),tag=apply(df2,2,justShow))

#Shaping the final dataframe
scoredDF<- as.data.frame(matrix(myObject,nrow=3))

  V1 V2 V3 V4   V5
1  1 ok  1  0 NULL
2 -1 ok  0  0    1
3  0  0 ok  1 NULL

将函数同时应用于来自两个数据帧的值，以生成第三个

1 个答案: