R中文本数据中两对组合的出现频率

我有一个包含多个字符串（文本）变量的文件，每个回答者都为每个变量写了一两句话。我希望能够找到每个词组合的频率（即“能力”与“性能”的频率）。 到目前为止，我的代码是：

#Setting up the data file 
data.text <- scan("C:/temp/tester.csv", what="char", sep="\n")

#Change everything to lower text
data.text <- tolower(data.text)

#Split the strings into separate words
data.words.list <- strsplit(data.text, "\\W+", perl=TRUE)
data.words.vector <- unlist(data.words.list)

#List each word and frequency
data.freq.list <- table(data.words.vector)

---

我不确定这是否是yu的意思，但与其在每两个单词的边界上拆分（我发现尝试和regex很痛苦），不如使用可靠的

头
和
尾
滑动技巧将每两个单词粘贴在一起

#  How I read your data
df <- read.table( text = 'ID   Reason_for_Dissatisfaction    Reason_for_Likelihood_to_Switch
1    "not happy with the service"  "better value at other place"
2    "poor customer service"       "tired of same old thing"
3    "they are overchanging me"    "bad service"
' , h = TRUE , stringsAsFactors = FALSE )


#  Split to words
wlist <- sapply( df[,-1] , strsplit , split = "\\W+", perl=TRUE)

#  Paste word pairs together
outl <- sapply( wlist , function(x) paste( head(x,-1) , tail(x,-1) , sep = " ") )

#  Table as per usual
table(unlist( outl ) )
are overchanging         at other      bad service     better value customer service 
               1                1                1                1                1 
      happy with        not happy          of same        old thing      other place 
               1                1                1                1                1 
 overchanging me    poor customer         same old      the service         they are 
               1                1                1                1                1 
        tired of         value at         with the 
               1                1                1

#我如何读取您的数据
df您的
数据.text
看起来像什么？您能提供几行的样本或一些具有代表性的示例数据吗？请参阅此处，了解如何最好地做到这一点的一些技巧：这条线是如何工作的？#将单词对粘贴在一起
#  How I read your data
df <- read.table( text = 'ID   Reason_for_Dissatisfaction    Reason_for_Likelihood_to_Switch
1    "not happy with the service"  "better value at other place"
2    "poor customer service"       "tired of same old thing"
3    "they are overchanging me"    "bad service"
' , h = TRUE , stringsAsFactors = FALSE )


#  Split to words
wlist <- sapply( df[,-1] , strsplit , split = "\\W+", perl=TRUE)

#  Paste word pairs together
outl <- sapply( wlist , function(x) paste( head(x,-1) , tail(x,-1) , sep = " ") )

#  Table as per usual
table(unlist( outl ) )
are overchanging         at other      bad service     better value customer service 
               1                1                1                1                1 
      happy with        not happy          of same        old thing      other place 
               1                1                1                1                1 
 overchanging me    poor customer         same old      the service         they are 
               1                1                1                1                1 
        tired of         value at         with the 
               1                1                1