Finding 2 & 3 word Phrases Using R TM Package

I am trying to find a code that actually works to find the most frequently used two and three word phrases in R text mining package (maybe there is another package for it that I do not know). I have been trying to use the tokenizer, but seem to have no luck.

If you worked on a similar situation in the past, could you post a code that is tested and actually works? Thank you so much!

You can pass in a custom tokenizing function to tm's DocumentTermMatrix function, so if you have package tau installed it's fairly straightforward.

library(tm); library(tau);

tokenize_ngrams <- function(x, n=3) return(rownames(as.data.frame(unclass(textcnt(x,method="string",n=n)))))

texts <- c("This is the first document.", "This is the second file.", "This is the third text.")
corpus <- Corpus(VectorSource(texts))
matrix <- DocumentTermMatrix(corpus,control=list(tokenize=tokenize_ngrams))

Where n in the tokenize_ngrams function is the number of words per phrase. This feature is also implemented in package RTextTools, which further simplifies things.

library(RTextTools)
texts <- c("This is the first document.", "This is the second file.", "This is the third text.")
matrix <- create_matrix(texts,ngramLength=3)

This returns a class of DocumentTermMatrix for use with package tm.

This is part 5 of the FAQ of the tm package:

5. Can I use bigrams instead of single tokens in a term-document matrix?

Yes. RWeka provides a tokenizer for arbitrary n-grams which can be directly passed on to the term-document matrix constructor. E.g.:

  library("RWeka")
  library("tm")

  data("crude")

  BigramTokenizer <- function(x) NGramTokenizer(x, Weka_control(min = 2, max = 2))
  tdm <- TermDocumentMatrix(crude, control = list(tokenize = BigramTokenizer))

  inspect(tdm[340:345,1:10])

This is my own made up creation for different purposes but I think may applicable to your needs too:

#User Defined Functions
Trim <- function (x) gsub("^\\s+|\\s+$", "", x)

breaker <- function(x) unlist(strsplit(x, "[[:space:]]|(?=[.!?*-])", perl=TRUE))

strip <- function(x, digit.remove = TRUE, apostrophe.remove = FALSE){
    strp <- function(x, digit.remove, apostrophe.remove){
        x2 <- Trim(tolower(gsub(".*?($|'|[^[:punct:]]).*?", "\\1", as.character(x))))
        x2 <- if(apostrophe.remove) gsub("'", "", x2) else x2
        ifelse(digit.remove==TRUE, gsub("[[:digit:]]", "", x2), x2)
    }
unlist(lapply(x, function(x) Trim(strp(x =x, digit.remove = digit.remove, 
    apostrophe.remove = apostrophe.remove)) ))
}

unblanker <- function(x)subset(x, nchar(x)>0)

#Fake Text Data
x <- "I like green eggs and ham.  They are delicious.  They taste so yummy.  I'm talking about ham and eggs of course"

#The code using Base R to Do what you want
breaker(x)
strip(x)
words <- unblanker(breaker(strip(x)))
textDF <- as.data.frame(table(words))
textDF$characters <- sapply(as.character(textDF$words), nchar)
textDF2 <- textDF[order(-textDF$characters, textDF$Freq), ]
rownames(textDF2) <- 1:nrow(textDF2)
textDF2
subset(textDF2, characters%in%2:3)

The corpus library has a function called term_stats that does what you want:

library(corpus)
corpus <- gutenberg_corpus(55) # Project Gutenberg #55, _The Wizard of Oz_
text_filter(corpus)$drop_punct <- TRUE # ignore punctuation
term_stats(corpus, ngrams = 2:3)
##    term             count support
## 1  of the             336       1
## 2  the scarecrow      208       1
## 3  to the             185       1
## 4  and the            166       1
## 5  said the           152       1
## 6  in the             147       1
## 7  the lion           141       1
## 8  the tin            123       1
## 9  the tin woodman    114       1
## 10 tin woodman        114       1
## 11 i am                84       1
## 12 it was              69       1
## 13 in a                64       1
## 14 the great           63       1
## 15 the wicked          61       1
## 16 wicked witch        60       1
## 17 at the              59       1
## 18 the little          59       1
## 19 the wicked witch    58       1
## 20 back to             57       1
## ⋮  (52511 rows total)

Here, count is the number of appearances, and support is the number of documents containing the term.