# Sample R commands in support of my previous answer
require(fortunes)
require(tm)
sentences <- NULL
for (i in 1:10) sentences <- c(sentences,fortune(i)$quote)
d <- data.frame(textCol =sentences )
ds <- DataframeSource(d)
dsc<-Corpus(ds)
dtm<- DocumentTermMatrix(dsc, control = list(weighting = weightTf, stopwords = TRUE))
dictC <- Dictionary(dtm)
# The query below is created from words in fortune(1) and fortune(2)
newQry <- data.frame(textCol = "lets stand up and be counted seems to work undocumented")
newQryC <- Corpus(DataframeSource(newQry))
dtmNewQry <- DocumentTermMatrix(newQryC, control = list(weighting=weightTf,stopwords=TRUE,dictionary=dict1))
dictQry <- Dictionary(dtmNewQry)
# Below does a naive similarity (number of features in common)
apply(dtm,1,function(x,y=dictQry){length(intersect(names(x)[x!= 0],y))})