t-score of collocates

[Earl Brown]

Corpus linguistics R-ists,

Hunston 2002:70 (Corpora in Applied Linguistics published by Cambridge) says: "The t-score is calculated by subtracting Expected from Observed and dividing the result by the standard deviation." I've googled around trying to find as clear a formula for t-score as Davies gives for Mutual Information here:

http://corpus.byu.edu/mutualInformation.asp

but have been unsuccessful. Hence, my post.

Can anyone supply me with the formula (preferably in R) for the t-score of a collocate, given:

•     the freq of the node word
•     the freq of the collocate
•     the freq of the collocation
•     the span around the node word
•     the size of the corpus in number of words

Here's what I've got in R for MI (given Davies' formula for MI):

freq.node <- 1262
freq.collocate <- 115
freq.collocation <- 24
size.corpus <- 96263399
span <- 3 # words on either side of node

log10((freq.collocation * size.corpus) / (freq.node * freq.collocate * (span * 2))) / log10(2)

Thanks in advance for any help. Best, Earl Brown

[Stefan Th. Gries]

Hi Earl

If you go to my research website, to appear h, you will find a formula for t that is easy to paraphrase into R.

Cheers,
STG
--
Stefan Th. Gries
----------------------------------
Univ. of California, Santa Barbara
http://tinyurl.com/stgries
----------------------------------

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[Earl Brown]

Thanks Stefan. I think I've implemented it correctly in R. I'll paste a script below that I wrote.

BTW, on page 6, in the paragraph above example 7, there's a typo of "100,00,000", which might be "99,966,209" anyway.

Thanks again. Earl Brown

##########
# Script to calculate raw freq, mutual information, and t-score for collocations

rm(list = ls(all = T))

# loads novel
novel <- tolower(scan("/Users/earlbrown/Corpora/Jane_Austen/Pride_and_Prejudice.txt", what = "char", sep = "\n"))

# gets words
words <- unlist(strsplit(novel, "[^-'a-zA-Z]+"))
words <- gsub("(^'|'$)", "", words)
words <- words[nchar(words) > 0]
freq.list.words <- sort(table(words), decreasing = T)

# gets bigrams
bigrams <- apply(mapply(seq, 1:(length(words) - 2 + 1), 2:length(words)), 2, function(x) paste(words[x], collapse = " "))
freq.list.bigrams <- sort(table(bigrams), decreasing = T)

# loops over bigrams
all.output <- c()
for (i in 1:length(freq.list.bigrams)) {
  cur.bigram <- names(freq.list.bigrams[i])
  if (i %% 100 == 0) cat("Working on bigram ", i, " of ", length(freq.list.bigrams), ": ", cur.bigram, "\n", sep = "")
 
  # gets bigram freq and expected freq
  wd1 <- unlist(strsplit(cur.bigram, " "))[1]
  wd2 <- unlist(strsplit(cur.bigram, " "))[2]
  aa <- freq.list.bigrams[names(freq.list.bigrams) == cur.bigram]
  bb <- freq.list.words[names(freq.list.words) == wd1] - aa
  cc <- freq.list.words[names(freq.list.words) == wd2] - aa
  exclude1 <- grep(sprintf("\\b%s\\b", wd1), names(freq.list.bigrams))
  exclude2 <- grep(sprintf("\\b%s\\b", wd2), names(freq.list.bigrams))
  exclude <- union(exclude1, exclude2)
  dd <- length(bigrams) - length(exclude)    
  aa.expected <- (aa + bb) * (aa + cc) / dd
 
  # gets MI and t-score
  mi.score <- log(aa/aa.expected, base = 2)
  t.score <- (aa - aa.expected) / sqrt(aa)
 
  # saves current bigram to collector
  cur.output <- paste(cur.bigram, aa, mi.score, t.score, sep = "\t")
  all.output <- append(all.output, cur.output)
 
} # next bigram

cat("BIGRAM\tRAW_FREQ\tMI\tT_SCORE", all.output, file = "bigrams.csv", sep = "\n")
cat("\aAll done!\n")

[Stefan Th. Gries]

Thanks, I'll keep my eyes open for that typo when the proofs arrive!

[Stefan Th. Gries]

> BTW, on page 6, in the paragraph above example 7, there's a typo of "100,00,000", which might be "99,966,209" anyway.

you sure? The computation of a_expected requires division by N, not d?

[Earl Brown]

I don't know. That would mind my R code is incorrect, right?

[Stefan Th. Gries]

It would :-) The expected frequency of the relevant 2x2 tables is
computed as in a chi-squared test so the denominator should be
N=a+b+c+d. Practically that will often make not much of a big
difference, but it's of course still wrong,