Hi Arezki,
In your example, np1 should be equal to np3 because it records
the frequency of the word2 as the 2nd word of the bigram. So,
in order to get the probability A, you can just use the
count(word1<> word2)/word2 freq.
Thanks,
Ying
On 2011/5/4 11:11, hammache wrote:
Hi all,
is that there is a measure that allowed me to have from a file
returned by count.pl
word1 <> word2 <> n11 n1p np1
...
word3 <> word2 <> n31 n3p np3
this result
word1 <> word2 <> A
A = count (word1 <> word2) / sum [count (word, word2)]
Thanks,
Arezki
