2
$\begingroup$

Consider a (somewhat nonsensical) sentence - "I see saw a see saw"

The observed bi-grams would be: "I see"

"see saw"

"saw a"

and,

"a see".

My aim is to smoothen out the probability mass of the bi-gram probabilities by using Good-Turing smoothing. For this, I need to find the count of unseen bi-grams, i.e., bi-grams with a frequency count of 0.

How do I do this?

1) Would this be a list of all bi-grams formed by using 2 non-consecutive words? For example, "I saw", "saw saw", "a I", etc.?

2) Would repetitions of the same word be included as bi-grams? Eg. "I I", "see see", etc.?

$\endgroup$
0
$\begingroup$

I just remembered that we create a table with all possible words as the header of each row and of each column. As a result, the list of all bi-grams would be all possible bi-grams formed by concatenating any 2 words.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.