How to compute unseen bi-grams in a corpus (for Good-Turing Smoothing)
$begingroup$
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.?
nlp
asked Sep 9 '18 at 18:08
rahsrahs
1213
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bumped to the homepage by Community♦ 9 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
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.?
nlp
asked Sep 9 '18 at 18:08
rahsrahs
1213
$endgroup$
bumped to the homepage by Community♦ 9 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
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.?
nlp
asked Sep 9 '18 at 18:08
rahsrahs
1213
$endgroup$
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.?
nlp
nlp
asked Sep 9 '18 at 18:08
rahsrahs
1213
asked Sep 9 '18 at 18:08
rahsrahs
1213
asked Sep 9 '18 at 18:08
rahsrahs
1213
asked Sep 9 '18 at 18:08
rahsrahs
1213
asked Sep 9 '18 at 18:08
rahsrahs
1213
1213
bumped to the homepage by Community♦ 9 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 9 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
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$begingroup$
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.
answered Sep 9 '18 at 18:21
rahsrahs
1213
$endgroup$
add a comment |
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$begingroup$
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.
answered Sep 9 '18 at 18:21
rahsrahs
1213
$endgroup$
add a comment |
$begingroup$
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.
answered Sep 9 '18 at 18:21
rahsrahs
1213
$endgroup$
add a comment |
$begingroup$
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.
answered Sep 9 '18 at 18:21
rahsrahs
1213
$endgroup$
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.
answered Sep 9 '18 at 18:21
rahsrahs
1213
answered Sep 9 '18 at 18:21
rahsrahs
1213
answered Sep 9 '18 at 18:21
rahsrahs
1213
answered Sep 9 '18 at 18:21
rahsrahs
1213
1213
add a comment |
add a comment |
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