nltk.lm.NgramCounter

Class for counting ngrams.

Will count any ngram sequence you give it ;)

First we need to make sure we are feeding the counter sentences of ngrams.

>>> text = [["a", "b", "c", "d"], ["a", "c", "d", "c"]]
>>> from nltk.util import ngrams
>>> text_bigrams = [ngrams(sent, 2) for sent in text]
>>> text_unigrams = [ngrams(sent, 1) for sent in text]

The counting itself is very simple.

>>> from nltk.lm import NgramCounter
>>> ngram_counts = NgramCounter(text_bigrams + text_unigrams)

You can conveniently access ngram counts using standard python dictionary notation. String keys will give you unigram counts.

>>> ngram_counts['a']
2
>>> ngram_counts['aliens']
0

If you want to access counts for higher order ngrams, use a list or a tuple. These are treated as "context" keys, so what you get is a frequency distribution over all continuations after the given context.

>>> sorted(ngram_counts[['a']].items())
[('b', 1), ('c', 1)]
>>> sorted(ngram_counts[('a',)].items())
[('b', 1), ('c', 1)]

This is equivalent to specifying explicitly the order of the ngram (in this case 2 for bigram) and indexing on the context. >>> ngram_counts[2][('a',)] is ngram_counts[['a']] True

Note that the keys in ConditionalFreqDist cannot be lists, only tuples! It is generally advisable to use the less verbose and more flexible square bracket notation.

To get the count of the full ngram "a b", do this:

>>> ngram_counts[['a']]['b']
1

Specifying the ngram order as a number can be useful for accessing all ngrams in that order.

>>> ngram_counts[2]
<ConditionalFreqDist with 4 conditions>

The keys of this ConditionalFreqDist are the contexts we discussed earlier. Unigrams can also be accessed with a human-friendly alias.

>>> ngram_counts.unigrams is ngram_counts[1]
True

Similarly to collections.Counter, you can update counts after initialization.

>>> ngram_counts['e']
0
>>> ngram_counts.update([ngrams(["d", "e", "f"], 1)])
>>> ngram_counts['e']
1