Sample usage for lm

Regression Tests

Issue 167

https://github.com/nltk/nltk/issues/167

>>> from nltk.corpus import brown
>>> from nltk.lm.preprocessing import padded_everygram_pipeline
>>> ngram_order = 3
>>> train_data, vocab_data = padded_everygram_pipeline(
...     ngram_order,
...     brown.sents(categories="news")
... )
>>> from nltk.lm import WittenBellInterpolated
>>> lm = WittenBellInterpolated(ngram_order)
>>> lm.fit(train_data, vocab_data)

Sentence containing an unseen word should result in infinite entropy because Witten-Bell is based ultimately on MLE, which cannot handle unseen ngrams. Crucially, it shouldn’t raise any exceptions for unseen words.

>>> from nltk.util import ngrams
>>> sent = ngrams("This is a sentence with the word aaddvark".split(), 3)
>>> lm.entropy(sent)
inf

If we remove all unseen ngrams from the sentence, we’ll get a non-infinite value for the entropy.

>>> sent = ngrams("This is a sentence".split(), 3)
>>> round(lm.entropy(sent), 14)
10.23701322869105

Issue 367

https://github.com/nltk/nltk/issues/367

Reproducing Dan Blanchard’s example: https://github.com/nltk/nltk/issues/367#issuecomment-14646110

>>> from nltk.lm import Lidstone, Vocabulary
>>> word_seq = list('aaaababaaccbacb')
>>> ngram_order = 2
>>> from nltk.util import everygrams
>>> train_data = [everygrams(word_seq, max_len=ngram_order)]
>>> V = Vocabulary(['a', 'b', 'c', ''])
>>> lm = Lidstone(0.2, ngram_order, vocabulary=V)
>>> lm.fit(train_data)

For doctest to work we have to sort the vocabulary keys.

>>> V_keys = sorted(V)
>>> round(sum(lm.score(w, ("b",)) for w in V_keys), 6)
1.0
>>> round(sum(lm.score(w, ("a",)) for w in V_keys), 6)
1.0
>>> [lm.score(w, ("b",)) for w in V_keys]
[0.05, 0.05, 0.8, 0.05, 0.05]
>>> [round(lm.score(w, ("a",)), 4) for w in V_keys]
[0.0222, 0.0222, 0.4667, 0.2444, 0.2444]

Here’s reproducing @afourney’s comment: https://github.com/nltk/nltk/issues/367#issuecomment-15686289

>>> sent = ['foo', 'foo', 'foo', 'foo', 'bar', 'baz']
>>> ngram_order = 3
>>> from nltk.lm.preprocessing import padded_everygram_pipeline
>>> train_data, vocab_data = padded_everygram_pipeline(ngram_order, [sent])
>>> from nltk.lm import Lidstone
>>> lm = Lidstone(0.2, ngram_order)
>>> lm.fit(train_data, vocab_data)

The vocabulary includes the “UNK” symbol as well as two padding symbols.

>>> len(lm.vocab)
6
>>> word = "foo"
>>> context = ("bar", "baz")

The raw counts.

>>> lm.context_counts(context)[word]
0
>>> lm.context_counts(context).N()
1

Counts with Lidstone smoothing.

>>> lm.context_counts(context)[word] + lm.gamma
0.2
>>> lm.context_counts(context).N() + len(lm.vocab) * lm.gamma
2.2

Without any backoff, just using Lidstone smoothing, P(“foo” | “bar”, “baz”) should be: 0.2 / 2.2 ~= 0.090909

>>> round(lm.score(word, context), 6)
0.090909

Issue 380

https://github.com/nltk/nltk/issues/380

Reproducing setup akin to this comment: https://github.com/nltk/nltk/issues/380#issue-12879030

For speed take only the first 100 sentences of reuters. Shouldn’t affect the test.
>>> from nltk.corpus import reuters
>>> sents = reuters.sents()[:100]
>>> ngram_order = 3
>>> from nltk.lm.preprocessing import padded_everygram_pipeline
>>> train_data, vocab_data = padded_everygram_pipeline(ngram_order, sents)
>>> from nltk.lm import Lidstone
>>> lm = Lidstone(0.2, ngram_order)
>>> lm.fit(train_data, vocab_data)
>>> lm.score("said", ("",)) < 1
True