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132 lines
3.7 KiB
Plaintext
132 lines
3.7 KiB
Plaintext
.. Copyright (C) 2001-2020 NLTK Project
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.. For license information, see LICENSE.TXT
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.. -*- coding: utf-8 -*-
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Regression Tests
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================
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Issue 167
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---------
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https://github.com/nltk/nltk/issues/167
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>>> from nltk.corpus import brown
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>>> from nltk.lm.preprocessing import padded_everygram_pipeline
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>>> ngram_order = 3
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>>> train_data, vocab_data = padded_everygram_pipeline(
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... ngram_order,
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... brown.sents(categories="news")
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... )
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>>> from nltk.lm import WittenBellInterpolated
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>>> lm = WittenBellInterpolated(ngram_order)
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>>> lm.fit(train_data, vocab_data)
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Sentence containing an unseen word should result in infinite entropy because
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Witten-Bell is based ultimately on MLE, which cannot handle unseen ngrams.
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Crucially, it shouldn't raise any exceptions for unseen words.
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>>> from nltk.util import ngrams
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>>> sent = ngrams("This is a sentence with the word aaddvark".split(), 3)
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>>> lm.entropy(sent)
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inf
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If we remove all unseen ngrams from the sentence, we'll get a non-infinite value
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for the entropy.
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>>> sent = ngrams("This is a sentence".split(), 3)
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>>> lm.entropy(sent)
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17.41365588455936
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Issue 367
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---------
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https://github.com/nltk/nltk/issues/367
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Reproducing Dan Blanchard's example:
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https://github.com/nltk/nltk/issues/367#issuecomment-14646110
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>>> from nltk.lm import Lidstone, Vocabulary
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>>> word_seq = list('aaaababaaccbacb')
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>>> ngram_order = 2
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>>> from nltk.util import everygrams
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>>> train_data = [everygrams(word_seq, max_len=ngram_order)]
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>>> V = Vocabulary(['a', 'b', 'c', ''])
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>>> lm = Lidstone(0.2, ngram_order, vocabulary=V)
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>>> lm.fit(train_data)
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For doctest to work we have to sort the vocabulary keys.
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>>> V_keys = sorted(V)
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>>> round(sum(lm.score(w, ("b",)) for w in V_keys), 6)
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1.0
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>>> round(sum(lm.score(w, ("a",)) for w in V_keys), 6)
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1.0
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>>> [lm.score(w, ("b",)) for w in V_keys]
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[0.05, 0.05, 0.8, 0.05, 0.05]
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>>> [round(lm.score(w, ("a",)), 4) for w in V_keys]
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[0.0222, 0.0222, 0.4667, 0.2444, 0.2444]
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Here's reproducing @afourney's comment:
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https://github.com/nltk/nltk/issues/367#issuecomment-15686289
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>>> sent = ['foo', 'foo', 'foo', 'foo', 'bar', 'baz']
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>>> ngram_order = 3
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>>> from nltk.lm.preprocessing import padded_everygram_pipeline
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>>> train_data, vocab_data = padded_everygram_pipeline(ngram_order, [sent])
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>>> from nltk.lm import Lidstone
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>>> lm = Lidstone(0.2, ngram_order)
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>>> lm.fit(train_data, vocab_data)
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The vocabulary includes the "UNK" symbol as well as two padding symbols.
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>>> len(lm.vocab)
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6
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>>> word = "foo"
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>>> context = ("bar", "baz")
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The raw counts.
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>>> lm.context_counts(context)[word]
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0
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>>> lm.context_counts(context).N()
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1
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Counts with Lidstone smoothing.
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>>> lm.context_counts(context)[word] + lm.gamma
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0.2
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>>> lm.context_counts(context).N() + len(lm.vocab) * lm.gamma
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2.2
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Without any backoff, just using Lidstone smoothing, P("foo" | "bar", "baz") should be:
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0.2 / 2.2 ~= 0.090909
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>>> round(lm.score(word, context), 6)
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0.090909
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Issue 380
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---------
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https://github.com/nltk/nltk/issues/380
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Reproducing setup akin to this comment:
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https://github.com/nltk/nltk/issues/380#issue-12879030
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For speed take only the first 100 sentences of reuters. Shouldn't affect the test.
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>>> from nltk.corpus import reuters
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>>> sents = reuters.sents()[:100]
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>>> ngram_order = 3
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>>> from nltk.lm.preprocessing import padded_everygram_pipeline
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>>> train_data, vocab_data = padded_everygram_pipeline(ngram_order, sents)
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>>> from nltk.lm import Lidstone
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>>> lm = Lidstone(0.2, ngram_order)
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>>> lm.fit(train_data, vocab_data)
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>>> lm.score("said", ("",)) < 1
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True
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