# Natural Language Toolkit
#
# Copyright (C) 2001-2023 NLTK Project
# Author: Ilia Kurenkov <ilia.kurenkov@gmail.com>
# URL: <https://www.nltk.org/>
# For license information, see LICENSE.TXT
"""Language Model Vocabulary"""
import sys
from collections import Counter
from collections.abc import Iterable
from functools import singledispatch
from itertools import chain
@singledispatch
def _dispatched_lookup(words, vocab):
raise TypeError(f"Unsupported type for looking up in vocabulary: {type(words)}")
@_dispatched_lookup.register(Iterable)
def _(words, vocab):
"""Look up a sequence of words in the vocabulary.
Returns an iterator over looked up words.
"""
return tuple(_dispatched_lookup(w, vocab) for w in words)
@_dispatched_lookup.register(str)
def _string_lookup(word, vocab):
"""Looks up one word in the vocabulary."""
return word if word in vocab else vocab.unk_label
[docs]class Vocabulary:
"""Stores language model vocabulary.
Satisfies two common language modeling requirements for a vocabulary:
- When checking membership and calculating its size, filters items
by comparing their counts to a cutoff value.
- Adds a special "unknown" token which unseen words are mapped to.
>>> words = ['a', 'c', '-', 'd', 'c', 'a', 'b', 'r', 'a', 'c', 'd']
>>> from nltk.lm import Vocabulary
>>> vocab = Vocabulary(words, unk_cutoff=2)
Tokens with counts greater than or equal to the cutoff value will
be considered part of the vocabulary.
>>> vocab['c']
3
>>> 'c' in vocab
True
>>> vocab['d']
2
>>> 'd' in vocab
True
Tokens with frequency counts less than the cutoff value will be considered not
part of the vocabulary even though their entries in the count dictionary are
preserved.
>>> vocab['b']
1
>>> 'b' in vocab
False
>>> vocab['aliens']
0
>>> 'aliens' in vocab
False
Keeping the count entries for seen words allows us to change the cutoff value
without having to recalculate the counts.
>>> vocab2 = Vocabulary(vocab.counts, unk_cutoff=1)
>>> "b" in vocab2
True
The cutoff value influences not only membership checking but also the result of
getting the size of the vocabulary using the built-in `len`.
Note that while the number of keys in the vocabulary's counter stays the same,
the items in the vocabulary differ depending on the cutoff.
We use `sorted` to demonstrate because it keeps the order consistent.
>>> sorted(vocab2.counts)
['-', 'a', 'b', 'c', 'd', 'r']
>>> sorted(vocab2)
['-', '<UNK>', 'a', 'b', 'c', 'd', 'r']
>>> sorted(vocab.counts)
['-', 'a', 'b', 'c', 'd', 'r']
>>> sorted(vocab)
['<UNK>', 'a', 'c', 'd']
In addition to items it gets populated with, the vocabulary stores a special
token that stands in for so-called "unknown" items. By default it's "<UNK>".
>>> "<UNK>" in vocab
True
We can look up words in a vocabulary using its `lookup` method.
"Unseen" words (with counts less than cutoff) are looked up as the unknown label.
If given one word (a string) as an input, this method will return a string.
>>> vocab.lookup("a")
'a'
>>> vocab.lookup("aliens")
'<UNK>'
If given a sequence, it will return an tuple of the looked up words.
>>> vocab.lookup(["p", 'a', 'r', 'd', 'b', 'c'])
('<UNK>', 'a', '<UNK>', 'd', '<UNK>', 'c')
It's possible to update the counts after the vocabulary has been created.
In general, the interface is the same as that of `collections.Counter`.
>>> vocab['b']
1
>>> vocab.update(["b", "b", "c"])
>>> vocab['b']
3
"""
[docs] def __init__(self, counts=None, unk_cutoff=1, unk_label="<UNK>"):
"""Create a new Vocabulary.
:param counts: Optional iterable or `collections.Counter` instance to
pre-seed the Vocabulary. In case it is iterable, counts
are calculated.
:param int unk_cutoff: Words that occur less frequently than this value
are not considered part of the vocabulary.
:param unk_label: Label for marking words not part of vocabulary.
"""
self.unk_label = unk_label
if unk_cutoff < 1:
raise ValueError(f"Cutoff value cannot be less than 1. Got: {unk_cutoff}")
self._cutoff = unk_cutoff
self.counts = Counter()
self.update(counts if counts is not None else "")
@property
def cutoff(self):
"""Cutoff value.
Items with count below this value are not considered part of vocabulary.
"""
return self._cutoff
[docs] def update(self, *counter_args, **counter_kwargs):
"""Update vocabulary counts.
Wraps `collections.Counter.update` method.
"""
self.counts.update(*counter_args, **counter_kwargs)
self._len = sum(1 for _ in self)
[docs] def lookup(self, words):
"""Look up one or more words in the vocabulary.
If passed one word as a string will return that word or `self.unk_label`.
Otherwise will assume it was passed a sequence of words, will try to look
each of them up and return an iterator over the looked up words.
:param words: Word(s) to look up.
:type words: Iterable(str) or str
:rtype: generator(str) or str
:raises: TypeError for types other than strings or iterables
>>> from nltk.lm import Vocabulary
>>> vocab = Vocabulary(["a", "b", "c", "a", "b"], unk_cutoff=2)
>>> vocab.lookup("a")
'a'
>>> vocab.lookup("aliens")
'<UNK>'
>>> vocab.lookup(["a", "b", "c", ["x", "b"]])
('a', 'b', '<UNK>', ('<UNK>', 'b'))
"""
return _dispatched_lookup(words, self)
def __getitem__(self, item):
return self._cutoff if item == self.unk_label else self.counts[item]
def __contains__(self, item):
"""Only consider items with counts GE to cutoff as being in the
vocabulary."""
return self[item] >= self.cutoff
def __iter__(self):
"""Building on membership check define how to iterate over
vocabulary."""
return chain(
(item for item in self.counts if item in self),
[self.unk_label] if self.counts else [],
)
def __len__(self):
"""Computing size of vocabulary reflects the cutoff."""
return self._len
def __eq__(self, other):
return (
self.unk_label == other.unk_label
and self.cutoff == other.cutoff
and self.counts == other.counts
)
def __str__(self):
return "<{} with cutoff={} unk_label='{}' and {} items>".format(
self.__class__.__name__, self.cutoff, self.unk_label, len(self)
)