NgdZddlZddlmZddlmZddlmZddlm Z edZ e edZ e e d ZGd d ZdS) zLanguage Model VocabularyN)Counter)Iterable)singledispatch)chainc@tdt|)Nz/Unsupported type for looking up in vocabulary: ) TypeErrortypewordsvocabs N/var/www/html/ai-engine/env/lib/python3.11/site-packages/nltk/lm/vocabulary.py_dispatched_lookuprs Sd5kkSS T TTc:tfd|DS)zcLook up a sequence of words in the vocabulary. Returns an iterator over looked up words. c38K|]}t|VdSNr).0wr s r z_..s.==!#Au--======r)tupler s `r _rs( ====u=== = ==rc||vr|n|jS)z$Looks up one word in the vocabulary.) unk_label)wordr s r _string_lookuprs5==44eo5rc`eZdZdZddZedZdZdZd Z d Z d Z d Z d Z dZdS) Vocabularya 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) ['-', '', 'a', 'b', 'c', 'd', 'r'] >>> sorted(vocab.counts) ['-', 'a', 'b', 'c', 'd', 'r'] >>> sorted(vocab) ['', '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 "". >>> "" 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") '' If given a sequence, it will return an tuple of the looked up words. >>> vocab.lookup(["p", 'a', 'r', 'd', 'b', 'c']) ('', 'a', '', 'd', '', '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 Nc||_|dkrtd|||_t|_|||nddS)aCreate 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. rz)Cutoff value cannot be less than 1. Got: N)r ValueError_cutoffrcountsupdate)selfr% unk_cutoffrs r __init__zVocabulary.__init__s^# >>UUUVV V! ii  f0FFb99999rc|jS)ziCutoff value. Items with count below this value are not considered part of vocabulary. )r$r's r cutoffzVocabulary.cutoffs |rcf|jj|i|td|D|_dS)zWUpdate vocabulary counts. Wraps `collections.Counter.update` method. c3K|]}dVdS)rN)rrs r rz$Vocabulary.update..s"((a((((((rN)r%r&sum_len)r' counter_argscounter_kwargss r r&zVocabulary.updates@  L;N;;;((4((((( rc"t||S)aLook up one or more words in the vocabulary. If passed one word as a string will return that word or `self.unk_label`. Otherwise will assume it was passed a sequence of words, will try to look each of them up and return an iterator over the looked up words. :param words: Word(s) to look up. :type words: Iterable(str) or str :rtype: generator(str) or str :raises: TypeError for types other than strings or iterables >>> from nltk.lm import Vocabulary >>> vocab = Vocabulary(["a", "b", "c", "a", "b"], unk_cutoff=2) >>> vocab.lookup("a") 'a' >>> vocab.lookup("aliens") '' >>> vocab.lookup(["a", "b", "c", ["x", "b"]]) ('a', 'b', '', ('', 'b')) r)r'r s r lookupzVocabulary.lookups,"%...rc@||jkr|jn |j|Sr)rr$r%r'items r __getitem__zVocabulary.__getitem__s!#t~55t||4;t;LLrc$|||jkS)zPOnly consider items with counts GE to cutoff as being in the vocabulary.)r,r7s r __contains__zVocabulary.__contains__sDzT[((rcdtfdjDjrjgngS)zKBuilding on membership check define how to iterate over vocabulary.c3$K|] }|v|V dSrr/)rr8r's r rz&Vocabulary.__iter__..s' : :dTT\\T\\\\ : :r)rr%rr+s`r __iter__zVocabulary.__iter__sD : : : :dk : : : $ 3T^     rc|jS)z1Computing size of vocabulary reflects the cutoff.)r1r+s r __len__zVocabulary.__len__s yrcb|j|jko|j|jko|j|jkSr)rr,r%)r'others r __eq__zVocabulary.__eq__s5 Neo - , u|+ , u|+ rctd|jj|j|jt |S)Nz/<{} with cutoff={} unk_label='{}' and {} items>)format __class____name__r,rlenr+s r __str__zVocabulary.__str__s2@GG N #T[$.#d))   r)Nrr )rG __module__ __qualname____doc__r)propertyr,r&r5r9r;r>r@rCrIr/rr rr%sWWr::::&X)))///0MMM)))            rr)rLsys collectionsrcollections.abcr functoolsr itertoolsrrregisterrstrrrr/rr rUs   $$$$$$$$$$$$UUUX&&>>'&>S!!66"!6 u u u u u u u u u u r