o rZh@s~dZddlZddlmZddlmZddlmZddlm Z eddZ e ed d Z e e d d ZGd ddZdS)zLanguage Model VocabularyN)Counter)Iterable)singledispatch)chaincCstdt|)Nz/Unsupported type for looking up in vocabulary: ) TypeErrortypewordsvocabr A/var/www/auris/lib/python3.10/site-packages/nltk/lm/vocabulary.py_dispatched_lookupsr cstfdd|DS)zcLook up a sequence of words in the vocabulary. Returns an iterator over looked up words. c3s|]}t|VqdSNr ).0wr r r sz_..)tuplerr rr _srcCs||vr|S|jS)z$Looks up one word in the vocabulary.) unk_label)wordr r r r _string_lookupsrc@sfeZdZdZdddZeddZd d Zd d Zd dZ ddZ ddZ ddZ ddZ ddZdS) 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 NcCsJ||_|dkrtd|||_t|_||dur |dSddS)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)selfrZ unk_cutoffrr r r __init__s   zVocabulary.__init__cC|jS)ziCutoff value. Items with count below this value are not considered part of vocabulary. )rr!r r r cutoffszVocabulary.cutoffcOs*|jj|i|tdd|D|_dS)zWUpdate vocabulary counts. Wraps `collections.Counter.update` method. css|]}dVqdS)rNr )rrr r r rsz$Vocabulary.update..N)rr sum_len)r!Z counter_argsZcounter_kwargsr r r r szVocabulary.updatecCs 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 r r r lookups zVocabulary.lookupcCs||jkr|jS|j|Sr)rrrr!itemr r r __getitem__szVocabulary.__getitem__cCs|||jkS)zPOnly consider items with counts GE to cutoff as being in the vocabulary.)r%r)r r r __contains__szVocabulary.__contains__cs*tfddjDjrjgSgS)zKBuilding on membership check define how to iterate over vocabulary.c3s|] }|vr|VqdSrr )rr*r$r r rsz&Vocabulary.__iter__..)rrrr$r r$r __iter__s  zVocabulary.__iter__cCr#)z1Computing size of vocabulary reflects the cutoff.)r'r$r r r __len__szVocabulary.__len__cCs$|j|jko|j|jko|j|jkSr)rr%r)r!otherr r r __eq__s   zVocabulary.__eq__cCsd|jj|j|jt|S)Nz/<{} with cutoff={} unk_label='{}' and {} items>)format __class____name__r%rlenr$r r r __str__szVocabulary.__str__)Nrr)r3 __module__ __qualname____doc__r"propertyr%r r(r+r,r-r.r0r5r r r r r%s Y   r)r8sys collectionsrcollections.abcr functoolsr itertoolsrr registerrstrrrr r r r s