(Clarification) Artificial Intelligence Word Classification (Python) [closed]

1

I was trying to do something that would meet this need:

  • Read the word
  • Tap into a dictionary
  • E Try to complete the word (if possible)

For example:

ENTRADA: CSA
DICIONARIO ["HOMEM", "CASA", "MULHER"]'
SAIDA: CASA

ENTRADA: OMEM
DICIONARIO ["HOMEM", "CASA", "MULHER"]'
SAIDA: HOMEM

ENTRADA: MUHER
DICIONARIO ["HOMEM", "CASA", "MULHER"]'
SAIDA: MULHER

ENTRADA: JOGO
DICIONARIO ["HOMEM", "CASA", "MULHER"]'
SAIDA: Palavra não definida e/ou desconhecida

I ended up finding some examples of scikit-learn and google, but I did not understand how to use them.

scikit-learn documentation

Google example

Please help me to clarify what is happening in this code:

import re
from collections import Counter

def words(text): return re.findall(r'\w+', text.lower())

WORDS = Counter(words(open('big.txt').read()))

def P(word, N=sum(WORDS.values())): 
//"Probability of 'word'."
return WORDS[word] / N

def correction(word): 
//"Most probable spelling correction for word."
return max(candidates(word), key=P)

def candidates(word): 
//"Generate possible spelling corrections for word."
return (known([word]) or known(edits1(word)) or known(edits2(word)) or [word])

def known(words): 
//"The subset of 'words' that appear in the dictionary of WORDS."
return set(w for w in words if w in WORDS)

def edits1(word):
//"All edits that are one edit away from 'word'."
letters    = 'abcdefghijklmnopqrstuvwxyz'
splits     = [(word[:i], word[i:])    for i in range(len(word) + 1)]
deletes    = [L + R[1:]               for L, R in splits if R]
transposes = [L + R[1] + R[0] + R[2:] for L, R in splits if len(R)>1]
replaces   = [L + c + R[1:]           for L, R in splits if R for c in letters]
inserts    = [L + c + R               for L, R in splits for c in letters]
return set(deletes + transposes + replaces + inserts)

def edits2(word): 
//"All edits that are two edits away from 'word'."
return (e2 for e1 in edits1(word) for e2 in edits1(e1))
    
asked by anonymous 14.12.2018 / 01:05

0 answers