# import necessary modules
from IPython.core.interactiveshell import InteractiveShell
InteractiveShell.ast_node_interactivity = "all"

%matplotlib inline
%config InlineBackend.figure_format = 'retina'

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd

from evaluation import test
from utils import load_data

from sklearn.naive_bayes import MultinomialNB

message0 = 'hello world hello hello world play'
message1 = 'test test test test one hello'

#Convert a collection of text documents to a matrix of token counts
from sklearn.feature_extraction.text import CountVectorizer
vectorizer = CountVectorizer()
bow = vectorizer.fit_transform([message0, message1])
print(bow)
print(type(bow))

  (0, 0)	3
  (0, 4)	2
  (0, 2)	1
  (1, 0)	1
  (1, 3)	4
  (1, 1)	1
<class 'scipy.sparse.csr.csr_matrix'>

As you can see, CountVectorizer returns a sparse matrix encoding our texts with the number of times a particular word occurs.

vocabulary = {v: k for k, v in vectorizer.vocabulary_.items()}
[vocabulary[i] for i in sorted([v for k,v in vectorizer.vocabulary_.items()])]
bow.toarray()

['hello', 'one', 'play', 'test', 'world']






array([[3, 0, 1, 0, 2],
       [1, 1, 0, 4, 0]])

Note that you can see how the encoding information is saved in a sparse matrix. For message0, on indices [0,4,2] you have values [3,2,1].

If we set binary=True when encoding messages, our encoder only records the whether the word is present or not, ignoring the numbber of occurance. For our first implementation of NaiveBayes, we will simply encode the presence of each word.

vectorizer_b = CountVectorizer(binary=True)
bow_b = vectorizer_b.fit_transform([message0, message1])
print(bow_b)
bow_b.toarray()

  (0, 0)	1
  (0, 4)	1
  (0, 2)	1
  (1, 0)	1
  (1, 3)	1
  (1, 1)	1





array([[1, 0, 1, 0, 1],
       [1, 1, 0, 1, 0]])
\[\phi_{j|y=1} = \frac{\sum_{i=1}^m 1\{x_j^{(i)} = 1 \wedge y^{(i)} = 1\}}{\sum_{i=1}^m1\{y^{(i)} = 1\}}\] \[\phi_{j|y=0} = \frac{\sum_{i=1}^m 1\{x_j^{(i)} = 1 \wedge y^{(i)} = 0\}}{\sum_{i=1}^m1\{y^{(i)} = 0\}}\] \[\phi_y = \frac{\sum_{i=1}^m 1\{y^{(i) = 1}\}}{m}\]