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Building a Naive Bayes Classifier
            Eric Wilson
          Search Engineer
           Manta Media
The problem: Undesirable Content
Recommended by 3 people:
Bob Perkins
It is a pleasure to work with Kim! Her work is beautiful and she is
professional, communicative, and friendly.

Fred
She lied and stole my money, STAY AWAY!!!!!

Jane Robinson
Very Quick Turn Around as asked - Synced up Perfectly Great Help!
Possible solutions
●   First approach: manually remove undesired
    content.
●   Attempt to filter based on lists of banned
    words.
●   Use a machine learning algorithm to identify
    undesirable content based on a small set of
    manually classified examples.
Using Naive Bayes isn't too hard!
●   We'll need a bit of probability, including the
    concept of conditional probability.
●   A few natural language processing ideas will
    be necessary.
●   Facility with any modern programming
    language.
●   Persistence with many details.
Probability 101
Suppose we choose a number from the set:
           U = {1,2,3,4,5,6,7,8,9,10}
Let A be the event that the number is even,
and B be the event that the number is prime.


Compute P(A), P(B), P(A|B), and P(B|A),
where P(A|B) is the probability of A given B.
Just count!

  1    9

                             3
           4                      7
                         2
                8
           6        10       5

               A             B


P(A) = 5/10 = 1/2
P(B) = 4/10 = 2/5
P(A|B) = 1/4
P(B|A) = 1/5
Bayes Theorem
P(A|B) = P(AB)/P(B)

P(B)P(A|B) = P(AB)

P(B)P(A|B) = P(A)P(B|A)

P(A|B) = P(A)P(B|A)/P(B)
A simplistic language model
Consider each document to be a set of words,
along with frequencies.
For example: “The premium quality for the
discount price” is viewed as:
{'the':2, 'premium':1, 'quality':1, 'for':1, 
'discount':1, 'price':1}

Same as “The discount quality for the premium
price,” since we don't care about order.
That seems … foolish
●   English is so complicated that we won't have
    any real hope of understanding semantics.
●   In many real-life scenarios, text that you want
    to classify is not exactly subtle.
●   If necessary, we can improve our language
    model later.
An example:
Type       Text                          Class
Training   Good happy good               Positive
Training   Good good service             Positive
Training   Good friendly                 Positive
Training   Lousy good cheat              Negative
Test       Good good good cheat lousy    ??

In order to be able to perform all calculations, we
will use an example with extremely small
documents.
What was the question?
We are trying to determine whether the last
recommendation was positive or negative.
We want to compute:
P(Pos|good good good lousy cheat)
By Bayes Theorem, this is equal to:
P(Pos)P(good good good lousy cheat|Pos)
   P(good good good lousy cheat)
What do we know?
P(Pos) = 3/4
P(good|Pos), P(cheat|Pos), P(lousy|Pos)
Are all easily computed by counting using the
training set.
Which is almost what we want ...
Wouldn't it be nice ...
Maybe we have all we need? Isn't
P(good good good lousy cheat|Pos) =
P(good|Pos)3P(lousy|Pos)P(cheat|Pos) ?


Well, yes, if these are independent events,
which almost certainly doesn't hold.
The “naive” assumption is that we can
consider these events independent.
The Naive Bayes Algorithm
If C1,C2,...,Cn are classes, and an instance has
features F1,F2,...,Fm, then the most likely class
for this instance is the one that maximizes the
following:
        P(Ci )P(F1|Ci )P(F2|Ci )...P(Fm|Ci )
Wasn't there a denominator?
If our goal was to compute the probability of
the most likely class, we should divide by:
               P(F1)P(F2)...P(Fm)
We can ignore this part because, we only care
about which class has the highest probability,
and this term is the same for each class.
Interesting theory but …
Won't this break as soon as we encounter a
word that isn't in our training set?
For example, if “goood” is not in our training
set, and occurs in our test set, then since
P(Pos|goood) = 0, so our product is zero for all
classes.
We need nonzero probabilities for all words,
even words that don't exist.
Plus-one smoothing
Just count every word one time more than it
actually occurs.
Since we are only concerned with relative
probabilities, this inaccuracy should be of no
concern.
P(word|C) = count(word|C) + 1
                count(C) + V
(V is the total vocabulary, so that our
probabilities sum to 1.)
Let's try it out:
 P(Pos) = ¾                         Type    Text                         Class

 P(Neg) = ¼                         Training Good happy good             Positive
                                            Good good service            Positive
                                            Good friendly                Positive
                                            Lousy good cheat             Negative
                                    Test    Good good good cheat lousy   ??
P(good|Pos) = (5+1)/(8+6) = 3/7
P(cheat|Pos) = (0+1)/(8+6) = 1/14          P(Pos|D5) ~ ¾ * (3/7)3*(1/14)*(1/14)
P(lousy|Pos) = (0+1)/(8+6) = 1/14           = 0.0003
P(good|Neg) = (1+1)/(3+6) = 2/9            P(Neg|D5) ~ ¼ * (2/9)3*(2/9)*(2/9)
P(cheat|Neg) = (1+1)/(3+6) = 2/9           = 0.0001
P(lousy|Neg) = (1+1)/(3+6) = 2/9
Training the classifier
●   Count instances of classes, store counts in a map.
●   Store counts of all words in a nested map:
    {'pos':
        {'good': 5, 'friendly': 1, 'service': 1, 'happy': 1},
    'neg':
        {'cheat': 1, 'lousy': 1, 'good': 1}
    }
●   Should be easy to compute probabilities.
●   Should be efficient (training time and memory.)
Some practical problems
●   Tokenization
●   Arithmetic
●   How to evaluate results?
Tokenization
●   Use whitespace?
    –   “food”, “food.”, food,” and “food!” all different.
●   Use whitespace and punctuation?
    –   “won't” tokenized to “won” and “t”
●   What about emails? Urls? Phone numbers?
    What about the things we haven't thought
    about yet?
●   Use a library. Lucene is a good choice.
Arithmetic
What happens when you multiply a large
amount of small numbers?
To prevent underflow, use sums of logs instead
of products of true probabilities.
Key properties of log:
   ●   log(AB) = log(A) + log(B)
   ●   x > y => log(x) > log(y)
   ●   Turns very small numbers into managable negative
       numbers
Evaluating a classifier
●   Precision and recall
●   Confusion matrix
●   Divide training set into nine “folds”, train
    classifier on nine folds, and check accuracy of
    classifying the tenth fold
Experiment
●   Tokenization strategies
    –   Stop words
    –   Capitalization
    –   Stemming
●   Language model
    –   Ignore multiplicities
    –   Smoothing
Contact me
●   wilson.eric.n@gmail.com
●   ewilson@manta.com
●   @wilsonericn
●   http://wilsonericn.wordpress.com

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Naive Bayes

  • 1. Building a Naive Bayes Classifier Eric Wilson Search Engineer Manta Media
  • 2. The problem: Undesirable Content Recommended by 3 people: Bob Perkins It is a pleasure to work with Kim! Her work is beautiful and she is professional, communicative, and friendly. Fred She lied and stole my money, STAY AWAY!!!!! Jane Robinson Very Quick Turn Around as asked - Synced up Perfectly Great Help!
  • 3. Possible solutions ● First approach: manually remove undesired content. ● Attempt to filter based on lists of banned words. ● Use a machine learning algorithm to identify undesirable content based on a small set of manually classified examples.
  • 4. Using Naive Bayes isn't too hard! ● We'll need a bit of probability, including the concept of conditional probability. ● A few natural language processing ideas will be necessary. ● Facility with any modern programming language. ● Persistence with many details.
  • 5. Probability 101 Suppose we choose a number from the set: U = {1,2,3,4,5,6,7,8,9,10} Let A be the event that the number is even, and B be the event that the number is prime. Compute P(A), P(B), P(A|B), and P(B|A), where P(A|B) is the probability of A given B.
  • 6. Just count! 1 9 3 4 7 2 8 6 10 5 A B P(A) = 5/10 = 1/2 P(B) = 4/10 = 2/5 P(A|B) = 1/4 P(B|A) = 1/5
  • 7. Bayes Theorem P(A|B) = P(AB)/P(B) P(B)P(A|B) = P(AB) P(B)P(A|B) = P(A)P(B|A) P(A|B) = P(A)P(B|A)/P(B)
  • 8. A simplistic language model Consider each document to be a set of words, along with frequencies. For example: “The premium quality for the discount price” is viewed as: {'the':2, 'premium':1, 'quality':1, 'for':1,  'discount':1, 'price':1} Same as “The discount quality for the premium price,” since we don't care about order.
  • 9. That seems … foolish ● English is so complicated that we won't have any real hope of understanding semantics. ● In many real-life scenarios, text that you want to classify is not exactly subtle. ● If necessary, we can improve our language model later.
  • 10. An example: Type Text Class Training Good happy good Positive Training Good good service Positive Training Good friendly Positive Training Lousy good cheat Negative Test Good good good cheat lousy ?? In order to be able to perform all calculations, we will use an example with extremely small documents.
  • 11. What was the question? We are trying to determine whether the last recommendation was positive or negative. We want to compute: P(Pos|good good good lousy cheat) By Bayes Theorem, this is equal to: P(Pos)P(good good good lousy cheat|Pos) P(good good good lousy cheat)
  • 12. What do we know? P(Pos) = 3/4 P(good|Pos), P(cheat|Pos), P(lousy|Pos) Are all easily computed by counting using the training set. Which is almost what we want ...
  • 13. Wouldn't it be nice ... Maybe we have all we need? Isn't P(good good good lousy cheat|Pos) = P(good|Pos)3P(lousy|Pos)P(cheat|Pos) ? Well, yes, if these are independent events, which almost certainly doesn't hold. The “naive” assumption is that we can consider these events independent.
  • 14. The Naive Bayes Algorithm If C1,C2,...,Cn are classes, and an instance has features F1,F2,...,Fm, then the most likely class for this instance is the one that maximizes the following: P(Ci )P(F1|Ci )P(F2|Ci )...P(Fm|Ci )
  • 15. Wasn't there a denominator? If our goal was to compute the probability of the most likely class, we should divide by: P(F1)P(F2)...P(Fm) We can ignore this part because, we only care about which class has the highest probability, and this term is the same for each class.
  • 16. Interesting theory but … Won't this break as soon as we encounter a word that isn't in our training set? For example, if “goood” is not in our training set, and occurs in our test set, then since P(Pos|goood) = 0, so our product is zero for all classes. We need nonzero probabilities for all words, even words that don't exist.
  • 17. Plus-one smoothing Just count every word one time more than it actually occurs. Since we are only concerned with relative probabilities, this inaccuracy should be of no concern. P(word|C) = count(word|C) + 1 count(C) + V (V is the total vocabulary, so that our probabilities sum to 1.)
  • 18. Let's try it out: P(Pos) = ¾ Type Text Class P(Neg) = ¼ Training Good happy good Positive Good good service Positive Good friendly Positive Lousy good cheat Negative Test Good good good cheat lousy ?? P(good|Pos) = (5+1)/(8+6) = 3/7 P(cheat|Pos) = (0+1)/(8+6) = 1/14 P(Pos|D5) ~ ¾ * (3/7)3*(1/14)*(1/14) P(lousy|Pos) = (0+1)/(8+6) = 1/14 = 0.0003 P(good|Neg) = (1+1)/(3+6) = 2/9 P(Neg|D5) ~ ¼ * (2/9)3*(2/9)*(2/9) P(cheat|Neg) = (1+1)/(3+6) = 2/9 = 0.0001 P(lousy|Neg) = (1+1)/(3+6) = 2/9
  • 19. Training the classifier ● Count instances of classes, store counts in a map. ● Store counts of all words in a nested map: {'pos': {'good': 5, 'friendly': 1, 'service': 1, 'happy': 1}, 'neg': {'cheat': 1, 'lousy': 1, 'good': 1} } ● Should be easy to compute probabilities. ● Should be efficient (training time and memory.)
  • 20. Some practical problems ● Tokenization ● Arithmetic ● How to evaluate results?
  • 21. Tokenization ● Use whitespace? – “food”, “food.”, food,” and “food!” all different. ● Use whitespace and punctuation? – “won't” tokenized to “won” and “t” ● What about emails? Urls? Phone numbers? What about the things we haven't thought about yet? ● Use a library. Lucene is a good choice.
  • 22. Arithmetic What happens when you multiply a large amount of small numbers? To prevent underflow, use sums of logs instead of products of true probabilities. Key properties of log: ● log(AB) = log(A) + log(B) ● x > y => log(x) > log(y) ● Turns very small numbers into managable negative numbers
  • 23. Evaluating a classifier ● Precision and recall ● Confusion matrix ● Divide training set into nine “folds”, train classifier on nine folds, and check accuracy of classifying the tenth fold
  • 24. Experiment ● Tokenization strategies – Stop words – Capitalization – Stemming ● Language model – Ignore multiplicities – Smoothing
  • 25. Contact me ● wilson.eric.n@gmail.com ● ewilson@manta.com ● @wilsonericn ● http://wilsonericn.wordpress.com