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Let's start with an example. In Harry Potter, Hogwarts has 4 houses with students sorted into each house. The same happens on my website and I don't know how many users are in each house. It could be 20 in one house 50 in another and 100 in the third and fourth.
Now, each student can earn points on the website and at the end of the year, the house with the most points will win.
But it's not fair to "only" do a sum of the points, as the house with a 100 students will have a much higher chance to win, as they have more users to earn points. So I need to come up with an algorithm which is fair.
You can see an example here: https://worldofpotter.dk/points
What I do now is to sum all the points for a house, and then divide it by the number of users who have earned more than 10 points. This is still not fair, though.
Any ideas on how to make this calculation more fair?
Things we need to take into account:
* The percent of users earning points in each house
* Few users earning LOTS of points
* Many users earning FEW points (It's not bad earning few points. It still counts towards the total points of the house)
Link to MySQL dump(with users, houses and points): https://worldofpotter.dk/wop_points_example.sql
Link to CSV of points only: https://worldofpotter.dk/points.csv
I'd use something like Discounted Cumulative Gain which is used for measuring the effectiveness of search engines.
The concept is as it follows:
FUNCTION evalHouseScore (0_INDEXED_SORTED_ARRAY scores):
score = 0;
FOR (int i = 0; i < scores.length; i++):
score += scores[i]/log2(i);
END_FOR
RETURN score;
END_FUNCTION;
This must be somehow modified as this way of measuring focuses on the first result. As this is subjective you should decide on your the way you would modify it. Below I'll post the code which some constants which you should try with different values:
FUNCTION evalHouseScore (0_INDEXED_SORTED_ARRAY scores):
score = 0;
FOR (int i = 0; i < scores.length; i++):
score += scores[i]/log2(i+K);
END_FOR
RETURN L*score;
END_FUNCTION
Consider changing the logarithm.
Tests:
int[] g = new int[] {758,294,266,166,157,132,129,116,111,88,83,74,62,60,60,52,43,40,28,26,25,24,18,18,17,15,15,15,14,14,12,10,9,5,5,4,4,4,4,3,3,3,2,1,1,1,1,1};
int[] s = new int[] {612,324,301,273,201,182,176,139,130,121,119,114,113,113,106,86,77,76,65,62,60,58,57,54,54,42,42,40,36,35,34,29,28,23,22,19,17,16,14,14,13,11,11,9,9,8,8,7,7,7,6,4,4,3,3,3,3,2,2,2,2,2,2,2,1,1,1};
int[] h = new int[] {813,676,430,382,360,323,265,235,192,170,107,103,80,70,60,57,43,41,21,17,15,15,12,10,9,9,9,8,8,6,6,6,4,4,4,3,2,2,2,1,1,1};
int[] r = new int[] {1398,1009,443,339,242,215,210,205,177,168,164,144,144,92,85,82,71,61,58,47,44,33,21,19,18,17,12,11,11,9,8,7,7,6,5,4,3,3,3,3,2,2,2,1,1,1,1};
The output is for different offsets:
1182
1543
1847
2286
904
1231
1421
1735
813
1120
1272
1557
It sounds like some sort of constraint between the houses may need to be introduced. I might suggest finding the person that earned the most points out of all the houses and using it as the denominator when rolling up the scores. This will guarantee the max value of a user's contribution is 1, then all the scores for a house can be summed and then divided by the number of users to normalize the house's score. That should give you a reasonable comparison. It does introduce issues with low numbers of users in a house that are high achievers in which you may want to consider lower limits to the number of house members. Another technique may be to introduce handicap scores for users to balance the scales. The algorithm will most likely flex over time based on the data you receive. To keep it fair it will take some responsive action after the initial iteration. Players can come up with some creative ways to make scoring systems work for them. Here is some pseudo-code in PHP that you may use:
<?php
$mostPointsEarned; // Find the user that earned the most points
$houseScores = [];
foreach ($houses as $house) {
$numberOfUsers = 0;
$normalizedScores = [];
foreach ($house->getUsers() as $user) {
$normalizedScores[] = $user->getPoints() / $mostPointsEarned;
$numberOfUsers++;
}
$houseScores[] = array_sum($normalizedScores) / $numberOfUsers;
}
var_dump($houseScores);
You haven't given any examples on what should be preferred state, and what are situations against which you want to be immune. (3,2,1,1 compared to 5,2 etc.)
It's also a pity you haven't provided us the dataset in some nice way to play.
scala> val input = Map( // as seen on 2016-09-09 14:10 UTC on https://worldofpotter.dk/points
'G' -> Seq(758,294,266,166,157,132,129,116,111,88,83,74,62,60,60,52,43,40,28,26,25,24,18,18,17,15,15,15,14,14,12,10,9,5,5,4,4,4,4,3,3,3,2,1,1,1,1,1),
'S' -> Seq(612,324,301,273,201,182,176,139,130,121,119,114,113,113,106,86,77,76,65,62,60,58,57,54,54,42,42,40,36,35,34,29,28,23,22,19,17,16,14,14,13,11,11,9,9,8,8,7,7,7,6,4,4,3,3,3,3,2,2,2,2,2,2,2,1,1,1),
'H' -> Seq(813,676,430,382,360,323,265,235,192,170,107,103,80,70,60,57,43,41,21,17,15,15,12,10,9,9,9,8,8,6,6,6,4,4,4,3,2,2,2,1,1,1),
'R' -> Seq(1398,1009,443,339,242,215,210,205,177,168,164,144,144,92,85,82,71,61,58,47,44,33,21,19,18,17,12,11,11,9,8,7,7,6,5,4,3,3,3,3,2,2,2,1,1,1,1)
) // and the results on the website were: 1. R 1951, 2. H 1859, 3. S 990, 4. G 954
Here is what I thought of:
def singleValuedScore(individualScores: Seq[Int]) = individualScores
.sortBy(-_) // sort from most to least
.zipWithIndex // add indices e.g. (best, 0), (2nd best, 1), ...
.map { case (score, index) => score * (1 + index) } // here is the 'logic'
.max
input.mapValues(singleValuedScore)
res: scala.collection.immutable.Map[Char,Int] =
Map(G -> 1044,
S -> 1590,
H -> 1968,
R -> 2018)
The overall positions would be:
Ravenclaw with 2018 aggregated points
Hufflepuff with 1968
Slytherin with 1590
Gryffindor with 1044
Which corresponds to the ordering on that web: 1. R 1951, 2. H 1859, 3. S 990, 4. G 954.
The algorithms output is maximal product of score of user and rank of the user within a house.
This measure is not affected by "long-tail" of users having low score compared to the active ones.
There are no hand-set cutoffs or thresholds.
You could experiment with the rank attribution (score * index or score * Math.sqrt(index) or score / Math.log(index + 1) ...)
I take it that the fair measure is the number of points divided by the number of house members. Since you have the number of points, the exercise boils down to estimate the number of members.
We are in short supply of data here as the only hint we have on member counts is the answers on the website. This makes us vulnerable to manipulation, members can trick us into underestimating their numbers. If the suggested estimation method to "count respondents with points >10" would be known, houses would only encourage the best to do the test to hide members from our count. This is a real problem and the only thing I will do about it is to present a "manipulation indicator".
How could we then estimate member counts? Since we do not know anything other than test results, we have to infer the propensity to do the test from the actual results. And we have little other to assume than that we would have a symmetric result distribution (of the logarithm of the points) if all members tested. Now let's say the strong would-be respondents are more likely to actually test than weak would-be respondents. Then we could measure the extra dropout ratio for the weak by comparing the numbers of respondents in corresponding weak and strong test-point quantiles.
To be specific, of the 205 answers, there are 27 in the worst half of the overall weakest quartile, while 32 in the strongest half of the best quartile. So an extra 5 respondents of the very weakest have dropped out from an assumed all-testing symmetric population, and to adjust for this, we are going to estimate member count from this quantile by multiplying the number of responses in it by 32/27=about 1.2. Similarly, we have 29/26 for the next less-extreme half quartiles and 41/50 for the two mid quartiles.
So we would estimate members by simply counting the number of respondents but multiplying the number of respondents in the weak quartiles mentioned above by 1.2, 1.1 and 0.8 respectively. If however any result distribution within a house would be conspicuously skewed, which is not the case now, we would have to suspect manipulation and re-design our member count.
For the sample at hand however, these adjustments to member counts are minor, and yields the same house ranks as from just counting the respondents without adjustments.
I got myself to amuse me a little bit with your question and some python programming with some random generated data. As some people mentioned in the comments you need to define what is fairness. If as you said you don't know the number of people in each of the houses, you can use the number of participations of each house, thus you motivate participation (it can be unfair depending on the number of people of each house, but as you said you don't have this data on the first place).
The important part of the code is the following.
import numpy as np
from numpy.random import randint # import random int
# initialize random seed
np.random.seed(4)
houses = ["Gryffindor","Slytherin", "Hufflepuff", "Ravenclaw"]
houses_points = []
# generate random data for each house
for _ in houses:
# houses_points.append(randint(0, 100, randint(60,100)))
houses_points.append(randint(0, 50, randint(2,10)))
# count participation
houses_participations = []
houses_total_points = []
for house_id in xrange(len(houses)):
houses_total_points.append(np.sum(houses_points[house_id]))
houses_participations.append(len(houses_points[house_id]))
# sum the total number of participations
total_participations = np.sum(houses_participations)
# proposed model with weighted total participation points
houses_partic_points = []
for house_id in xrange(len(houses)):
tmp = houses_total_points[house_id]*houses_participations[house_id]/total_participations
houses_partic_points.append(tmp)
The results of this method are the following:
House Points per Participant
Gryffindor: [46 5 1 40]
Slytherin: [ 8 9 39 45 30 40 36 44 38]
Hufflepuff: [42 3 0 21 21 9 38 38]
Ravenclaw: [ 2 46]
House Number of Participations per House
Gryffindor: 4
Slytherin: 9
Hufflepuff: 8
Ravenclaw: 2
House Total Points
Gryffindor: 92
Slytherin: 289
Hufflepuff: 172
Ravenclaw: 48
House Points weighted by a participation factor
Gryffindor: 16
Slytherin: 113
Hufflepuff: 59
Ravenclaw: 4
You'll find the complete file with printing results here (https://gist.github.com/silgon/5be78b1ea0b55a20d90d9ec3e7c515e5).
You should enter some more rules to define the fairness.
Idea 1
You could set up the rule that anyone has to earn at least 10 points to enter the competition.
Then you can calculate the average points for each house.
Positive: Everyone needs to show some motivation.
Idea 2
Another approach would be to set the rule that from each house only the 10 best students will count for the competition.
Positive: Easy rule to calculate the points.
Negative: Students might become uninterested if they see they can't reach the top 10 places of their house.
From my point of view, your problem is diveded in a few points:
The best thing to do would be to re - assignate the player in the different Houses so that each House has the same number of players. (as explain by #navid-vafaei)
If you don't want to do that because you believe that it may affect your game popularity with player whom are in House that they don't want because you can change the choice of the Sorting Hat at least in the movie or books.
In that case, you can sum the point of the student's house and divide by the number of students. You may just remove the number of student with a very low score. You may remove as well the student with a very low activity because students whom skip school might be fired.
The most important part for me n your algorithm is weather or not you give points for all valuables things:
In the Harry Potter's story, the students earn point on the differents subjects they chose at school and get point according to their score.
At the end of the year, there is a special award event. At that moment, the Director gave points for valuable things which cannot be evaluated in the subject at school suche as the qualites (bravery for example).
Imagine I have person 1, 2, 3 and 4, then I have shirt styles A, B, C, D and I want to distribute the shirt styles to the people such that 25% of them get style A, 25% get style B, 25% get style C and 25% get style D but some of the people refuse to wear certain styles, these people are represented by Fs. How can I randomly match all the people with the styles they are willing to wear to get the approximate distribution?
A B C D
1 T F T T
2 T F F F
3 T T T T
4 T T T F
In this case this is easy and 25% is can be fully achieved, just give each person a different style. However, I intend to take this problem beyond this simple situation, my solution has to be generic. The number or styles, the number of people, and the distribution is all variable. Sometimes, the distribution will be impossible to create 100% accurately, approximate/close/best effor is expected. The selection process should be random and attempt to maintain the distribution.
I'm pretty agnostic to the language here, I'm just seeking the algorithm. Though preferably it would be able to be distributed.
Finding a solution when you are hampered by the Fs is https://en.wikipedia.org/wiki/Assignment_problem. One way to select an arbitrary assignment when there are many would be to set random costs where a style is acceptable to a person and then let it find the assignment with lowest possible cost. However it is not obvious that this will fit any natural definition of random. One (very inefficient) natural definition of random would be to select from all possible assignments at random until you get one that is acceptable to everybody. The distribution you get from this might not be the same as the one you would get by setting up random costs and then solving the resulting assignment problem.
You are using the term 'randomly match' which should be used with caution. The correct interpretation, I believe, is a random selection from the set of all valid solutions, so basically if we could enumerate all valid solution - we could trivially solve the problem.
You are looking for a close-enough solution, so we need to better define what a valid solution is. I suggest to define some threshold (say 1% error at most).
In your example, there are 4 groups (those assigned with shirt style A/B/C/D). Therefore, there are 2^4-1 possible person archetypes (love/hate each of A/B/C/D, with the assumption that anyone loves at least one shirt style). Each archetype population has a given population size, and each archetype can be assigned to some of the 4 groups (1 or more).
The goal is to divide the population of each archetype between the 4 groups, such that say, the size of each group is between L and H.
Lets formalize it.
Problem statement:
Denote A(0001),...,A(1111): the population size of each of the 15 archetypes
Denote G1(0001): the size of A(0001) assigned to G1, etc.
Given
L,H: constants
A(0001),...,A(1111): 15 constants
Our goal is to find all integer solutions for
G1(0001),G1(0011),G1(0101),G1(0111),G1(1001),G1(1011),G1(1101),G1(1111),
G2(0010),G2(0011),G2(0110),G2(0111),G2(1010),G2(1011),G2(1110),G2(1111),
G3(0100),G3(0101),G3(0110),G3(0111),G3(1100),G3(1101),G3(1110),G3(1111),
G4(1000),G4(1001),G4(1010),G4(1011),G4(1100),G4(1101),G4(1110),G4(1111)
subject to:
G1(0001) = A(0001)
G2(0010) = A(0010)
G2(0011) + G1(0011) = A(0011)
G3(0100) = A(0100)
G3(0101) + G1(0101) = A(0101)
G3(0110) + G2(0110) = A(0110)
G3(0111) + G2(0101) + G1(0101) = A(0111)
G4(1000) = A(1000)
G4(1001) + G1(1001) = A(1001)
G4(1010) + G2(1010) = A(1010)
G4(1011) + G2(1011) + G1(1011) = A(1011)
G4(1100) + G3(1100) = A(1100)
G4(1101) + G3(1101) + G1(1101) = A(1101)
G4(1110) + G3(1110) + G2(1110) = A(1110)
G4(1111) + G3(1111) + G2(1111) + G1(1111) = A(1111)
L < G1(0001)+G1(0011)+G1(0101)+G1(0111)+G1(1001)+G1(1011)+G1(1101)+G1(1111) <
H
L < G2(0010)+G2(0011)+G2(0110)+G2(0111)+G2(1010)+G2(1011)+G2(1110)+G2(1111) <
H
L < G3(0100)+G3(0101)+G3(0110)+G3(0111)+G3(1100)+G3(1101)+G3(1110)+G3(1111) <
H
L < G4(1000)+G4(1001)+G4(1010)+G4(1011)+G4(1100)+G4(1101)+G4(1110)+G4(1111) <
H
Now we can use an integer programming solver for the job.
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I am finding it hard to understand the process of Naive Bayes, and I was wondering if someone could explain it with a simple step by step process in English. I understand it takes comparisons by times occurred as a probability, but I have no idea how the training data is related to the actual dataset.
Please give me an explanation of what role the training set plays. I am giving a very simple example for fruits here, like banana for example
training set---
round-red
round-orange
oblong-yellow
round-red
dataset----
round-red
round-orange
round-red
round-orange
oblong-yellow
round-red
round-orange
oblong-yellow
oblong-yellow
round-red
The accepted answer has many elements of k-NN (k-nearest neighbors), a different algorithm.
Both k-NN and NaiveBayes are classification algorithms. Conceptually, k-NN uses the idea of "nearness" to classify new entities. In k-NN 'nearness' is modeled with ideas such as Euclidean Distance or Cosine Distance. By contrast, in NaiveBayes, the concept of 'probability' is used to classify new entities.
Since the question is about Naive Bayes, here's how I'd describe the ideas and steps to someone. I'll try to do it with as few equations and in plain English as much as possible.
First, Conditional Probability & Bayes' Rule
Before someone can understand and appreciate the nuances of Naive Bayes', they need to know a couple of related concepts first, namely, the idea of Conditional Probability, and Bayes' Rule. (If you are familiar with these concepts, skip to the section titled Getting to Naive Bayes')
Conditional Probability in plain English: What is the probability that something will happen, given that something else has already happened.
Let's say that there is some Outcome O. And some Evidence E. From the way these probabilities are defined: The Probability of having both the Outcome O and Evidence E is:
(Probability of O occurring) multiplied by the (Prob of E given that O happened)
One Example to understand Conditional Probability:
Let say we have a collection of US Senators. Senators could be Democrats or Republicans. They are also either male or female.
If we select one senator completely randomly, what is the probability that this person is a female Democrat? Conditional Probability can help us answer that.
Probability of (Democrat and Female Senator)= Prob(Senator is Democrat) multiplied by Conditional Probability of Being Female given that they are a Democrat.
P(Democrat & Female) = P(Democrat) * P(Female | Democrat)
We could compute the exact same thing, the reverse way:
P(Democrat & Female) = P(Female) * P(Democrat | Female)
Understanding Bayes Rule
Conceptually, this is a way to go from P(Evidence| Known Outcome) to P(Outcome|Known Evidence). Often, we know how frequently some particular evidence is observed, given a known outcome. We have to use this known fact to compute the reverse, to compute the chance of that outcome happening, given the evidence.
P(Outcome given that we know some Evidence) = P(Evidence given that we know the Outcome) times Prob(Outcome), scaled by the P(Evidence)
The classic example to understand Bayes' Rule:
Probability of Disease D given Test-positive =
P(Test is positive|Disease) * P(Disease)
_______________________________________________________________
(scaled by) P(Testing Positive, with or without the disease)
Now, all this was just preamble, to get to Naive Bayes.
Getting to Naive Bayes'
So far, we have talked only about one piece of evidence. In reality, we have to predict an outcome given multiple evidence. In that case, the math gets very complicated. To get around that complication, one approach is to 'uncouple' multiple pieces of evidence, and to treat each of piece of evidence as independent. This approach is why this is called naive Bayes.
P(Outcome|Multiple Evidence) =
P(Evidence1|Outcome) * P(Evidence2|outcome) * ... * P(EvidenceN|outcome) * P(Outcome)
scaled by P(Multiple Evidence)
Many people choose to remember this as:
P(Likelihood of Evidence) * Prior prob of outcome
P(outcome|evidence) = _________________________________________________
P(Evidence)
Notice a few things about this equation:
If the Prob(evidence|outcome) is 1, then we are just multiplying by 1.
If the Prob(some particular evidence|outcome) is 0, then the whole prob. becomes 0. If you see contradicting evidence, we can rule out that outcome.
Since we divide everything by P(Evidence), we can even get away without calculating it.
The intuition behind multiplying by the prior is so that we give high probability to more common outcomes, and low probabilities to unlikely outcomes. These are also called base rates and they are a way to scale our predicted probabilities.
How to Apply NaiveBayes to Predict an Outcome?
Just run the formula above for each possible outcome. Since we are trying to classify, each outcome is called a class and it has a class label. Our job is to look at the evidence, to consider how likely it is to be this class or that class, and assign a label to each entity.
Again, we take a very simple approach: The class that has the highest probability is declared the "winner" and that class label gets assigned to that combination of evidences.
Fruit Example
Let's try it out on an example to increase our understanding: The OP asked for a 'fruit' identification example.
Let's say that we have data on 1000 pieces of fruit. They happen to be Banana, Orange or some Other Fruit.
We know 3 characteristics about each fruit:
Whether it is Long
Whether it is Sweet and
If its color is Yellow.
This is our 'training set.' We will use this to predict the type of any new fruit we encounter.
Type Long | Not Long || Sweet | Not Sweet || Yellow |Not Yellow|Total
___________________________________________________________________
Banana | 400 | 100 || 350 | 150 || 450 | 50 | 500
Orange | 0 | 300 || 150 | 150 || 300 | 0 | 300
Other Fruit | 100 | 100 || 150 | 50 || 50 | 150 | 200
____________________________________________________________________
Total | 500 | 500 || 650 | 350 || 800 | 200 | 1000
___________________________________________________________________
We can pre-compute a lot of things about our fruit collection.
The so-called "Prior" probabilities. (If we didn't know any of the fruit attributes, this would be our guess.) These are our base rates.
P(Banana) = 0.5 (500/1000)
P(Orange) = 0.3
P(Other Fruit) = 0.2
Probability of "Evidence"
p(Long) = 0.5
P(Sweet) = 0.65
P(Yellow) = 0.8
Probability of "Likelihood"
P(Long|Banana) = 0.8
P(Long|Orange) = 0 [Oranges are never long in all the fruit we have seen.]
....
P(Yellow|Other Fruit) = 50/200 = 0.25
P(Not Yellow|Other Fruit) = 0.75
Given a Fruit, how to classify it?
Let's say that we are given the properties of an unknown fruit, and asked to classify it. We are told that the fruit is Long, Sweet and Yellow. Is it a Banana? Is it an Orange? Or Is it some Other Fruit?
We can simply run the numbers for each of the 3 outcomes, one by one. Then we choose the highest probability and 'classify' our unknown fruit as belonging to the class that had the highest probability based on our prior evidence (our 1000 fruit training set):
P(Banana|Long, Sweet and Yellow)
P(Long|Banana) * P(Sweet|Banana) * P(Yellow|Banana) * P(banana)
= _______________________________________________________________
P(Long) * P(Sweet) * P(Yellow)
= 0.8 * 0.7 * 0.9 * 0.5 / P(evidence)
= 0.252 / P(evidence)
P(Orange|Long, Sweet and Yellow) = 0
P(Other Fruit|Long, Sweet and Yellow)
P(Long|Other fruit) * P(Sweet|Other fruit) * P(Yellow|Other fruit) * P(Other Fruit)
= ____________________________________________________________________________________
P(evidence)
= (100/200 * 150/200 * 50/200 * 200/1000) / P(evidence)
= 0.01875 / P(evidence)
By an overwhelming margin (0.252 >> 0.01875), we classify this Sweet/Long/Yellow fruit as likely to be a Banana.
Why is Bayes Classifier so popular?
Look at what it eventually comes down to. Just some counting and multiplication. We can pre-compute all these terms, and so classifying becomes easy, quick and efficient.
Let z = 1 / P(evidence). Now we quickly compute the following three quantities.
P(Banana|evidence) = z * Prob(Banana) * Prob(Evidence1|Banana) * Prob(Evidence2|Banana) ...
P(Orange|Evidence) = z * Prob(Orange) * Prob(Evidence1|Orange) * Prob(Evidence2|Orange) ...
P(Other|Evidence) = z * Prob(Other) * Prob(Evidence1|Other) * Prob(Evidence2|Other) ...
Assign the class label of whichever is the highest number, and you are done.
Despite the name, Naive Bayes turns out to be excellent in certain applications. Text classification is one area where it really shines.
Your question as I understand it is divided in two parts, part one being you need a better understanding of the Naive Bayes classifier & part two being the confusion surrounding Training set.
In general all of Machine Learning Algorithms need to be trained for supervised learning tasks like classification, prediction etc. or for unsupervised learning tasks like clustering.
During the training step, the algorithms are taught with a particular input dataset (training set) so that later on we may test them for unknown inputs (which they have never seen before) for which they may classify or predict etc (in case of supervised learning) based on their learning. This is what most of the Machine Learning techniques like Neural Networks, SVM, Bayesian etc. are based upon.
So in a general Machine Learning project basically you have to divide your input set to a Development Set (Training Set + Dev-Test Set) & a Test Set (or Evaluation set). Remember your basic objective would be that your system learns and classifies new inputs which they have never seen before in either Dev set or test set.
The test set typically has the same format as the training set. However, it is very important that the test set be distinct from the training corpus: if we simply
reused the training set as the test set, then a model that simply memorized its input, without learning how to generalize to new examples, would receive misleadingly high scores.
In general, for an example, 70% of our data can be used as training set cases. Also remember to partition the original set into the training and test sets randomly.
Now I come to your other question about Naive Bayes.
To demonstrate the concept of Naïve Bayes Classification, consider the example given below:
As indicated, the objects can be classified as either GREEN or RED. Our task is to classify new cases as they arrive, i.e., decide to which class label they belong, based on the currently existing objects.
Since there are twice as many GREEN objects as RED, it is reasonable to believe that a new case (which hasn't been observed yet) is twice as likely to have membership GREEN rather than RED. In the Bayesian analysis, this belief is known as the prior probability. Prior probabilities are based on previous experience, in this case the percentage of GREEN and RED objects, and often used to predict outcomes before they actually happen.
Thus, we can write:
Prior Probability of GREEN: number of GREEN objects / total number of objects
Prior Probability of RED: number of RED objects / total number of objects
Since there is a total of 60 objects, 40 of which are GREEN and 20 RED, our prior probabilities for class membership are:
Prior Probability for GREEN: 40 / 60
Prior Probability for RED: 20 / 60
Having formulated our prior probability, we are now ready to classify a new object (WHITE circle in the diagram below). Since the objects are well clustered, it is reasonable to assume that the more GREEN (or RED) objects in the vicinity of X, the more likely that the new cases belong to that particular color. To measure this likelihood, we draw a circle around X which encompasses a number (to be chosen a priori) of points irrespective of their class labels. Then we calculate the number of points in the circle belonging to each class label. From this we calculate the likelihood:
From the illustration above, it is clear that Likelihood of X given GREEN is smaller than Likelihood of X given RED, since the circle encompasses 1 GREEN object and 3 RED ones. Thus:
Although the prior probabilities indicate that X may belong to GREEN (given that there are twice as many GREEN compared to RED) the likelihood indicates otherwise; that the class membership of X is RED (given that there are more RED objects in the vicinity of X than GREEN). In the Bayesian analysis, the final classification is produced by combining both sources of information, i.e., the prior and the likelihood, to form a posterior probability using the so-called Bayes' rule (named after Rev. Thomas Bayes 1702-1761).
Finally, we classify X as RED since its class membership achieves the largest posterior probability.
Naive Bayes comes under supervising machine learning which used to make classifications of data sets.
It is used to predict things based on its prior knowledge and independence assumptions.
They call it naive because it’s assumptions (it assumes that all of the features in the dataset are equally important and independent) are really optimistic and rarely true in most real-world applications.
It is classification algorithm which makes the decision for the unknown data set. It is based on Bayes Theorem which describe the probability of an event based on its prior knowledge.
Below diagram shows how naive Bayes works
Formula to predict NB:
How to use Naive Bayes Algorithm ?
Let's take an example of how N.B woks
Step 1: First we find out Likelihood of table which shows the probability of yes or no in below diagram.
Step 2: Find the posterior probability of each class.
Problem: Find out the possibility of whether the player plays in Rainy condition?
P(Yes|Rainy) = P(Rainy|Yes) * P(Yes) / P(Rainy)
P(Rainy|Yes) = 2/9 = 0.222
P(Yes) = 9/14 = 0.64
P(Rainy) = 5/14 = 0.36
Now, P(Yes|Rainy) = 0.222*0.64/0.36 = 0.39 which is lower probability which means chances of the match played is low.
For more reference refer these blog.
Refer GitHub Repository Naive-Bayes-Examples
Ram Narasimhan explained the concept very nicely here below is an alternative explanation through the code example of Naive Bayes in action
It uses an example problem from this book on page 351
This is the data set that we will be using
In the above dataset if we give the hypothesis = {"Age":'<=30', "Income":"medium", "Student":'yes' , "Creadit_Rating":'fair'} then what is the probability that he will buy or will not buy a computer.
The code below exactly answers that question.
Just create a file called named new_dataset.csv and paste the following content.
Age,Income,Student,Creadit_Rating,Buys_Computer
<=30,high,no,fair,no
<=30,high,no,excellent,no
31-40,high,no,fair,yes
>40,medium,no,fair,yes
>40,low,yes,fair,yes
>40,low,yes,excellent,no
31-40,low,yes,excellent,yes
<=30,medium,no,fair,no
<=30,low,yes,fair,yes
>40,medium,yes,fair,yes
<=30,medium,yes,excellent,yes
31-40,medium,no,excellent,yes
31-40,high,yes,fair,yes
>40,medium,no,excellent,no
Here is the code the comments explains everything we are doing here! [python]
import pandas as pd
import pprint
class Classifier():
data = None
class_attr = None
priori = {}
cp = {}
hypothesis = None
def __init__(self,filename=None, class_attr=None ):
self.data = pd.read_csv(filename, sep=',', header =(0))
self.class_attr = class_attr
'''
probability(class) = How many times it appears in cloumn
__________________________________________
count of all class attribute
'''
def calculate_priori(self):
class_values = list(set(self.data[self.class_attr]))
class_data = list(self.data[self.class_attr])
for i in class_values:
self.priori[i] = class_data.count(i)/float(len(class_data))
print "Priori Values: ", self.priori
'''
Here we calculate the individual probabilites
P(outcome|evidence) = P(Likelihood of Evidence) x Prior prob of outcome
___________________________________________
P(Evidence)
'''
def get_cp(self, attr, attr_type, class_value):
data_attr = list(self.data[attr])
class_data = list(self.data[self.class_attr])
total =1
for i in range(0, len(data_attr)):
if class_data[i] == class_value and data_attr[i] == attr_type:
total+=1
return total/float(class_data.count(class_value))
'''
Here we calculate Likelihood of Evidence and multiple all individual probabilities with priori
(Outcome|Multiple Evidence) = P(Evidence1|Outcome) x P(Evidence2|outcome) x ... x P(EvidenceN|outcome) x P(Outcome)
scaled by P(Multiple Evidence)
'''
def calculate_conditional_probabilities(self, hypothesis):
for i in self.priori:
self.cp[i] = {}
for j in hypothesis:
self.cp[i].update({ hypothesis[j]: self.get_cp(j, hypothesis[j], i)})
print "\nCalculated Conditional Probabilities: \n"
pprint.pprint(self.cp)
def classify(self):
print "Result: "
for i in self.cp:
print i, " ==> ", reduce(lambda x, y: x*y, self.cp[i].values())*self.priori[i]
if __name__ == "__main__":
c = Classifier(filename="new_dataset.csv", class_attr="Buys_Computer" )
c.calculate_priori()
c.hypothesis = {"Age":'<=30', "Income":"medium", "Student":'yes' , "Creadit_Rating":'fair'}
c.calculate_conditional_probabilities(c.hypothesis)
c.classify()
output:
Priori Values: {'yes': 0.6428571428571429, 'no': 0.35714285714285715}
Calculated Conditional Probabilities:
{
'no': {
'<=30': 0.8,
'fair': 0.6,
'medium': 0.6,
'yes': 0.4
},
'yes': {
'<=30': 0.3333333333333333,
'fair': 0.7777777777777778,
'medium': 0.5555555555555556,
'yes': 0.7777777777777778
}
}
Result:
yes ==> 0.0720164609053
no ==> 0.0411428571429
I try to explain the Bayes rule with an example.
What is the chance that a random person selected from the society is a smoker?
You may reply 10%, and let's assume that's right.
Now, what if I say that the random person is a man and is 15 years old?
You may say 15 or 20%, but why?.
In fact, we try to update our initial guess with new pieces of evidence ( P(smoker) vs. P(smoker | evidence) ). The Bayes rule is a way to relate these two probabilities.
P(smoker | evidence) = P(smoker)* p(evidence | smoker)/P(evidence)
Each evidence may increase or decrease this chance. For example, this fact that he is a man may increase the chance provided that this percentage (being a man) among non-smokers is lower.
In the other words, being a man must be an indicator of being a smoker rather than a non-smoker. Therefore, if an evidence is an indicator of something, it increases the chance.
But how do we know that this is an indicator?
For each feature, you can compare the commonness (probability) of that feature under the given conditions with its commonness alone. (P(f | x) vs. P(f)).
P(smoker | evidence) / P(smoker) = P(evidence | smoker)/P(evidence)
For example, if we know that 90% of smokers are men, it's not still enough to say whether being a man is an indicator of being smoker or not. For example if the probability of being a man in the society is also 90%, then knowing that someone is a man doesn't help us ((90% / 90%) = 1. But if men contribute to 40% of the society, but 90% of the smokers, then knowing that someone is a man increases the chance of being a smoker (90% / 40%) = 2.25, so it increases the initial guess (10%) by 2.25 resulting 22.5%.
However, if the probability of being a man was 95% in the society, then regardless of the fact that the percentage of men among smokers is high (90%)! the evidence that someone is a man decreases the chance of him being a smoker! (90% / 95%) = 0.95).
So we have:
P(smoker | f1, f2, f3,... ) = P(smoker) * contribution of f1* contribution of f2 *...
=
P(smoker)*
(P(being a man | smoker)/P(being a man))*
(P(under 20 | smoker)/ P(under 20))
Note that in this formula we assumed that being a man and being under 20 are independent features so we multiplied them, it means that knowing that someone is under 20 has no effect on guessing that he is man or woman. But it may not be true, for example maybe most adolescence in a society are men...
To use this formula in a classifier
The classifier is given with some features (being a man and being under 20) and it must decide if he is an smoker or not (these are two classes). It uses the above formula to calculate the probability of each class under the evidence (features), and it assigns the class with the highest probability to the input. To provide the required probabilities (90%, 10%, 80%...) it uses the training set. For example, it counts the people in the training set that are smokers and find they contribute 10% of the sample. Then for smokers checks how many of them are men or women .... how many are above 20 or under 20....In the other words, it tries to build the probability distribution of the features for each class based on the training data.