AA is not only where the alcoholics get help but also where were data miners go when they need to find a basic algorithm to locate all the combinations that appear a certain number of times in a set of information.

The Apriori algorithm, takes an initial data set, beats it to a pulp, strains it, and gives you all the possible ways of cooking a hot poket. With cheese, no cheese, with meat, or no meat, this algorithm gets you every possible cobination and the number of times it happens in the database set.

Confused? Good! Since i like examples to understand things lets use one. Lets say you go to the market. You by some milk, a can of decaf and some dog food. The supermarket chain records this info for not only your trasaction but for everyone (talk about big brother looking over you). After about a month the super market chain has all this info on the things people buy. they see how some people buy milk with cookies and how some people buy squeeky toys with condoms (dont ask). But they want to find more info. They pop open their computer run the Apriori algorithm and determine that they want to find all combination that occure in the data set more than 2 times. After running the algorihtm they see that milk and tomatoes sold alot together. Good market info if you want to have a sale on tomatoes so you can get rid of your milk ðŸ™‚ They also found that cookies with toothpaste were also items that sold in hgh volumes aswell. In data mining terms we use the words “frequent” to state how well a itemset (the combination of items) was old.

So in closing, this its 1 am and the algorithm is coded below it will allow you to stick in a dataset of information and receive the number of times the combination occures that meet your expected minimum supporting threshold, the minimum number of times you want the combination to appear.

Before i get to the code, let me go over it first. The below code reads a filepath that we need to use. I have two files you can use. One is the

test.txt file and the other is the retail.txt file. When running the test.txt file the system finishes very quickly and stores the itemsets in a output file of your choice. On the other hand, when we run retail.txt and set the minimum_sup to less than 2000 the aplication runs for 2 hours, a sign of either a very poor implementation or a very bad langauage to use on data mining.

Algorithm

1. Get all the unique items in the dataset

2. For each unique item count the number of records that contain the combination.

3. If the number of times it appears is greater than the min_sup keep it and prune the other unique values that didnt meet the min_sup.

4. with the new set of values join them to form your new list.

5. check if new set has any subsets that are not frequent.

6. trash the itemsets that have subsets that are not frequent.

7. with new clean set go to #2.

8. continue until new list is null, this happens when the new list has all itemsets which have subsets as not being frequent.

Ruby Code:

##########

# Apriori Algorithm Implementation

#

# @author Armando Padilla, mando81@prodigy.net

# @homepage http://www.armando.ws

#

# v0.0.2 month.week.day

##########

#require ‘profiler’

#Profiler__::start_profile

start = Time.now

puts “Started – “+start.to_s

#############

# Create the new list (Lk) for the next iteration

#

# @param itemSets Lk-1 list that has been cleared out of all

# none frequent sets.

#############

def createList(itemSets)

###

# HOLDS THE NEW LIST

###

listX = Array.new()

if !itemSets.nil?

#Check if the initial elements are the same

#if they are the same then

#Check if the last item is less than the next row’s last item.

#if its less then join both items (union on both items)

k = 1

itemSetSize = itemSets.size

itemSets.each do |list1|

n=k

until n == itemSetSize

#Check if we can combine the lists

#if initial run yes lets combine the list

#otherwise follow the steps

if itemSets[n].size == 1

if !list1.eql?(itemSets[n])

@row = Array.new()

@row.push(list1,itemSets[n])

@row = @row.sort

@row = @row.flatten!

###

# DO NOT ADD IF WE ALREADY HAVE THE SET

###

if !listX.include?(@row)

listX.push(@row)

end

end

else

####

#if the all the indexes before the last are the same

#then combine ONLY IF the last indexes are the same

####

toIndex = list1.size-2

###

# GET ALL THE VALUES UP TO THE LAST INDEX

# EG. [1,2,3,4,5] and [1,2,3,4,6]

# BECOMES [1,2,3,4] and [1,2,3,4]

###

arrayA = list1[0..toIndex]

arrayB = itemSets[n][0..toIndex]

if arrayA.eql?(arrayB)

totalCountA = list1.size-1

totalCountB = itemSets[n].size-1

if list1[totalCountA] < itemSets[n][totalCountB]

@row = Array.new()

@row.push(list1|itemSets[n])

@row = @row.sort.uniq

@row = @row.flatten!

###

# DO NOT ADD IF WE ALREADY HAVE THE SET

###

if !listX.include?(@row)

listX.push(@row)

end

end

end

#puts “=========”

end

n = n+1

end

k = k+1

end

end

return listX

end

##############

# Returns the item sets that support the minimum supporting threshold.

#

##############

def getFrequentItemSets(itemSets, minSup, fileInputPath)

########

# RUNS THROUGH THE DATA SET AND GETS THE COUNT

######

list = Array.new()

##

# FOREACH ITEM CHECK HOW MANY TIMES IT APPEARS IN THE DATA SET

##

itemSets.each do |itemSet|

itemSetSize = itemSet.size

count = 0

###

# FOR EACH ROW IN THE ITEM SET

# RUN THROUGH EACH ROW IN THE DATA SET

# AND DETERMINE IF THE ITEM SET APPEARS IN THE RECORD

###

@dataSet.each do |set|

currentRow = set

if itemSetSize == 1

if currentRow.index(itemSet[0])

count = count+1

end

else

###

# FOR EACH ROW CHECK IF THE ITEMSET IS PRESENT

#

# ARMANDO – USING INTERSECTION TO COMPARE

# =======================================

# ITEMSET = A

# CURRENT ROW = B

# TAKE THE INTERSECTION OF A AND B

# IF SIZE OF INTERSECTION IS EQUAL TO SIZE OF ITEMSET THEN

# WE LOCATED THE ITEM SET IN THE ROW

###

intersection = []

itemSet.each do |a|

set.each do |b|

if a == b

intersection.push(a)

end

end

end

if intersection.size >= itemSetSize

#if (itemSet¤tRow).size >= itemSet.size

count = count+1

end

end

end #end – run through data set

###

# DETERMINE IF IT MEETS THE MIN_SUP

# if the calculatedValue is equal to or greater than the

# minimum supporting threshold ad it to our new List (L k)

###

if count >= minSup

@row = Array.new()

@row.push(itemSet)

list.push(itemSet)

string = “”

itemSet.each do |x|

string = string+” “+x.to_s

end

@file.puts string+” :”+count.to_s

@file.flush

else

irow = Array.new()

irow.push(itemSet)

@infrequentSets.push(irow)

end

p(itemSet)

end #end – run though each item

return list.sort

end #end – getFrequentItemSets

##############

# Returns boolean when the item set is either

# frequent or infrequent

##############

def isInfrequentSets(itemSets)

globalCount = 0

###

# FOREACH ITEMSET CHECK IF ITS IN THE INFREQUENT ITEM SET

###

itemSets.each do |itemSet|

count = 0

@infrequentSets.each do |itemSetX|

###

# IF WE FOUND A ITEM SET WITH at least 2 sub items

# REMOVE IT

###

if(itemSet&itemSetX[0]).size >= 2

count = count+1

end

end

if count != 0

globalCount = globalCount + 1

end

end

if globalCount == itemSets.size

return true

end

end

#####

# Reads the input from a text file and create an array

# Each line is an array of elements.

# Elements broken out by white space.

#

# @param fileInputPath Path to data set file.

#

####

def getDataSet(fileInputPath)

dataSet = []

if File.exists?(fileInputPath)

###

# GET THE FILE CONTENT

###

dataContainer = File.open(fileInputPath, “r”)

###

# FOREACH LINE BREAK APART BY WHITESPACE

# AND PLACE INTO ARRAY

###

flist = []

dataContainer.readlines.each do |line|

flist.push(line)

end

flist.each do |line|

elements = line.split(” “)

row = Array.new()

elements.each do |element|

row.push(element.to_i)

if @globalHash[element.to_i] != nil

@globalHash[element.to_i] = @globalHash[element.to_i]+1

else

@globalHash[element.to_i] = 1

end

end

###

# ADD THE ROW AS AN ARRAY

# TO THE DATA SET

###

dataSet.push(row)

end

else

puts “Error: File, “+fileInputPath+”, does not exist”

end

return dataSet

end

#########################################################

#########################################################

#########################################################

if ARGV.size == 3

###

# INITIALIZE VARIABLES

###

min_sup = ARGV[2].to_i

fileInputPath = ARGV[0]

outputFilePath = ARGV[1]

@infrequentSets = Array.new()

@globalHash = Hash.new() #holds out initial List with counts

#####

# GET THE DATA SET FROM FILE

#####

@dataSet = Array.new()

@dataSet = getDataSet(fileInputPath)

####

# WRITE OUT THE OUTPUT

####

@file = File.open(outputFilePath, “w+”)

#####

# GET ALL THE ITEMS WITH MIN_SUP > N – List1

#####

@itemsContainer = []

@globalHash.keys.each do |element|

if @globalHash[element].to_i >= min_sup

@itemsContainer.push(element)

string = element.to_s

@file.puts string+” :”+@globalHash[element].to_s

@file.flush

end

end

@itemsContainer.sort!

####

#MAKE ALL ITEMS IN CONTAINER AN ARRAY

####

@items = Array.new()

@itemsContainer.each do |x|

@row = Array.new()

@row.push(x)

@items.push(@row)

end

@items = createList(@items)

######

# RUN THROUGH THE DATA SET UNTIL WE HAVE REACHED

# ALL FREQUENT ITEMSETS.

######

k = 1

@listK = @items

foundAllFrequentSets = FALSE

while foundAllFrequentSets == FALSE

###

# GET ALL THE ITEMS THAT PASS THE MIN_SUP THRESHOLD

###

@frequentSet = getFrequentItemSets(@listK, min_sup, fileInputPath)

p(@frequentSet)

#######

# CREATE THE NEW LIST FROM THE FRQUENT ITEMSET

# THAT MET MIN_SUP

#######

@listK = createList(@frequentSet)

#check if the result all contain infrequent sub sets

if isInfrequentSets(@listK)

foundAllFrequentSets = TRUE

end

k = k+1

end #while foundAllFrequentSets == FALSE

@file.close

else

puts “Error: Must Be Format Type :: apriori.rb

endTime = Time.now

puts “Ended – “+endTime.to_s+”\n”

puts “Total Time: “+(endTime-start).to_s+”\n”

#Profiler__::stop_profile

#Profiler__::print_profile($stderr)

BTW: the data sets were created by my prof Dr. Sun.

Armando Padilla