Association Rule Mining
Association rule learning is a popular and well researched method for discovering interesting relations between variables in large databases. Let I = {i1, i2,..., in} be a set of n binary attributes called items. Let D = {t1, t2,..., tm} be a set of transactions called the database. Each transaction in D has a unique transaction ID and contains a subset of the items in I. An association rule is defined as an implication of the form X ⇒ Y where X, Y ⊆ I and X ∩ Y = Ø. The item sets X and Y are called antecedent (left-hand-side or LHS) and consequent (right-hand-side or RHS) of the rule, respectively. The support supp(X) of an item set X is defined as the proportion of transactions in the database which contain the item set. Note that the support of an association rule X ⇒ Y is supp(X ∪ Y). The confidence of a rule is defined conf(X ⇒ Y) = supp(X ∪ Y) / supp(X). Confidence can be interpreted as an estimate of the probability P(Y | X), the probability of finding the RHS of the rule in transactions under the condition that these transactions also contain the LHS.
For example, the rule {onions, potatoes} ⇒ {burger} found in the sales data of a supermarket would indicate that if a customer buys onions and potatoes together, he or she is likely to also buy burger. Such information can be used as the basis for decisions about marketing activities such as promotional pricing or product placements.
Association rules are usually required to satisfy a user-specified minimum support and a user-specified minimum confidence at the same time. Association rule generation is usually split up into two separate steps:
- First, minimum support is applied to find all frequent item sets in a database (i.e. frequent item set mining).
- Second, these frequent item sets and the minimum confidence constraint are used to form rules.
Frequent Itemset Mining
Finding all frequent item sets in a database is difficult since it involves searching all possible item sets (item combinations). The set of possible item sets is the power set over I (the set of items) and has size 2n - 1 (excluding the empty set which is not a valid item set). Although the size of the power set grows exponentially in the number of items n in I, efficient search is possible using the downward-closure property of support (also called anti-monotonicity) which guarantees that for a frequent item set also all its subsets are frequent and thus for an infrequent item set, all its supersets must be infrequent.
In practice, we may only consider the frequent item set that has the maximum number of items bypassing all the sub item sets. An item set is maximal frequent if none of its immediate supersets is frequent.
For a maximal frequent item set, even though we know that all the sub item sets are frequent, we don't know the actual support of those sub item sets, which are very important to find the association rules within the item sets. If the final goal is association rule mining, we would like to discover closed frequent item sets. An item set is closed if none of its immediate supersets has the same support as the item set.
Some well known algorithms of frequent item set mining are Apriori, Eclat, and FP-Growth. Apriori is the best-known algorithm to mine association rules. It uses a breadth-first search strategy to counting the support of item sets and uses a candidate generation function which exploits the downward closure property of support. Eclat is a depth-first search algorithm using set intersection.
FP-growth (frequent pattern growth) algorithm employs an extended prefix-tree (FP-tree) structure to store the database in a compressed form. The FP-growth algorithm is currently one of the fastest approaches to discover frequent item sets. FP-growth adopts a divide-and-conquer approach to decompose both the mining tasks and the databases. It uses a pattern fragment growth method to avoid the costly process of candidate generation and testing used by Apriori.
The basic idea of the FP-growth algorithm can be described as a recursive elimination scheme: in a preprocessing step delete all items from the transactions that are not frequent individually, i.e., do not appear in a user-specified minimum number of transactions. Then select all transactions that contain the least frequent item (least frequent among those that are frequent) and delete this item from them. Recurse to process the obtained reduced (also known as projected) database, remembering that the item sets found in the recursion share the deleted item as a prefix. On return, remove the processed item from the database of all transactions and start over, i.e., process the second frequent item etc. In these processing steps the prefix tree, which is enhanced by links between the branches, is exploited to quickly find the transactions containing a given item and also to remove this item from the transactions after it has been processed.
When the input itemsets are already in memory, the below methods can be used.
The parameter itemsets is the item set database,
where each row is an item set, which may have different length.
The item identifiers have to be in [0, n), where n is the number of items.
Item set should NOT contain duplicated items. Note that it is reordered after the call.
The parameter minSupport is the required minimum support of item sets in terms
of frequency. The return can be the list of frequent item sets. Often the output
is too big to be stored in memory. In that case, the user may provide
a PrintStream or file path to save the output. The return number is
the number of discovered frequent item sets.
def fpgrowth(minSupport: Int, itemsets: Array[Array[Int]]): Stream[ItemSet]
public class FPTree {
public static FPTree of(int minSupport, int[][] itemsets);
public static FPTree of(double minSupport, int[][] itemsets);
}
public class FPGrowth {
public static Stream<ItemSet> apply(FPTree tree);
}
fun fpgrowth(minSupport: Int, itemsets: Array<IntArray>): Stream<ItemSet>
In practice, even the raw input is often too big to fit into the memory. To conquer this challenge, the below methods scan the input file twice. We first scan the database to obtains the frequency of single items. Then we scan the data again to construct the FP-Tree, which is a compressed form of data. In this way, we don't need load the whole database into the main memory.
def fptree(minSupport: Int, supplier: Supplier[Stream[Array[Int]]]): FPTree
def fpgrowth(tree: FPTree): Stream[ItemSet]
public class FPTree {
public static FPTree of(int minSupport, Supplier<Stream<int[]>> supplier);
public static FPTree of(double minSupport, Supplier<Stream<int[]>> supplier);
}
fun fptree(minSupport: Int, supplier: Supplier<Stream<IntArray>>): FPTree
fun fpgrowth(tree: FPTree): Stream<ItemSet>
In the below example, we apply FP-Growth to a toy data.
smile> val itemsets = Array(
Array(1, 3),
Array(2),
Array(4),
Array(2, 3, 4),
Array(2, 3),
Array(2, 3),
Array(1, 2, 3, 4),
Array(1, 3),
Array(1, 2, 3),
Array(1, 2, 3)
)
smile> fpgrowth(3, itemsets).forEach { itemset => println(itemset) }
4 (3)
1 (5)
2 1 (3)
3 2 1 (3)
3 1 (5)
2 (7)
3 2 (6)
3 (8)
smile> import smile.association.*
smile> int[][] itemsets = {
{1, 3},
{2},
{4},
{2, 3, 4},
{2, 3},
{2, 3},
{1, 2, 3, 4},
{1, 3},
{1, 2, 3},
{1, 2, 3}
}
itemsets ==> int[10][] { int[2] { 1, 3 }, int[1] { 2 }, int[1] ... 3 }, int[3] { 1, 2, 3 } }
smile> var tree = FPTree.of(3, itemsets)
tree ==> smile.association.FPTree@2a7b6f69
smile> FPGrowth.apply(tree).forEach(itemset -> System.out.println(itemset))
4 (3)
1 (5)
2 1 (3)
3 2 1 (3)
3 1 (5)
2 (7)
3 2 (6)
3 (8)
>>> import smile.association.*
>>> val itemsets = arrayOf(
intArrayOf(1, 3),
intArrayOf(2),
intArrayOf(4),
intArrayOf(2, 3, 4),
intArrayOf(2, 3),
intArrayOf(2, 3),
intArrayOf(1, 2, 3, 4),
intArrayOf(1, 3),
intArrayOf(1, 2, 3),
intArrayOf(1, 2, 3)
)
>>> fpgrowth(3, itemsets).forEach { println(it) }
4 (3)
1 (5)
2 1 (3)
3 2 1 (3)
3 1 (5)
2 (7)
3 2 (6)
3 (8)
Each row is a frequent item set, ending with the frequency in parentheses.
The sample data directory data/transaction contains a couple of
large benchmark datasets.
smile> val tree = fptree(1000, () => {
smile.util.Paths.getTestDataLines("transaction/kosarak.dat").map { line =>
line.split(" ").map(_.toInt).toArray
}
})
smile> fpgrowth(tree).limit(10).forEach { itemset => println(itemset) }
5634 (1000)
3805 (1001)
3376 (1001)
2279 (1001)
6333 (1002)
243 (1002)
808 (1003)
3875 (1004)
2265 (1004)
996 (1004)
smile> var tree = FPTree.of(1000, () -> {
try {
return smile.util.Paths.getTestDataLines("transaction/kosarak.dat").
map(line -> Arrays.stream(line.split(" ")).
mapToInt(Integer::valueOf).
toArray()
);
} catch (IOException ex) {
ex.printStackTrace();
}
return Stream.empty();
})
tree ==> smile.association.FPTree@3dddbe65
smile> FPGrowth.apply(tree).limit(10).forEach(itemset -> System.out.println(itemset))
5634 (1000)
3805 (1001)
3376 (1001)
2279 (1001)
6333 (1002)
243 (1002)
808 (1003)
3875 (1004)
2265 (1004)
996 (1004)
>>> import smile.util.*;
import java.util.function.*;
import java.util.stream.*;
class Parser : Supplier<Stream<IntArray>> {
override fun get(): Stream<IntArray> {
return smile.util.Paths.getTestDataLines("transaction/kosarak.dat").map { line ->
line.split(" ").map({ w -> w.toInt() }).toIntArray()
}
}
}
val tree = fptree(1000, Parser())
>>> fpgrowth(tree).limit(10).forEach { println(it) }
5634 (1000)
3805 (1001)
3376 (1001)
2279 (1001)
6333 (1002)
243 (1002)
808 (1003)
3875 (1004)
2265 (1004)
996 (1004)
The data kosarak.data has 990,002 transactions.
With minimum support 1000, our implementation generates 711,424
frequent item sets in about 30 seconds on a modern laptop.
Association Rules
After mining frequent itemsets, it is straightforward to form association rules that meet minimum confidence constraint. Our implementation generates association rules in a storage efficient way by employing the total support tree that is a kind of compressed set enumeration tree.
def arm(minSupport: Int, confidence: Double, itemsets: Array[Array[Int]]): Stream[AssociationRule]
def arm(confidence: Double, tree: FPTree): Stream[AssociationRule]
public class ARM {
public static Stream<AssociationRule> apply(double confidence, FPTree tree);
}
def arm(minSupport: Int, confidence: Double, itemsets: Array<IntArray>): Stream<AssociationRule>
def arm(confidence: Double, tree: FPTree): Stream<AssociationRule>
The API is similar to fpgrowth except the additional parameter confidence.
smile> arm(0.6, tree).limit(10).forEach { rule => println(rule) }
(11) => (6) support = 32.73% confidence = 89.00% lift = 1.47 leverage = 0.1039
(3, 11) => (6) support = 14.51% confidence = 89.09% lift = 1.47 leverage = 0.0462
(1) => (6) support = 13.34% confidence = 66.89% lift = 1.10 leverage = 0.0123
(3, 1) => (6) support = 5.84% confidence = 68.28% lift = 1.12 leverage = 0.0064
(6, 1) => (11) support = 8.70% confidence = 65.17% lift = 1.77 leverage = 0.0379
(11, 1) => (6) support = 8.70% confidence = 93.70% lift = 1.54 leverage = 0.0306
(6, 3, 1) => (11) support = 3.81% confidence = 65.30% lift = 1.78 leverage = 0.0167
(3, 11, 1) => (6) support = 3.81% confidence = 93.73% lift = 1.54 leverage = 0.0134
(218) => (6) support = 7.85% confidence = 87.67% lift = 1.44 leverage = 0.0241
(3, 218) => (6) support = 3.43% confidence = 87.86% lift = 1.45 leverage = 0.0106
smile> ARM.apply(0.6, tree).limit(10).forEach(rule -> System.out.println(rule))
(11) => (6) support = 32.73% confidence = 89.00% lift = 1.47 leverage = 0.1039
(3, 11) => (6) support = 14.51% confidence = 89.09% lift = 1.47 leverage = 0.0462
(1) => (6) support = 13.34% confidence = 66.89% lift = 1.10 leverage = 0.0123
(3, 1) => (6) support = 5.84% confidence = 68.28% lift = 1.12 leverage = 0.0064
(6, 1) => (11) support = 8.70% confidence = 65.17% lift = 1.77 leverage = 0.0379
(11, 1) => (6) support = 8.70% confidence = 93.70% lift = 1.54 leverage = 0.0306
(6, 3, 1) => (11) support = 3.81% confidence = 65.30% lift = 1.78 leverage = 0.0167
(3, 11, 1) => (6) support = 3.81% confidence = 93.73% lift = 1.54 leverage = 0.0134
(218) => (6) support = 7.85% confidence = 87.67% lift = 1.44 leverage = 0.0241
(3, 218) => (6) support = 3.43% confidence = 87.86% lift = 1.45 leverage = 0.0106
>>> arm(0.6, tree).limit(10).forEach { println(it) }
(11) => (6) support = 32.73% confidence = 89.00% lift = 1.47 leverage = 0.1039
(3, 11) => (6) support = 14.51% confidence = 89.09% lift = 1.47 leverage = 0.0462
(1) => (6) support = 13.34% confidence = 66.89% lift = 1.10 leverage = 0.0123
(3, 1) => (6) support = 5.84% confidence = 68.28% lift = 1.12 leverage = 0.0064
(6, 1) => (11) support = 8.70% confidence = 65.17% lift = 1.77 leverage = 0.0379
(11, 1) => (6) support = 8.70% confidence = 93.70% lift = 1.54 leverage = 0.0306
(6, 3, 1) => (11) support = 3.81% confidence = 65.30% lift = 1.78 leverage = 0.0167
(3, 11, 1) => (6) support = 3.81% confidence = 93.73% lift = 1.54 leverage = 0.0134
(218) => (6) support = 7.85% confidence = 87.67% lift = 1.44 leverage = 0.0241
(3, 218) => (6) support = 3.43% confidence = 87.86% lift = 1.45 leverage = 0.0106
