Apriori Algorithm is used in finding frequent itemsets. Identifying associations between items in a dataset of transactions can be useful in various data mining tasks. For example, a supermarket can make better shelf arrangement if they know which items are purchased together frequently. The challenge is that given a dataset D having T transactions each with n number of attributes, how to find itemsets that appear frequently in D? This can be trivially solved by generating all possible itemsets (and checking each of the candidate itemset against support threshold.) which is computationally expensive. Apriori algorithm effectively eliminates majority of itemsets without counting their support value. Non-frequent itemsets are known prior to calculation of support count, thus the name Apriori alogrithm. It works on the following principle which is also known as apriori property:  

If an itemset is frequent, then all of its subsets must also be frequent.

  Suppose{c, d, e}is a frequent itemset. Clearly, any transaction that contains {c, d, e} must also contain its subsets, {c, d}, {c, e}, {d, e}, {c}, {d}, and {e}. As a result, if {c, d, e} is frequent, then all subsets of {c, d, e} must also be frequent. The algorithm works on bottom-up approach, that is generate 1-itemsets, eliminate those below support threshold then generate 2-itemsets, eliminate … and so on. This way only frequent _k_-itemsets are used to generate _k+1_-itemsets significantly reducing number of candidates. Let’s see an example: Our dataset contains four transactions with following particulars:

TID

Items

100

1, 3, 4

200

2, 3, 5

300

1, 2, 3, 5

400

2, 5

Support value is calculated as support (a -> b) = Number of transactions a and b appear in / total transactions. But for the sake of simplicity we use support value as number of times each transaction appears. We also assume support threshold = 2. Step 1 : Generate 1-itemsets, calculate support for each and mark itemsets below support threshold

Itemset

Support

{1}

2

{2}

3

{3}

3

{4}

1

{5}

3

Now, mark itemsets where support is below threshold

Itemset

Support

{1}

2

{2}

3

{3}

3

{4}

1

{5}

3

Step 1 : Generate 2-itemsets, calculate support for each and mark itemsets below support threshold. Remember for generating 2-itemsets we do not consider eliminated 1-itemsets.

Itemset

Support

{1, 2}

1

{1, 3}

2

{1, 5}

1

{2, 3}

2

{2, 5}

3

{3, 5}

2

Marking 2-itemsets below support threshold

Itemset

Support

{1, 2}

1

{1, 3}

2

{1, 5}

1

{2, 3}

2

{2, 5}

3

{3, 5}

2

Step 3 : Generate 3-itemsets, calculate support for each and mark itemsets below support threshold

Itemset

Support

{2, 3, 5}

2

  Stop: We stop here because 4-itemsets cannot be generated as there are only three items left! Following are the frequent itemsets that we have generated and which are above support threshold: {1}, {2}, {3}, {5}, {1, 3}, {2, 3}, {2, 5}, {3, 5} and {2, 3, 5}. Itemsets generated but not qualified are : {4}, {1, 2} and {1, 5}. If we do not use Apriori and instead rely on brute-force approach then number of candidate itemsets generated will be much larger! Feel free to ask your questions in comments section below!

Introduction

Intuitive partitioning or natural partitioning is used in data discretization. Data discretization is the process of converting continuous values of an attribute into categorical data or partitions or intervals. This helps reducing data size by reducing number of possible values, so instead of storing every observation, we store partition range in which each observation falls. One of the easiest ways to partition numeric values is using intuitive (natural) partitioning.

Intuitive partitioning for data discretization

1. If an interval covers 3, 6, 7 or 9 distinct values at most significant digit, then create 3 intervals. Here, there can be 3 equal width intervals for 3, 6, 9; and 3 intervals in the grouping of 2-3-2 each for 7.
2. If it covers 2, 4 or 8 distinct values at most significant digit, then create 4 sub-intervals of equal-width.
3. If it covers 1, 5, or 10 distinct values at the most significant digit, then partition range into 5 equal-width intervals.

Let’s understand with an example:

Part I : The Data

Assume that we have records showing profits made in each sale throughout a financial year. Profit data range is -3,51,976 to +4,70,00,896. Negative profit value is loss ;)

Part II : Dealing with noisy data

For purpose of avoiding noise, extremely high or extremely low values are not considered. So first we need to smooth out our data so let’s discard bottom 5% and top 5% values.

Part III : Finding MSD and interval range

• Suppose after discarding above data new values for LOW = -159876 and HIGH = 1838761. Here, most Significant Digit or MSD is at million position.
• Next step is to round down LOW and round up HIGH at MSD. So LOW = -1000000 and HIGH = 2000000. -1000000 is nearest down million to -159876 and 2000000 is nearest up million to 1838761.
• Next we identify range of this interval. Range = HIGH - LOW that is 2000000 - (-1000000) = 3000000. We consider only MSD here, so range of this interval is: 3.

Part IV : Applying rules

• Now that we know range = 3, we can apply rule #1.
• Rule #1 states that we can divide this interval into three equal size intervals:
  • Interval 1 : -1000000 to 0
  • Interval 2 : 0 to 1000000
  • Interval 3 : 1000000 to 2000000
• You should be thinking how 0 can be part of multiple intervals? You’re right! We should represent it as follows:
  • Interval 1 : (-1000000 … 0]
  • Interval 2 : (0 … 1000000]
  • Interval 3 : (1000000 … 2000000]
  • Here (a … b] denotes range that excludes a but includes b. ( , ]  is notation for half-open interval.

Conclusion

Now that we have partitions, we would want to replace profit data points with partition value in which each of it falls. This will save us storage space and complexity.

References

1. Han, J., Kamber M. (2006), Data Mining : Concepts and Techniques, Second Edition,  91-94.
2. The Range (Statistics), MathIsFun.com