Data Mining Tutorial
What is it?
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Large datasets
Fast methods
Not significance testing
Topics
– Trees (recursive splitting)
– Nearest Neighbor
– Neural Networks
– Clustering
– Association Analysis
Trees
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A “divisive” method (splits)
Start with “root node” – all in one group
Get splitting rules
Response often binary
Result is a “tree”
Example: Loan Defaults
Example: Framingham Heart Study
Recursive Splitting
Pr{default} =0.007
Pr{default} =0.012
Pr{default} =0.006
X1=Debt
To
Income
Ratio
Pr{default} =0.0001
Pr{default} =0.003
No default
Default
X2 = Age
Some Actual Data
• Framingham Heart
Study
• First Stage Coronary
Heart Disease
– P{CHD} = Function of:
• Age - no drug yet! 
• Cholesterol
• Systolic BP
Import
Example of a “tree”
All 1615 patients
Split # 1: Age
Systolic BP
“terminal node”
How to make splits?
• Which variable to use?
• Where to split?
– Cholesterol > ____
– Systolic BP > _____
• Goal: Pure “leaves” or “terminal nodes”
• Ideal split: Everyone with BP>x has
problems, nobody with BP<x has
problems
Where to Split?
• First review Chi-square tests
• Contingency tables
Heart Disease
No
Yes
Low
BP
High
BP
Heart Disease
No
Yes
95
5
100
75
25
55
45
100
75
25
DEPENDENT
INDEPENDENT
c2 Test Statistic
• Expect 100(150/200)=75 in upper left if
independent (etc. e.g. 100(50/200)=25)
Heart Disease
No
Yes
Low
BP
High
BP
2
(
observed

exp
ected
)
c 2  allcells
exp ected
95
(75)
55
(75)
5
(25)
45
(25)
100
150
50
200
100
WHERE IS HIGH BP CUTOFF???
2(400/75)+
2(400/25) =
42.67
Compare to
Tables –
Significant!
Measuring “Worth” of a Split
• P-value is probability of Chi-square as
great as that observed if independence is
true. (Pr {c2>42.67} is 6.4E-11)
• P-values all too small.
• Logworth = -log10(p-value) = 10.19
• Best Chi-square  max logworth.
Logworth for Age Splits
Age 47 maximizes logworth
How to make splits?
• Which variable to use?
• Where to split?
– Cholesterol > ____
– Systolic BP > _____
• Idea – Pick BP cutoff to minimize p-value
for c2
• What does “signifiance” mean now?
Multiple testing
• 50 different BPs in data, 49 ways to split
• Sunday football highlights always look
good!
• If he shoots enough baskets, even 95%
free throw shooter will miss.
• Jury trial analogy
• Tried 49 splits, each has 5% chance of
declaring significance even if there’s no
relationship.
Multiple testing
a=
Pr{ falsely reject hypothesis 2}
a=
Pr{ falsely reject hypothesis 1}
Pr{ falsely reject one or the other} < 2a
Desired: 0.05 probabilty or less
Solution: use a = 0.05/2
Or – compare 2(p-value) to 0.05
Multiple testing
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50 different BPs in data, m=49 ways to split
Multiply p-value by 49
Bonferroni – original idea
Kass – apply to data mining (trees)
Stop splitting if minimum p-value is large.
For m splits, logworth becomes
-log10(m*p-value)
Other Split Evaluations
• Gini Diversity Index
– { A A A A B A B B C B}
– Pick 2, Pr{different} = 1-Pr{AA}-Pr{BB}-Pr{CC}
• 1-[10+6+0]/45=29/45=0.64
– {AABCBAABCC}
• 1-[6+3+3]/45 = 33/45 = 0.73  MORE DIVERSE, LESS PURE
• Shannon Entropy
– Larger  more diverse (less pure)
–
-Si pi log2(pi)
{0.5, 0.4, 0.1}  1.36
{0.4, 0.2, 0.3}  1.51
(more diverse)
Goals
• Split if diversity in parent “node” > summed
diversities in child nodes
• Observations should be
– Homogeneous (not diverse) within leaves
– Different between leaves
– Leaves should be diverse
• Framingham tree used Gini for splits
Cross validation
• Traditional stats – small dataset, need all
observations to estimate parameters of
interest.
• Data mining – loads of data, can afford
“holdout sample”
• Variation: n-fold cross validation
– Randomly divide data into n sets
– Estimate on n-1, validate on 1
– Repeat n times, using each set as holdout.
Pruning
• Grow bushy tree on the “fit data”
• Classify holdout data
• Likely farthest out branches do not
improve, possibly hurt fit on holdout data
• Prune non-helpful branches.
• What is “helpful”? What is good
discriminator criterion?
Goals
• Want diversity in parent “node” > summed
diversities in child nodes
• Goal is to reduce diversity within leaves
• Goal is to maximize differences between
leaves
• Use same evaluation criteria as for splits
• Costs (profits) may enter the picture for
splitting or evaluation.
Accounting for Costs
• Pardon me (sir, ma’am) can you spare
some change?
• Say “sir” to male +$2.00
• Say “ma’am” to female +$5.00
• Say “sir” to female -$1.00 (balm for
slapped face)
• Say “ma’am” to male -$10.00 (nose splint)
Including Probabilities
Leaf has Pr(M)=.7, Pr(F)=.3.
You say:
M
F
True
Gender
M
0.7 (2)
0.7 (-10)
0.3 (5)
F
Expected profit is 2(0.7)-1(0.3) = $1.10 if I say “sir”
Expected profit is -7+1.5 = -$5.50 (a loss) if I say “Ma’am”
Weight leaf profits by leaf size (# obsns.) and sum
Prune (and split) to maximize profits.
Additional Ideas
• Forests – Draw samples with replacement
(bootstrap) and grow multiple trees.
• Random Forests – Randomly sample the
“features” (predictors) and build multiple
trees.
• Classify new point in each tree then
average the probabilities, or take a
plurality vote from the trees
• “Bagging” – Bootstrap aggregation
• “Boosting” – Similar, iteratively reweights points
that were misclassified to produce sequence of
more accurate trees.
* Lift Chart
- Go from leaf of most
to least response.
- Lift is cumulative
proportion responding.
Regression Trees
• Continuous response (not just class)
• Predicted response constant in regions
Predict 80
Predict 50
X2
Predict
130
Predict 100
X1
Predict
20
• Predict Pi in cell i.
• Yij jth response in cell i.
• Split to minimize Si Sj (Yij-Pi)2
Predict 80
Predict 50
Predict
130
Predict 100
Predict
20
• Predict Pi in cell i.
• Yij jth response in cell i.
• Split to minimize Si Sj (Yij-Pi)2
Logistic Regression
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“Trees” seem to be main tool.
Logistic – another classifier
Older – “tried & true” method
Predict probability of response from input
variables (“Features”)
• Linear regression gives infinite range of
predictions
• 0 < probability < 1 so not linear regression.
• Logistic idea: Map p in (0,1) to L in whole
real line
• Use L = ln(p/(1-p))
• Model L as linear in temperature
• Predicted L = a + b(temperature)
• Given temperature X, compute a+bX then p
= eL/(1+eL)
• p(i) = ea+bXi/(1+ea+bXi)
• Write p(i) if response, 1-p(i) if not
• Multiply all n of these together, find a,b to
maximize
Example: Ignition
• Flame exposure time = X
• Ignited Y=1, did not ignite Y=0
– Y=0, X= 3, 5, 9 10 ,
13,
16
– Y=1, X =
11, 12 14, 15, 17, 25, 30
• Q=(1-p)(1-p)(1-p)(1-p)pp(1-p)pp(1-p)ppp
• P’s all different p=f(exposure)
• Find a,b to maximize Q(a,b)
Generate Q for array of (a,b) values
DATA LIKELIHOOD;
ARRAY Y(14) Y1-Y14; ARRAY X(14) X1-X14;
DO I=1 TO 14; INPUT X(I) y(I) @@; END;
DO A = -3 TO -2 BY .025;
DO B = 0.2 TO 0.3 BY .0025;
Q=1;
DO i=1 TO 14;
L=A+B*X(i); P=EXP(L)/(1+EXP(L));
IF Y(i)=1 THEN Q=Q*P; ELSE Q=Q*(1-P);
END; IF Q<0.0006 THEN Q=0.0006; OUTPUT; END;END;
CARDS;
3 0 5 0 7 1 9 0 10 0 11 1 12 1 13 0 14 1 15 1 16 0 17 1
25 1 30 1
;
Likelihood function (Q)
-2.6
0.23
IGNITION DATA
The LOGISTIC Procedure
Analysis of Maximum Likelihood Estimates
Parameter
Intercept
TIME
DF
1
1
Estimate
-2.5879
0.2346
Standard
Error
1.8469
0.1502
Wald
Chi-Square
1.9633
2.4388
Pr > ChiSq
0.1612
0.1184
Association of Predicted Probabilities and Observed Responses
Percent Concordant
Percent Discordant
Percent Tied
Pairs
79.2
20.8
0.0
48
Somers' D
Gamma
Tau-a
c
0.583
0.583
0.308
0.792
4 right,
1 wrong
5 right,
4 wrong
Example: Framingham
• X=age
• Y=1 if heart trouble, 0 otherwise
Framingham
The LOGISTIC Procedure
Analysis of Maximum Likelihood Estimates
Parameter
DF
Intercept
age
1
1
Standard
Wald
Estimate
Error Chi-Square
-5.4639
0.0630
0.5563
0.0110
96.4711
32.6152
Pr>ChiSq
<.0001
<.0001
Example:
Shuttle Missions
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O-rings failed in Challenger disaster
Low temperature
Prior flights “erosion” and “blowby” in O-rings
Feature: Temperature at liftoff
Target: problem (1) - erosion or blowby vs. no
problem (0)
Neural Networks
• Very flexible functions
• “Hidden Layers”
• “Multilayer Perceptron”
output
inputs
Logistic function of
Logistic functions
Of data
Arrows represent linear
combinations of “basis
functions,” e.g. logistics
b1
Example:
Y = a + b1 p1 + b2 p2 + b3 p3
Y = 4 + p1+ 2 p2 - 4 p3
• Should always use holdout sample
• Perturb coefficients to optimize fit (fit data)
– Nonlinear search algorithms
• Eliminate unnecessary arrows using
holdout data.
• Other basis sets
– Radial Basis Functions
– Just normal densities (bell shaped) with
adjustable means and variances.
Terms
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Train: estimate coefficients
Bias: intercept a in Neural Nets
Weights: coefficients b
Radial Basis Function: Normal density
Score: Predict (usually Y from new Xs)
Activation Function: transformation to target
Supervised Learning: Training data has
response.
Hidden Layer
L1 = -1.87 - .27*Age – 0.20*SBP22
H11=exp(L1)/(1+exp(L1))
L2 = -20.76 -21.38*H11
Pr{first_chd} = exp(L2)/(1+exp(L2))
“Activation Function”
Unsupervised Learning
• We have the “features” (predictors)
• We do NOT have the response even on a
training data set (UNsupervised)
• Clustering
– Agglomerative
• Start with each point separated
– Divisive
• Start with all points in one cluster then spilt
EM  PROC FASTCLUS
• Step 1 – find “seeds” as separated as
possible
• Step 2 – cluster points to nearest seed
– Drift: As points are added, change seed
(centroid) to average of each coordinate
– Alternatively: Make full pass then recompute
seed and iterate.
Clusters as Created
As Clustered
Cubic Clustering Criterion
(to decide # of Clusters)
• Divide random scatter of (X,Y) points into
4 quadrants
• Pooled within cluster variation much less
than overall variation
• Large variance reduction
• Big R-square despite no real clusters
• CCC compares random scatter R-square
to what you got to decide #clusters
• 3 clusters for “macaroni” data.
Association Analysis
• Market basket analysis
– What they’re doing when they scan your “VIP”
card at the grocery
– People who buy diapers tend to also buy
_________ (beer?)
– Just a matter of accounting but with new
terminology (of course  )
– Examples from SAS Appl. DM Techniques, by
Sue Walsh:
Termnilogy
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Baskets: ABC ACD BCD ADE BCE
Rule Support
Confidence
X=>Y Pr{X and Y} Pr{Y|X}
A=>D
2/5
2/3
C=>A
2/5
2/4
B&C=>D
1/5
1/3
Don’t be Fooled!
• Lift = Confidence /Expected Confidence if Independent
Checking->
Saving V
No
(1500)
Yes
(8500)
(10000)
No
500
3500
4000
Yes
1000
5000
6000
SVG=>CHKG Expect 8500/10000 = 85% if independent
Observed Confidence is 5000/6000 = 83%
Lift = 83/85 < 1.
Savings account holders actually LESS likely than others to
have checking account !!!
Summary
• Data mining – a set of fast stat methods for
large data sets
• Some new ideas, many old or extensions of old
• Some methods:
– Decision Trees
– Nearest Neighbor
– Neural Nets
– Clustering
– Association