For written notes on this lecture, please read chapter 3 of The Practical Bioinformatician,
http://www.wspc.com/books/lifesci/5547.html
A Practical Introduction
to Bioinformatics
Limsoon Wong
Institute for Infocomm Research
Lecture 1, May 2004
Copyright 2004 limsoon wong
Course Plan
• How to do experiment and feature
generation
• DNA feature recognition
• Microarray analysis
• Sequence homology interpretation
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Essence of Bioinformatics
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What is Bioinformatics?
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Themes of Bioinformatics
Bioinformatics =
Data Mgmt +
Knowledge Discovery +
Sequence Analysis +
Physical Modelling + ….
Knowledge Discovery =
Statistics + Algorithms + Databases
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Benefits of Bioinformatics
To the patient:
Better drug, better treatment
To the pharma:
Save time, save cost, make more $
To the scientist:
Better science
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Essence of Knowledge Discovery
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What is Knowledge Discovery?
Jonathan’s blocks
Jessica’s blocks
Whose block
is this?
Jonathan’s rules
Jessica’s rules
: Blue or Circle
: All the rest
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What is Knowledge Discovery?
Question: Can you explain how?
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The Steps of Knowledge Discovery
• Training data gathering
• Feature generation
 k-grams, colour, texture, domain know-how, ...
• Feature selection
 Entropy, 2, CFS, t-test, domain know-how...
• Feature integration
 SVM, ANN, PCL, CART, C4.5, kNN, ...
Some
classifier/
methods
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What is Accuracy?
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What is Accuracy?
No. of correct predictions
Accuracy =
No. of predictions
TP + TN
=
TP + TN + FP + FN
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Examples (Balanced Population)
classifier
A
B
C
D
•
•
•
•
TP
25
50
25
37
TN
25
25
50
37
FP
25
25
0
13
FN Accuracy
25
50%
0
75%
25
75%
13
74%
Clearly, B, C, D are all better than A
Is B better than C, D?
Is C better than B, D?
Accuracy may not
Is D better than B, C?
tell the whole story
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Examples (Unbalanced Population)
classifier
A
B
C
D
TP
25
0
50
30
TN
75
150
0
100
FP
75
0
150
50
FN Accuracy
25
50%
50
75%
0
25%
20
65%
• Clearly, D is better than A
• Is B better than A, C, D?
high accuracy is meaningless if population is unbalanced
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What is Sensitivity (aka Recall)?
No. of correct positive predictions
Sensitivity =
wrt positives
No. of positives
TP
=
TP + FN
Sometimes sensitivity wrt negatives is termed specificity
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What is Precision?
No. of correct positive predictions
Precision =
wrt positives
No. of positives predictions
TP
=
TP + FP
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Abstract Model of a Classifier
•
•
•
•
Given a test sample S
Compute scores p(S), n(S)
Predict S as negative if p(S) < t * n(s)
Predict S as positive if p(S)  t * n(s)
t is the decision threshold of the classifier
changing t affects the recall and precision,
and hence accuracy, of the classifier
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Precision-Recall Trade-off
• A predicts better than
B if A has better
recall and precision
than B
• There is a trade-off
between recall and
precision
precision
• In some applications,
once you reach a
satisfactory
precision, you
optimize for recall
• In some applications,
once you reach a
satisfactory recall,
you optimize for
precision
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Comparing Prediction Performance
• Accuracy is the obvious measure
– But it conveys the right intuition only when
the positive and negative populations are
roughly equal in size
• Recall and precision together form a
better measure
– But what do you do when A has better recall
than B and B has better precision than A?
So let us look at some alternate measures ….
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F-Measure (Used By Info Extraction Folks)
• Take the harmonic mean of recall and
precision
F=
classifier
A
B
C
D
2 * recall * precision
recall + precision
TP TN FP
25 75 75
0 150
0
50
0 150
30 100 50
(wrt positives)
FN Accuracy F-measure
25
50%
33%
50
75% undefined
0
25%
40%
20
65%
46%
Does not accord with intuition
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Adjusted Accuracy
• Weigh by the importance of the classes
Adjusted accuracy =  * Sensitivity +
 * Specificity
where  +  = 1
typically,  =  = 0.5
classifier
A
B
C
D
TP TN FP FN Accuracy Adj Accuracy
25 75 75 25
50%
50%
0 150
0 50
75%
50%
50
0 150 0
25%
50%
30 100 50 20
65%
63%
But people can’t always agree on values for , 
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ROC Curves
• By changing t, we get a
range of sensitivities
and specificities of a
classifier
• A predicts better than B
if A has better
sensitivities than B at
most specificities
• Leads to ROC curve
that plots sensitivity vs.
(1 – specificity)
• Then the larger the
area under the ROC
curve, the better
1 – specificity
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What is Cross Validation?
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Construction of a Classifier
Training
samples
Build Classifier
Classifier
Test
instance
Apply Classifier
Prediction
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Estimate Accuracy: Wrong Way
Training
samples
Build
Classifier
Classifier
Apply
Classifier
Predictions
Estimate
Accuracy
Accuracy
Why is this way of estimating accuracy wrong?
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Recall ...
…the abstract model of a classifier
•
•
•
•
Given a test sample S
Compute scores p(S), n(S)
Predict S as negative if p(S) < t * n(s)
Predict S as positive if p(S)  t * n(s)
t is the decision threshold of the classifier
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K-Nearest Neighbour Classifier (k-NN)
• Given a sample S, find the k observations
Si in the known data that are “closest” to it,
and average their responses.
• Assume S is well approximated by its
neighbours
n(S) =  1
p(S) =  1
Si Nk(S)  DP
Si Nk(S)  DN
where Nk(S) is the neighbourhood of S
defined by the k nearest samples to it.
Assume distance between samples is Euclidean distance for now
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Estimate Accuracy: Wrong Way
Training
samples
Build
1-NN
1-NN
Apply
1-NN
Predictions
Estimate
Accuracy
100%
Accuracy
For sure k-NN (k = 1) has 100% accuracy in the
“accuracy estimation” procedure above. But does
this accuracy generalize to new test instances?
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Estimate Accuracy: Right Way
Training
samples
Build
Classifier
Classifier
Testing
samples
Apply
Classifier
Predictions
Estimate
Accuracy
Accuracy
Testing samples are NOT to be used
during “Build Classifier”
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How Many Training and Testing
Samples?
• No fixed ratio between training and
testing samples; but typically 2:1 ratio
• Proportion of instances of different
classes in testing samples should be
similar to proportion in training samples
• What if there are insufficient samples to
reserve 1/3 for testing?
• Ans: Cross validation
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Cross Validation
1.Test 2.Train 3.Train 4.Train 5.Train • Divide samples
1.Train 2.Test
1.Train 2.Train
1.Train 2.Train
1.Train 2.Train
into k roughly
equal parts
3.Train 4.Train 5.Train
• Each part has
3.Test 4.Train 5.Train
similar proportion
of samples from
3.Train 4.Test 5.Train
different classes
3.Train 4.Train 5.Test • Use each part to
testing other parts
• Total up accuracy
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How Many Fold?
Accuracy
• If samples are
divided into k parts,
we call this k-fold
cross validation
• Choose k so that
– the k-fold cross
validation accuracy
does not change
much from k-1 fold
– each part within the kfold cross validation
has similar accuracy
• k = 5 or 10 are
popular choices for k.
Size of training set
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Bias-Variance Trade-Off
• In k-fold cross
validation,
Accuracy
– small k tends to under
estimate accuracy
(i.e., large bias
downwards)
– Large k has smaller
bias, but can have
high variance
Size of training set
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Curse of Dimensionality
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Recall ...
…the abstract model of a classifier
•
•
•
•
Given a test sample S
Compute scores p(S), n(S)
Predict S as negative if p(S) < t * n(s)
Predict S as positive if p(S)  t * n(s)
t is the decision threshold of the classifier
Copyright 2004 limsoon wong
K-Nearest Neighbour Classifier (k-NN)
• Given a sample S, find the k observations
Si in the known data that are “closest” to it,
and average their responses.
• Assume S is well approximated by its
neighbours
n(S) =  1
p(S) =  1
Si Nk(S)  DP
Si Nk(S)  DN
where Nk(S) is the neighbourhood of S
defined by the k nearest samples to it.
Assume distance between samples is Euclidean distance for now
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Curse of Dimensionality
• How much of each
dimension is needed
to cover a proportion
r of total sample
space?
• Calculate by ep(r) = r1/p
• So, to cover 1% of a
15-D space, need 85%
of each dimension!
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
r=0.01
r=0.1
p=3
p=6
p=9
p=12
p=15
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Consequence of the Curse
• Suppose the number of samples given to
us in the total sample space is fixed
• Let the dimension increase
• Then the distance of the k nearest
neighbours of any point increases
• Then the k nearest neighbours are less
and less useful for prediction, and can
confuse the k-NN classifier
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What is Feature Selection?
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Tackling the Curse
• Given a sample space of p dimensions
• It is possible that some dimensions are
irrelevant
• Need to find ways to separate those
dimensions (aka features) that are
relevant (aka signals) from those that are
irrelevant (aka noise)
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Signal Selection (Basic Idea)
• Choose a feature w/ low intra-class distance
• Choose a feature w/ high inter-class distance
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Signal Selection (eg., t-statistics)
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Self-fulfilling Oracle
• Construct artificial
dataset with 100
samples, each with
100,000 randomly
generated features
and randomly
assigned class labels
• select 20 features with
the best t-statistics (or
other methods)
• Evaluate accuracy by
cross validation using
only the 20 selected
features
• The resultant estimated
accuracy can be ~90%
• But the true accuracy
should be 50%, as the
data were derived
randomly
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What Went Wrong?
• The 20 features were selected from the
whole dataset
• Information in the held-out testing
samples has thus been “leaked” to the
training process
• The correct way is to re-select the 20
features at each fold; better still, use a
totally new set of samples for testing
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A Popular classifier: C4.5
Based on the slides of Ozge Guler
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What is C4.5?
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What is C4.5?
• C4.5 was introduced by Quinlan for inducing
classification models (viz. decision trees) from
data
• C4.5 builds a decision tree using a top-down
induction of decision trees approach, recursively
partitioning the data into smaller subsets, based
on the value of an attribute
• At each step in the construction of the decision
tree, C4.5 selects the attribute that maximizes the
information gain ratio
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Golfing Example
• Given records on weather for playing golf
• Decides to play or not to play
• The non-goal attributes are:
attributes
outlook
temperature
humidity
windy
possible values
sunny, overcast, rain
continuous
continuous
true, false
• The goal attribute is:
attribute
play
possible values
play, don't
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Traning Data
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Golfing Example
• In the decision tree each node corresponds to a
non-goal attribute and each arc to a possible
value of that attribute
• In the decision tree at each node should be
associated the non-goal attribute which is most
informative among the attributes not yet
considered in the path from the root
• Entropy is used to measure how informative is a
node
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Entropy
• If we are given a probability distribution
P = (p1, p2, .., pn)
• then the Information conveyed by this
distribution, also called the Entropy of P, is:
I(P)=-(p1*log(p1)+p2*log(p2)+..+pn*log(pn))
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Information
• If a set T of records is partitioned into disjoint
exhaustive classes C1, C2, .., Ck on basis of
value of the goal attribute, then info needed to
identify the class of an element of T is
Info(T) = I(P),
• where P is the probability distribution of the
partition (C1, C2, .., Ck):
P = (|C1|/|T|, |C2|/|T|, ..., |Ck|/|T|)
• In our golfing example,
Info(T) = I(9/14, 5/14) = 0.94
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Information of an attribute
• If we partition T on basis of value of a non-goal
attribute X into sets T1, T2, .., Tn then info
needed to identify class of an element of T is
weighted ave of info needed to identify the
class of an element of Ti,
Info(X,T) = Σ(|Ti|/|T|) * Info(Ti)
• In our golfing example, for attribute Outlook
Info(Outlook,T)=5/14*I(2/5,3/5)+
4/14*I(4/4,0)+
5/14*I(3/5,2/5)
= 0.694
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Information Gain
• Consider the quantity Gain(X,T) defined as
Gain(X,T) = Info(T) – Info(X,T)
• which is the diff betw info needed to identify an
element of T and info needed to identify an
element of T after value of attribute X has been
obtained, i.e., this is the gain in information due
to attribute X
• In our golfing example, for Outlook attribute
Gain(Outlook,T) = Info(T) – Info(Outlook,T)
=0.94 – 0.694 = 0.246
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Information Gain
• If we instead consider the attribute Windy
Info(Windy,T) = 0.892
Gain(Windy,T) = 0.94 – 0.892 = 0.048
• So Outlook offers more info gain than Windy
 Use this notion of gain to rank attributes
and to build decision trees where at each
node is located the attribute with greatest
gain among the attributes not yet
considered in the path from the root
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Decision Tree
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Using Gain Ratios
• The notion of Gain introduced earlier tends to
favor attributes that have a large number of
values.
• For example, if we have an attribute D that has
a distinct value for each record, then Info(D,T)
is 0, thus Gain(D,T) is maximal.
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Using Gain Ratios
• To compensate for this Quinlan suggested
using the following ratio:
GainRatio(D,T) = Gain(D,T)/SplitInfo(D,T)
• where SplitInfo(D,T) is info due to split of T on
basis of value of goal attribute D
SplitInfo(D,T) = I(|T1|/|T|, |T2|/|T|, .., |Tm|/|T|)
• where {T1, T2, .. Tm} is the partition of T
induced by value of D
Copyright 2004 limsoon wong
Using Gain Ratios
• In our golfing example
SplitInfo(Outlook,T)= – 5/14*log(5/14)
– 4/14*log(4/14)
– 5/14*log(5/14)
= 1.577
GainRatio(Outlook, T) = 0.246/1.577 = 0.156
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Deriving Rule Set
• It is easy to derive a rule set from a
decision tree:
– Write a rule for each path in the decision
tree from the root to a leaf.
– In that rule the left-hand side is easily built
from the label of the nodes and the labels of
the arcs.
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Notes
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References
• John A. Swets, “Measuring the accuracy of diagnostic
systems”, Science 240:1285--1293, 1988
• Trevor Hastie, Robert Tibshirani, Jerome Friedman, The
Elements of Statistical Learning: Data Mining, Inference,
and Prediction, Springer, 2001. Chapters 1, 7
• Lance D. Miller et al., “Optimal gene expression analysis
by microarrays”, Cancer Cell 2:353--361, 2002
• Ross Quinlan, C4.5: Programming for Machine Learning,
Morgan Kaufmann, 1993
• Limsoon Wong, The Practical Bioinformatician, World
Scientific, 2004
Copyright 2004 limsoon wong