Distance Measures • Remember K-Nearest Neighbor are determined on the bases of some kind of “distance” between points. • Two major classes of distance measure: 1. Euclidean : based on position of points in some k -dimensional space. 2. Noneuclidean : not related to position or space.
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Distance Measures
• Remember K-Nearest Neighbor are determined on the bases of some kind of “distance” between points.
• Two major classes of distance measure:1. Euclidean : based on position of points in some k
-dimensional space.2. Noneuclidean : not related to position or space.
Scales of Measurement
• Applying a distance measure largely depends on the type of
input data
• Major scales of measurement:1. Nominal Data (aka Nominal Scale Variables)
• Typically classification data, e.g. m/f • no ordering, e.g. it makes no sense to state that M > F • Binary variables are a special case of Nominal scale variables.
2. Ordinal Data (aka Ordinal Scale)• ordered but differences between values are not important • e.g., political parties on left to right spectrum given labels 0, 1, 2 • e.g., Likert scales, rank on a scale of 1..5 your degree of satisfaction • e.g., restaurant ratings
Scales of Measurement
• Applying a distance function largely depends on the type of
input data
• Major scales of measurement:
3. Numeric type Data (aka interval scaled)• Ordered and equal intervals. Measured on a linear scale. • Differences make sense• e.g., temperature (C,F), height, weight, age, date
Scales of Measurement• Only certain operations can be performed
on certain scales of measurement.
Nominal Scale
Ordinal Scale
Interval Scale
1. Equality2. Count
3. Rank(Cannot quantify difference)
4. Quantify the difference
Axioms of a Distance Measure• d is a distance measure if it is a function from
pairs of points to reals such that:1. d(x,x) = 0.2. d(x,y) = d(y,x).3. d(x,y) > 0.
Some Euclidean Distances• L2 norm (also common or Euclidean distance):
– The most common notion of “distance.”
• L1 norm (also Manhattan distance)
– distance if you had to travel along coordinates only.
)||...|||(|),( 22
22
2
11 pp jx
ix
jx
ix
jx
ixjid
||...||||),(2211 pp jxixjxixjxixjid
Examples L1 and L2 norms
x = (5,5)
y = (9,8)L2-norm:dist(x,y) = (42+32) = 5
L1-norm:dist(x,y) = 4+3 = 7
4
35
Another Euclidean Distance• L∞ norm : d(x,y) = the maximum of the
differences between x and y in any dimension.
Non-Euclidean Distances• Jaccard measure for binary vectors
• Cosine measure = angle between vectors from the origin to the points in question.
• Edit distance = number of inserts and deletes to change one string into another.
Jaccard Measure• A note about Binary variables first – Symmetric binary variable
• If both states are equally valuable and carry the same weight, that is, there is no preference on which outcome should be coded as 0 or 1.
• Like “gender” having the states male and female– Asymmetric binary variable:
• If the outcomes of the states are not equally important, such as the positive and negative outcomes of a disease test.
• We should code the rarest one by 1 (e.g., HIV positive), and the other by 0 (HIV negative).
– Given two asymmetric binary variables, the agreement of two 1s (a positive match) is then considered more important than that of two 0s (a negative match).
Jaccard Measure• A contingency table for binary data
• Simple matching coefficient (invariant, if the binary variable is
symmetric):
• Jaccard coefficient (noninvariant if the binary variable is
asymmetric):
dcbacb jid
),(
cbacb jid
),(
pdbcasum
dcdc
baba
sum
0
1
01
Object i
Object j
Jaccard Measure Example• Example
– All attributes are asymmetric binary– let the values Y and P be set to 1, and the value N be set to 0
cbacb jid
),(
Name Fever Cough Test-1 Test-2 Test-3 Test-4 Jack Y N P N N N Mary Y N P N P N Jim Y P N N N N
75.0211
21),(
67.0111
11),(
33.0102
10),(
maryjimd
jimjackd
maryjackd
pdbcasum
dcdc
baba
sum
0
1
01
Cosine Measure
• Think of a point as a vector from the origin (0,0,…,0) to its location.
• Two points’ vectors make an angle, whose cosine is the normalized dot-product of the vectors.– Example:– p1.p2 = 2; |p1| = |p2| = 3.– cos() = 2/3; is about 48 degrees.
p1
p2p1.p2
|p2|dist(p1, p2) = = arccos(p1.p2/|p2||p1|)
Edit Distance• The edit distance of two strings is the
number of inserts and deletes of characters needed to turn one into the other.
• Equivalently, d(x,y) = |x| + |y| -2|LCS(x,y)|.– LCS = longest common subsequence = longest
string obtained both by deleting from x and deleting from y.
Example• x = abcde ; y = bcduve.
• LCS(x,y) = bcde.• D(x,y) = |x| + |y| - 2|LCS(x,y)| = 5 + 6 –2*4 =
3.
• What left?• Normalize it in the range [0-1]. We will study
normalization formulas in next lecture.
Back to k-Nearest Neighbor (Pseudo-code)
• Missing values Imputation using k-NN.• Input: Dataset (D), size of K
• for each record (x) with at least on missing value in D.– for each data object (y) in D.
• Take the Distance (x,y)• Save the distance and y in array Similarity (S) array.
– Sort the array S in descending order– Pick the top K data objects from S
• Impute the missing attribute value (s) of x on the basic of known values of S (use Mean/Median or MOD).
K-Nearest Neighbor Drawbacks
• The major drawbacks of this approach are the– Choice of selecting exact distance functions.– Considering all attributes when attempting to
retrieve the similar type of examples.– Searching through all the dataset for finding the
same type of instances.– Algorithm Cost: ?
Noisy Data• Noise: Random error, Data Present but not correct.– Data Transmission error– Data Entry problem
• Removing noise– Data Smoothing (rounding, averaging within a window).– Clustering/merging and Detecting outliers.
• Data Smoothing– First sort the data and partition it into (equi-depth) bins.– Then the values in each bin using Smooth by Bin Means,
Smooth by Bin Median, Smooth by Bin Boundaries, etc.
Noisy Data (Binning Methods)Sorted data for price (in dollars): 4, 8, 9, 15, 21, 21, 24, 25, 26, 28, 29, 34* Partition into (equi-depth) bins: - Bin 1: 4, 8, 9, 15 - Bin 2: 21, 21, 24, 25 - Bin 3: 26, 28, 29, 34* Smoothing by bin means: - Bin 1: 9, 9, 9, 9 - Bin 2: 23, 23, 23, 23 - Bin 3: 29, 29, 29, 29* Smoothing by bin boundaries: - Bin 1: 4, 4, 4, 15 - Bin 2: 21, 21, 25, 25 - Bin 3: 26, 26, 26, 34
Noisy Data (Clustering)• Outliers may be detected by clustering, where similar values are
organized into groups or “clusters”.
• Values which falls outside of the set of clusters may be considered outliers.
Data Discretization• The task of attribute (feature)-discretization techniques is to
discretize the values of continuous features into a small number of intervals, where each interval is mapped to a discrete symbol.
• Advantages:- – Simplified data description and easy-to-understand data and final data-
mining results.– Only Small interesting rules are mined.– End-results processing time decreased.– End-results accuracy improved.
Effect of Continuous Data on Results Accuracy
age income age buys_computer<=30 medium 9 ?<=30 medium 11 ?<=30 medium 13 ?
age income age buys_computer<=30 medium 9 no<=30 medium 10 no<=30 medium 11 no<=30 medium 12 no
Data Mining
• If ‘age <= 30’ and income = ‘medium’ and age = ‘9’ then buys_computer = ‘no’
• If ‘age <= 30’ and income = ‘medium’ and age = ‘10’ then buys_computer = ‘no’
• If ‘age <= 30’ and income = ‘medium’ and age = ‘11’ then buys_computer = ‘no’
• If ‘age <= 30’ and income = ‘medium’ and age = ‘12’ then buys_computer = ‘no’
Discover only those rules which contain support (frequency) greater >= 1
Due to the missing value in training dataset, the accuracy of prediction decreases and becomes “66.7%”
Entropy-Based Discretization• Given a set of samples S, if S is partitioned into two intervals S1
and S2 using boundary T, the entropy after partitioning is
• Where pi is the probability of class i in S1, determined by dividing the number of samples of class i in S1 by the total number of samples in S1.
Example 1IDID 1 2 3 4 5 6 7 8 9
AgeAge 21 22 24 25 27 27 27 35 41
GradeGrade F F P F P P P P P
• Let Grade be the class attribute. Use entropy-based discretization to divide the range of ages into different discrete intervals.
• There are 6 possible boundaries. They are 21.5, 23, 24.5, 26, 31, and 38.
• Let us consider the boundary at T = 21.5. Let S1 = {21} Let S2 = {22, 24, 25, 27, 27, 27, 35, 41}
(21+22) / 2 = 21.5
(22+24) / 2 = 23
Example 1 (cont’)
• The number of elements in S1 and S2 are:|S1| = 1|S2| = 8
• The entropy of S1 is
• The entropy of S2 is
ID 1 2 3 4 5 6 7 8 9
Age 21 22 24 25 27 27 27 35 41
Grade F F P F P P P P P
)0(log)0()1(log)1(
)P(log)P()F(log)F()(
22
221 GradePGradePGradePGradePSEnt
)6(log)6()2(log)2(
)P(log)P()F(log)F()(
22
222 GradePGradePGradePGradePSEnt
Example 1 (cont’)
• Hence, the entropy after partitioning at T = 21.5 is
...
)(|9|
|8|)(
|9|
|1|
)(||
||)(
||
||),(
21
22
11
SEntSEnt
SEntS
SSEnt
S
STSE
Example 1 (cont’)• The entropies after partitioning for all the boundaries are:
T = 21.5 = E(S,21.5) T = 23 = E(S,23) . . T = 38 = E(S,38)
Select the boundary with the smallest entropySuppose best is T = 23ID 1 2 3 4 5 6 7 8 9
Age 21 22 24 25 27 27 27 35 41
Grade F F P F P P P P P
Now recursively apply entropy discretization upon both partitions
References– G. Batista and M. Monard, “The study of K-Nearest
Neighbor as a Imputation Method”, 2002 . (I will placed at the course folder)
– “CS345 --- Lecture Notes”, by Jeff D Ullman at Stanford. http://www-db.stanford.edu/~ullman/cs345-notes.html
– Vipin Kumar’s course in data mining offered at University of Minnesota
– official text book slides of Jiawei Han and Micheline Kamber, “Data Mining: Concepts and Techniques”, Morgan Kaufmann Publishers, August 2000.
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