KEMBAR78
Clustering Methods with R | PDF
Clustering Methods with R
Akira Murakami
Department of English Language and Applied Linguistics
University of Birmingham
a.murakami@bham.ac.uk
Cluster Analysis
• Cluster analysis finds groups in data.
• Objects in the same cluster are similar to each other.
• Objects in different clusters are dissimilar.
• A variety of algorithms have been proposed.
• Saying “I ran a cluster analysis” does not mean much.
• Used in data mining or as a statistical analysis.
• Unsupervised machine learning technique.
2
Cluster Analysis in SLA
• In SLA, clustering has been applied to identify the typology of
learners’
• motivational profiles (Csizér & Dörnyei, 2005),
• ability/aptitude profiles (Rysiewicz, 2008),
• developmental profiles based on international posture, L2
willingness to communicate, and frequency of communication
in L2 (Yashima & Zenuk-Nishide, 2008),
• cognitive and achievement profiles based on L1 achievement,
intelligence, L2 aptitude, and L2 proficiency (Sparks, Patton,
& Ganschow, 2012).
3
Similarity Measure
• Cluster analysis groups the observations that are
“similar”. But how do we measure similarity?
• Let’s suppose that we are interested in clustering L1
groups according to their accuracy of different
linguistic features (i.e., accuracy profile of L1
groups).
• As the measure of accuracy, we use an index that
takes the value between 0 and 1, such as the TLU
score.
4
│ │ │ │ │ │ │ │ │ │ │
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Mathematical Distance
5
│ │ │ │ │ │ │ │ │ │ │
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
L1 Korean
Mathematical Distance
6
│ │ │ │ │ │ │ │ │ │ │
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
L1 Korean
L1 German
Mathematical Distance
7
│ │ │ │ │ │ │ │ │ │ │
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
L1 Korean
L1 German
Distance = 0.2
Mathematical Distance
8
│ │ │ │ │ │ │ │ │ │ │
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
L1 Korean
L1 German
Distance = 0.2
L1 Japanese
Distance = 0.1
Mathematical Distance
9
(Dis)Similarity Matrix
10
L1 Korean L1 German L1 Japanese
L1 Korean 0.0
L1 German 0.2 0.0
L1 Japanese 0.1 0.3 0.0
Distance Measures
• Things are simple in 1D, but get more complicated in 2D or above.
• Different measures of distance
• Euclidean distance
• Manhattan distance
• Maximum distance
• Mahalanobis distance
• Hamming distance
• etc
11
Distance Measures
• Things are simple in 1D, but get more complicated in 2D or above.
• Different measures of distance
• Euclidean distance
• Manhattan distance
• Maximum distance
• Mahalanobis distance
• Hamming distance
• etc
12
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Euclidean Distance
13
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
Euclidean Distance
14
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
Euclidean Distance
15
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
(0.4−0.8)2
+(0.8−0.6)2
Euclidean Distance
16
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
0.45
Euclidean Distance
17
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
0.45
L1 Japanese (0.6, 0.5)
Euclidean Distance
18
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
0.45
L1 Japanese (0.6, 0.5)
0.36
0.22
Euclidean Distance
19
(Dis)Similarity Matrix
20
L1 Korean L1 German L1 Japanese
L1 Korean 0.00
L1 German 0.45 0.00
L1 Japanese 0.36 0.22 0.00
0.0 0.2 0.4 0.6 0.8 1.0
0.00.20.40.60.81.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
Plural−sAccuracy
L1 German (0.3, 0.6, 0.9)
L1 Korean (0.6, 0.9, 0.6)
L1 Japanese (0.9, 0.4, 0.5)
Euclidean Distance (3D)
21
0.0 0.2 0.4 0.6 0.8 1.0
0.00.20.40.60.81.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
Plural−sAccuracy
L1 German (0.3, 0.6, 0.9)
L1 Korean (0.6, 0.9, 0.6)
L1 Japanese (0.9, 0.4, 0.5)
0.75
0.52
0.59
Euclidean Distance (3D)
22
(Dis)Similarity Matrix
23
L1 Korean L1 German L1 Japanese
L1 Korean 0.00
L1 German 0.52 0.00
L1 Japanese 0.59 0.75 0.00
Distance Measures
• Things are simple in 1D, but get more complicated in 2D or above.
• Different measures of distance
• Euclidean distance
• Manhattan distance
• Maximum distance
• Mahalanobis distance
• Hamming distance
• etc
24
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
Manhattan Distance
25
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
Manhattan Distance
26
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
0.4
0.2
Manhattan Distance
27
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
L1 German
(0.8, 0.6)
L1 Korean
(0.4, 0.8)
0.4
0.2
Manhattan Distance
28
→ Distance = 0.4 + 0.2 = 0.6
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
(0.1, 0.4)
(0.9, 0.3)
(0.6, 0.9)
Manhattan Distance
29
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
(0.1, 0.4)
(0.9, 0.3)
(0.6, 0.9)
0.5
0.5
0.71
0.1
0.8
0.81
Manhattan Distance
30
Article Accuracy
Pasttense−edAccuracy
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
(0.1, 0.4)
(0.9, 0.3)
(0.6, 0.9)
0.5
0.5
0.71
0.1
0.8
0.81
Manhattan Distance
31
Euclidean: 0.71
Manhattan: 0.5 + 0.5 = 1.00
Euclidean: 0.81
Manhattan: 0.1 + 0.8 = 0.90
dist()
• In R, dist function is used to obtain dissimilarity
matrices.
• Practicals
32
Clustering Methods
• Now that we know the concept of similarity, we
move on to the clustering of objects based on the
similarity.
• A number of methods have been proposed for
clustering. We will look at the following two:
• agglomerative hierarchical cluster analysis
• k-means
33
Clustering Methods
• Now that we know the concept of similarity, we
move on to the clustering of objects based on the
similarity.
• A number of methods have been proposed for
clustering. We will look at the following two:
• agglomerative hierarchical cluster analysis
• k-means
34
Agglomerative Hierarchical Cluster Analysis
• In agglomerative hierarchical clustering,
observations are clustered in a bottom-up manner.
1. Each observation forms an independent cluster
at the beginning.
2. The two clusters that are most similar are
clustered together.
3. 2 is repeated until all the observations are
clustered in a single cluster.
35
Linkage Criteria
• How do we calculate the similarity between clusters
that each includes multiple observations?
• Ward’s criterion (Ward’s method)
• complete-linkage
• single-linkage
• etc.
36
Linkage Criteria
• How do we calculate the similarity between clusters
that each includes multiple observations?
• Ward’s criterion (Ward’s method)
• complete-linkage
• single-linkage
• etc.
37
Ward’s Method
• Ward’s method leads to the smallest within-cluster
variance.
• At each iteration, two clusters are merged so that it
yields the smallest increase of the sum of squared
errors.
• Sum of Squared Errors (SSE): the sum of the
squared difference between the mean of the cluster
and individual data points.
38
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
Ward’s Method
39
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
Ward’s Method
40
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
mean (0.3, 0.6)
Ward’s Method
41
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
mean (0.3, 0.6)
0.22
0.22
Ward’s Method
42
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
mean (0.3, 0.6)
0.05
0.05
Ward’s Method
43
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
mean (0.3, 0.6)
0.05
0.05
Ward’s Method
44→ 0.05 + 0.05 = 0.10
• This procedure is repeated for all of the pairs.
45
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
Ward’s Method
46
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
x
Ward’s Method
47
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
x
(0.3, 0.3)
(0.6, 0.8)
Ward’s Method
48
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
x
(0.3, 0.3)
(0.6, 0.8)
( 0.1
2
+0.1
2
)
2
= 0.02
0.2
2
= 0.04
Ward’s Method
49
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
x
(0.3, 0.3)
(0.6, 0.8)
( 0.1
2
+0.1
2
)
2
= 0.02
0.2
2
= 0.04
Ward’s Method
SSE = 0.02 + 0.02 + 0.04 + 0.04 = 0.12
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x (0.45, 0.55)
Ward’s Method
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x (0.45, 0.55)
0.12
0.08
0.06
0.18
Ward’s Method
SSE = 0.12 + 0.08 + 0.06 + 0.18 = 0.46
ΔSSE
• SSE before the merger: 0.12
• SSE after the merger: 0.46
• Difference (ΔSSE): 0.46 - 0.12 = 0.34
53
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
x
x
Ward’s Method
54
Dendrogram
55
1
2
5
3
4
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Cluster Dendrogram
hclust (*, "ward.D2")
dd.dist
Height
56
Practicals
Linkage Criteria
• How do we know the similarity between clusters
that each includes multiple observations?
• Ward’s criterion (Ward’s method)
• complete-linkage
• single-linkage
• etc.
57
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
Complete Linkage
58
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
Complete Linkage
59
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
0.7
Complete Linkage
60
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
0.4
Single Linkage
61
Potential Pitfall of Hierarchical Clustering
• It assumes hierarchical structure in the clustering.
• Let us say that our data included two L1 groups over three
proficiency levels.
• If we group the data into two clusters, the best split may be
between the two L1 groups.
• If we group them into three clusters, the best groups may be by
proficiency groups.
• In this case, three-cluster solution is not nested within two-
cluster solution, and hierarchical clustering may fail to identify
the two clusters.
62
63
k-means Clustering
k-means Clustering
• K-means clustering does not assume a hierarchical
structure of clusters.
• i.e., no parent/child clusters
• Analysts need to specify the number of clusters.
64
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1 (0.4, 0.2)
2 (0.2, 0.4)
3 (0.4, 0.8) 4 (0.8, 0.8)
5 (0.9, 0.4)
k-means Clustering
65
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
x
x
1
2
3 4
5
(Centroid 1)
(Centroid 2)
k-means Clustering
66
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
x
x
1
2
3 4
5
(Centroid 1)
(Centroid 2)
0.28
0.60
0.45 0.72
0.72
0.64
0.70
k-means Clustering
67
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1
2
3 4
5
x
x
Centroid 1
Centroid 2
k-means Clustering
68
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1
2
3 4
5
x
x
Centroid 1
Centroid 2
0.40
0.41
0.50
0.22
0.45
0.22
0.28
0.42
0.21
0.63
k-means Clustering
69
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
Article Accuracy
Pasttense−edAccuracy
1
2
3 4
5
x
x
Centroid 1
Centroid 2
k-means Clustering
70
k-Means Clustering
• The optimal number of clusters depends on the intended use.
• There is no “correct” or “wrong” choice in the number of
clusters.
• NP hard
• The algorithm only approximates solutions.
• Randomness is involved in the solution. You get different
solutions every time you run it.
• It assumes convex clusters.
71
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
y1 Concave
72
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
y1
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Concave
73
74
Practicals
Within-Learner Centering
• The mean accuracy value of each learner was subtracted from all the
data points of the learner.
• For example, let's suppose the mean sentence length (MSL) of
Learner A over 10 writings was
• {4.0, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8} 



and that of Learner B was
• {8.0, 8.2, 8.4, 8.6, 8.8, 9.0, 9.2, 9.4, 9.6, 9.8}
• The difference in MSL is identical in the two learners (+0.2 per writing).
• But the absolute MSL is widely different.
75
Within-Learner Centering
• The mean value of Learner A (4.9) is subtracted from all the data
points of Learner A:
• → {-0.90, -0.70, -0.50, -0.30, -0.10, 0.10, 0.30, 0.50, 0.70,
0.90}.
• Similarly, the mean value of Learner B (8.90) is subtracted from
all the data points of Learner B:
• → {-0.90, -0.70, -0.50, -0.30, -0.10, 0.10, 0.30, 0.50, 0.70,
0.90}.
• It is guaranteed that these two learners are clustered into the
same group as they have exactly the same set of values.
76
77
Cluster Validation
Cluster Validation/Evaluation
• We got clusters and explored them, but how do we
know how good the clusters are, or whether they
indeed capture signal and not just noise?
• Are the clusters ‘real’?
• Is it the difference in the true learning curve that
the earlier clustering captured or is it just the
random noise?
78
Two Types of Validation
• External Validation
• Internal Validation
79
External Validation
• If there is a a systematic pattern between clusters
and some external criteria, such as the proficiency
or L1 of learners, then what the cluster analysis
captured is unlikely to be just noise.
80
Internal Validation
• Measures of goodness of clusters
• silhouette width
• Davies–Bouldin index
• Dunn index
• etc.
81
Internal Validation
• Measures of goodness of clusters
• silhouette width
• Davies–Bouldin index
• Dunn index
• etc.
82
Silhouette Width
• Intuitively, the silhouette value is large if within-
cluster dissimilarity is small (i.e., learners within
each cluster have similar developmental
trajectories) and between-cluster dissimilarity is
large (i.e., learners in different clusters have
different learning curves).
• The silhouette is given to each data point (i.e.,
learner), and all the silhouette values are averaged
to measure the cluster distinctiveness of a cluster
analysis.
83
• Let’s say there are three clusters, A through C.
• Let’s further say that i is a member of Cluster A.
• Let a(i) be the average distance between that learner and all the
other learners that belong to the same cluster.
• We also calculate the average distances
1. between the learner and all the other learners that belong to
Cluster B
2. between the learner and all the other learners that belong to
Cluster C
• Let b(i) be the smaller of the two above (1-2).
• s(i) = (b(i) - a(i)) / max(a(i), b(i))
84
Silhouette Width
0.0 0.2 0.4 0.6 0.8 1.0
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Silhouette Width
85
0.0 0.2 0.4 0.6 0.8 1.0
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Silhouette Width
86
0.0 0.2 0.4 0.6 0.8 1.0
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Silhouette Width
87
→ Average = 0.022 (the value of a(i))
0.0 0.2 0.4 0.6 0.8 1.0
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Silhouette Width
88
→ Average = 0.191
0.0 0.2 0.4 0.6 0.8 1.0
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Silhouette Width
89
→ Average = 0.240
Silhouette Width
• a(i) = 0.022
• b(i) = 0.191 (the smaller of the other two)
• s(i) = (b(i) - a(i)) / max(a(i), b(i))
• s(i) = (0.191 - 0.022) / 0.191 = 0.882
• This is repeated for all the data points.
• Goodness of clustering: mean silhouette width across
all the data points.
90
Bootstrapping
• Now that we have a measure of how good our
clustering is, the next question is whether it is good
enough to be considered non-random.
• We can address this question through the technique
called bootstrapping.
• The idea is similar to the usual hypothesis-testing
procedure.
• We obtain the null distribution of the silhouette value
and see where our value falls.
91
• More specific procedure is as follows:
1. For each learner, we sample 30 writings (with replacement).
2. We run a k-means cluster analysis with the data obtained in
1 and calculate the mean silhouette value.
3. 1 and 2 are repeated e.g., 10,000 times, resulting in 10,000
mean silhouette values which we consider as the null
distribution.
4. We examine whether the 95% range of 3 includes our
observed mean silhouette value.
92
Bootstrapping
• The idea here is that we practically randomize the order
of the writings within individual learners and follow the
same procedure as our main analysis.
• Since the order of writings is random, there should not
be any systematic pattern of development observed.
• The clusters obtained in this manner thus captures noise
alone. We calculate the mean silhouette value on the
noise-only, random clusters, and obtain its distribution by
repeating the whole procedure a large number of times.
93
Bootstrapping
94
langtest.jp
langtest.jp
95
http://langtest.jp
Paper Introducing langtest.jp
96
http://applij.oxfordjournals.org/content/early/2015/06/24/applin.amv025.abstract
langtest.jp
97
98
Demo

Clustering Methods with R

  • 1.
    Clustering Methods withR Akira Murakami Department of English Language and Applied Linguistics University of Birmingham a.murakami@bham.ac.uk
  • 2.
    Cluster Analysis • Clusteranalysis finds groups in data. • Objects in the same cluster are similar to each other. • Objects in different clusters are dissimilar. • A variety of algorithms have been proposed. • Saying “I ran a cluster analysis” does not mean much. • Used in data mining or as a statistical analysis. • Unsupervised machine learning technique. 2
  • 3.
    Cluster Analysis inSLA • In SLA, clustering has been applied to identify the typology of learners’ • motivational profiles (Csizér & Dörnyei, 2005), • ability/aptitude profiles (Rysiewicz, 2008), • developmental profiles based on international posture, L2 willingness to communicate, and frequency of communication in L2 (Yashima & Zenuk-Nishide, 2008), • cognitive and achievement profiles based on L1 achievement, intelligence, L2 aptitude, and L2 proficiency (Sparks, Patton, & Ganschow, 2012). 3
  • 4.
    Similarity Measure • Clusteranalysis groups the observations that are “similar”. But how do we measure similarity? • Let’s suppose that we are interested in clustering L1 groups according to their accuracy of different linguistic features (i.e., accuracy profile of L1 groups). • As the measure of accuracy, we use an index that takes the value between 0 and 1, such as the TLU score. 4
  • 5.
    │ │ ││ │ │ │ │ │ │ │ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Mathematical Distance 5
  • 6.
    │ │ ││ │ │ │ │ │ │ │ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 L1 Korean Mathematical Distance 6
  • 7.
    │ │ ││ │ │ │ │ │ │ │ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 L1 Korean L1 German Mathematical Distance 7
  • 8.
    │ │ ││ │ │ │ │ │ │ │ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 L1 Korean L1 German Distance = 0.2 Mathematical Distance 8
  • 9.
    │ │ ││ │ │ │ │ │ │ │ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 L1 Korean L1 German Distance = 0.2 L1 Japanese Distance = 0.1 Mathematical Distance 9
  • 10.
    (Dis)Similarity Matrix 10 L1 KoreanL1 German L1 Japanese L1 Korean 0.0 L1 German 0.2 0.0 L1 Japanese 0.1 0.3 0.0
  • 11.
    Distance Measures • Thingsare simple in 1D, but get more complicated in 2D or above. • Different measures of distance • Euclidean distance • Manhattan distance • Maximum distance • Mahalanobis distance • Hamming distance • etc 11
  • 12.
    Distance Measures • Thingsare simple in 1D, but get more complicated in 2D or above. • Different measures of distance • Euclidean distance • Manhattan distance • Maximum distance • Mahalanobis distance • Hamming distance • etc 12
  • 13.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Euclidean Distance 13
  • 14.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) Euclidean Distance 14
  • 15.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) Euclidean Distance 15
  • 16.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) (0.4−0.8)2 +(0.8−0.6)2 Euclidean Distance 16
  • 17.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) 0.45 Euclidean Distance 17
  • 18.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) 0.45 L1 Japanese (0.6, 0.5) Euclidean Distance 18
  • 19.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) 0.45 L1 Japanese (0.6, 0.5) 0.36 0.22 Euclidean Distance 19
  • 20.
    (Dis)Similarity Matrix 20 L1 KoreanL1 German L1 Japanese L1 Korean 0.00 L1 German 0.45 0.00 L1 Japanese 0.36 0.22 0.00
  • 21.
    0.0 0.2 0.40.6 0.8 1.0 0.00.20.40.60.81.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy Plural−sAccuracy L1 German (0.3, 0.6, 0.9) L1 Korean (0.6, 0.9, 0.6) L1 Japanese (0.9, 0.4, 0.5) Euclidean Distance (3D) 21
  • 22.
    0.0 0.2 0.40.6 0.8 1.0 0.00.20.40.60.81.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy Plural−sAccuracy L1 German (0.3, 0.6, 0.9) L1 Korean (0.6, 0.9, 0.6) L1 Japanese (0.9, 0.4, 0.5) 0.75 0.52 0.59 Euclidean Distance (3D) 22
  • 23.
    (Dis)Similarity Matrix 23 L1 KoreanL1 German L1 Japanese L1 Korean 0.00 L1 German 0.52 0.00 L1 Japanese 0.59 0.75 0.00
  • 24.
    Distance Measures • Thingsare simple in 1D, but get more complicated in 2D or above. • Different measures of distance • Euclidean distance • Manhattan distance • Maximum distance • Mahalanobis distance • Hamming distance • etc 24
  • 25.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) Manhattan Distance 25
  • 26.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) Manhattan Distance 26
  • 27.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) 0.4 0.2 Manhattan Distance 27
  • 28.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 L1 German (0.8, 0.6) L1 Korean (0.4, 0.8) 0.4 0.2 Manhattan Distance 28 → Distance = 0.4 + 0.2 = 0.6
  • 29.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 (0.1, 0.4) (0.9, 0.3) (0.6, 0.9) Manhattan Distance 29
  • 30.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 (0.1, 0.4) (0.9, 0.3) (0.6, 0.9) 0.5 0.5 0.71 0.1 0.8 0.81 Manhattan Distance 30
  • 31.
    Article Accuracy Pasttense−edAccuracy 0.0 0.20.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 (0.1, 0.4) (0.9, 0.3) (0.6, 0.9) 0.5 0.5 0.71 0.1 0.8 0.81 Manhattan Distance 31 Euclidean: 0.71 Manhattan: 0.5 + 0.5 = 1.00 Euclidean: 0.81 Manhattan: 0.1 + 0.8 = 0.90
  • 32.
    dist() • In R,dist function is used to obtain dissimilarity matrices. • Practicals 32
  • 33.
    Clustering Methods • Nowthat we know the concept of similarity, we move on to the clustering of objects based on the similarity. • A number of methods have been proposed for clustering. We will look at the following two: • agglomerative hierarchical cluster analysis • k-means 33
  • 34.
    Clustering Methods • Nowthat we know the concept of similarity, we move on to the clustering of objects based on the similarity. • A number of methods have been proposed for clustering. We will look at the following two: • agglomerative hierarchical cluster analysis • k-means 34
  • 35.
    Agglomerative Hierarchical ClusterAnalysis • In agglomerative hierarchical clustering, observations are clustered in a bottom-up manner. 1. Each observation forms an independent cluster at the beginning. 2. The two clusters that are most similar are clustered together. 3. 2 is repeated until all the observations are clustered in a single cluster. 35
  • 36.
    Linkage Criteria • Howdo we calculate the similarity between clusters that each includes multiple observations? • Ward’s criterion (Ward’s method) • complete-linkage • single-linkage • etc. 36
  • 37.
    Linkage Criteria • Howdo we calculate the similarity between clusters that each includes multiple observations? • Ward’s criterion (Ward’s method) • complete-linkage • single-linkage • etc. 37
  • 38.
    Ward’s Method • Ward’smethod leads to the smallest within-cluster variance. • At each iteration, two clusters are merged so that it yields the smallest increase of the sum of squared errors. • Sum of Squared Errors (SSE): the sum of the squared difference between the mean of the cluster and individual data points. 38
  • 39.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) Ward’s Method 39
  • 40.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) Ward’s Method 40
  • 41.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x mean (0.3, 0.6) Ward’s Method 41
  • 42.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x mean (0.3, 0.6) 0.22 0.22 Ward’s Method 42
  • 43.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x mean (0.3, 0.6) 0.05 0.05 Ward’s Method 43
  • 44.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x mean (0.3, 0.6) 0.05 0.05 Ward’s Method 44→ 0.05 + 0.05 = 0.10
  • 45.
    • This procedureis repeated for all of the pairs. 45
  • 46.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x Ward’s Method 46
  • 47.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x x Ward’s Method 47
  • 48.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x x (0.3, 0.3) (0.6, 0.8) Ward’s Method 48
  • 49.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x x (0.3, 0.3) (0.6, 0.8) ( 0.1 2 +0.1 2 ) 2 = 0.02 0.2 2 = 0.04 Ward’s Method 49
  • 50.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x x (0.3, 0.3) (0.6, 0.8) ( 0.1 2 +0.1 2 ) 2 = 0.02 0.2 2 = 0.04 Ward’s Method SSE = 0.02 + 0.02 + 0.04 + 0.04 = 0.12
  • 51.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x (0.45, 0.55) Ward’s Method
  • 52.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x (0.45, 0.55) 0.12 0.08 0.06 0.18 Ward’s Method SSE = 0.12 + 0.08 + 0.06 + 0.18 = 0.46
  • 53.
    ΔSSE • SSE beforethe merger: 0.12 • SSE after the merger: 0.46 • Difference (ΔSSE): 0.46 - 0.12 = 0.34 53
  • 54.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) x x Ward’s Method 54
  • 55.
  • 56.
  • 57.
    Linkage Criteria • Howdo we know the similarity between clusters that each includes multiple observations? • Ward’s criterion (Ward’s method) • complete-linkage • single-linkage • etc. 57
  • 58.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) Complete Linkage 58
  • 59.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) Complete Linkage 59
  • 60.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) 0.7 Complete Linkage 60
  • 61.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) 0.4 Single Linkage 61
  • 62.
    Potential Pitfall ofHierarchical Clustering • It assumes hierarchical structure in the clustering. • Let us say that our data included two L1 groups over three proficiency levels. • If we group the data into two clusters, the best split may be between the two L1 groups. • If we group them into three clusters, the best groups may be by proficiency groups. • In this case, three-cluster solution is not nested within two- cluster solution, and hierarchical clustering may fail to identify the two clusters. 62
  • 63.
  • 64.
    k-means Clustering • K-meansclustering does not assume a hierarchical structure of clusters. • i.e., no parent/child clusters • Analysts need to specify the number of clusters. 64
  • 65.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 (0.4, 0.2) 2 (0.2, 0.4) 3 (0.4, 0.8) 4 (0.8, 0.8) 5 (0.9, 0.4) k-means Clustering 65
  • 66.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy x x 1 2 3 4 5 (Centroid 1) (Centroid 2) k-means Clustering 66
  • 67.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy x x 1 2 3 4 5 (Centroid 1) (Centroid 2) 0.28 0.60 0.45 0.72 0.72 0.64 0.70 k-means Clustering 67
  • 68.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 2 3 4 5 x x Centroid 1 Centroid 2 k-means Clustering 68
  • 69.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 2 3 4 5 x x Centroid 1 Centroid 2 0.40 0.41 0.50 0.22 0.45 0.22 0.28 0.42 0.21 0.63 k-means Clustering 69
  • 70.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Article Accuracy Pasttense−edAccuracy 1 2 3 4 5 x x Centroid 1 Centroid 2 k-means Clustering 70
  • 71.
    k-Means Clustering • Theoptimal number of clusters depends on the intended use. • There is no “correct” or “wrong” choice in the number of clusters. • NP hard • The algorithm only approximates solutions. • Randomness is involved in the solution. You get different solutions every time you run it. • It assumes convex clusters. 71
  • 72.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 y1 Concave 72
  • 73.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 y1 x x xx x x x x x xx x x x x x x x x xx x xx x x x xx x x xx x x x xx x x x x x x x x x x xx x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x xx x xx x x x x x x x x x x x x x x x x x x xx x x xx x x x x x x x x x x x x x x x x x x xx x xx x x x x x x x x x x x x x x x x x x x x x x x xx x x x x xx x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Concave 73
  • 74.
  • 75.
    Within-Learner Centering • Themean accuracy value of each learner was subtracted from all the data points of the learner. • For example, let's suppose the mean sentence length (MSL) of Learner A over 10 writings was • {4.0, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8} 
 
 and that of Learner B was • {8.0, 8.2, 8.4, 8.6, 8.8, 9.0, 9.2, 9.4, 9.6, 9.8} • The difference in MSL is identical in the two learners (+0.2 per writing). • But the absolute MSL is widely different. 75
  • 76.
    Within-Learner Centering • Themean value of Learner A (4.9) is subtracted from all the data points of Learner A: • → {-0.90, -0.70, -0.50, -0.30, -0.10, 0.10, 0.30, 0.50, 0.70, 0.90}. • Similarly, the mean value of Learner B (8.90) is subtracted from all the data points of Learner B: • → {-0.90, -0.70, -0.50, -0.30, -0.10, 0.10, 0.30, 0.50, 0.70, 0.90}. • It is guaranteed that these two learners are clustered into the same group as they have exactly the same set of values. 76
  • 77.
  • 78.
    Cluster Validation/Evaluation • Wegot clusters and explored them, but how do we know how good the clusters are, or whether they indeed capture signal and not just noise? • Are the clusters ‘real’? • Is it the difference in the true learning curve that the earlier clustering captured or is it just the random noise? 78
  • 79.
    Two Types ofValidation • External Validation • Internal Validation 79
  • 80.
    External Validation • Ifthere is a a systematic pattern between clusters and some external criteria, such as the proficiency or L1 of learners, then what the cluster analysis captured is unlikely to be just noise. 80
  • 81.
    Internal Validation • Measuresof goodness of clusters • silhouette width • Davies–Bouldin index • Dunn index • etc. 81
  • 82.
    Internal Validation • Measuresof goodness of clusters • silhouette width • Davies–Bouldin index • Dunn index • etc. 82
  • 83.
    Silhouette Width • Intuitively,the silhouette value is large if within- cluster dissimilarity is small (i.e., learners within each cluster have similar developmental trajectories) and between-cluster dissimilarity is large (i.e., learners in different clusters have different learning curves). • The silhouette is given to each data point (i.e., learner), and all the silhouette values are averaged to measure the cluster distinctiveness of a cluster analysis. 83
  • 84.
    • Let’s saythere are three clusters, A through C. • Let’s further say that i is a member of Cluster A. • Let a(i) be the average distance between that learner and all the other learners that belong to the same cluster. • We also calculate the average distances 1. between the learner and all the other learners that belong to Cluster B 2. between the learner and all the other learners that belong to Cluster C • Let b(i) be the smaller of the two above (1-2). • s(i) = (b(i) - a(i)) / max(a(i), b(i)) 84 Silhouette Width
  • 85.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 y1 x x x x x x x x x x x x x x x x x x x x Silhouette Width 85
  • 86.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 y1 x x x x x x x x x x x x x x x x x x x x Silhouette Width 86
  • 87.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 y1 x x x x x x x x x x x x x x x x x x x x Silhouette Width 87 → Average = 0.022 (the value of a(i))
  • 88.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 y1 x x x x x x x x x x x x x x x x x x x x Silhouette Width 88 → Average = 0.191
  • 89.
    0.0 0.2 0.40.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 y1 x x x x x x x x x x x x x x x x x x x x Silhouette Width 89 → Average = 0.240
  • 90.
    Silhouette Width • a(i)= 0.022 • b(i) = 0.191 (the smaller of the other two) • s(i) = (b(i) - a(i)) / max(a(i), b(i)) • s(i) = (0.191 - 0.022) / 0.191 = 0.882 • This is repeated for all the data points. • Goodness of clustering: mean silhouette width across all the data points. 90
  • 91.
    Bootstrapping • Now thatwe have a measure of how good our clustering is, the next question is whether it is good enough to be considered non-random. • We can address this question through the technique called bootstrapping. • The idea is similar to the usual hypothesis-testing procedure. • We obtain the null distribution of the silhouette value and see where our value falls. 91
  • 92.
    • More specificprocedure is as follows: 1. For each learner, we sample 30 writings (with replacement). 2. We run a k-means cluster analysis with the data obtained in 1 and calculate the mean silhouette value. 3. 1 and 2 are repeated e.g., 10,000 times, resulting in 10,000 mean silhouette values which we consider as the null distribution. 4. We examine whether the 95% range of 3 includes our observed mean silhouette value. 92 Bootstrapping
  • 93.
    • The ideahere is that we practically randomize the order of the writings within individual learners and follow the same procedure as our main analysis. • Since the order of writings is random, there should not be any systematic pattern of development observed. • The clusters obtained in this manner thus captures noise alone. We calculate the mean silhouette value on the noise-only, random clusters, and obtain its distribution by repeating the whole procedure a large number of times. 93 Bootstrapping
  • 94.
  • 95.
  • 96.
  • 97.
  • 98.