STAT 503 Case Study: Clustering of music clips
1
Description
This data was collected by Dr Cook from her own CDs. Using a Mac she read the track into the music
editing software Amadeus II, snipped and saved the first 40 seconds as a WAV file. (WAV is an audio format
developed by Microsoft, commonly used on Windows but it is getting less popular.) These files were read
into R using the package tuneR. This converts the audio file into numeric data. All of the CDs contained left
and right channels, and variables were calculated on both channels. The resulting data has 62 rows (cases)
and 7 columns (variables).
• LVar, LAve, LMax: average, variance, maximum of the frequencies of the left channel.
• LFEner: an indicator of the amplitude or loudness of the sound.
• LFreq: Median of the location of the 15 highest peak in the periodogram.
There are 11 tracks by Abba, 11 from the Beatles and 10 the Eels, which would be considered to be
Rock, and 13 tracks by Vivaldi, 6 of Mozart and 8 of Beethoven, considered to be Classical. There are also
3 tracks from Enya, considered to be New Wave. The main question we want to answer is:
“Can we group the tracks into a small number of clusters according to their similarity on audio charactieristics?”
This information might be used to arrange tracks on a digital music player. Other questions of interest might
be:
• Do the rock tracks have different characteristics than classical tracks?
• How does Enya compare to rock and classical tracks?
• Are there differences between the tracks of different artists?
1
2
Plan for Analysis
Approach
Summary
statistics
(marginal and conditional)
Reason
extract location/scale information
Plots
explore data distributions
Numerical clustering
Grouping the tracks into clusters of
similar audio attributes. Use hierarchical, k-means, model-based and
self-organizing maps.
2
Type of questions addressed
“How are rock tracks different on average than classical tracks?” “What
is the average LAve for Abba relative
to other Artists?”
“Are there unusual tracks?”, “Is
there any obvious clustering of the
tracks?”
“Which tracks might be considered
alike?”
3
3.1
Results
Summary Statistics
LVar
1.99×107
2.64×107
Mean
SD
LAve
-7.81
47.22
LMax
2.25×104
8.76×103
LFEner
104.03
5.48
LFreq
231.39
176.69
Table 1: Overall means and standard deviations of the variables.
Artist
Abba
Beatles
Eels
LVar
8.52×106
4.45×107
5.11×107
LAve
-81.5
-5.99
4.59
LMax
2.35×104
2.76×104
3.13×104
LFEner
103
108
108
LFreq
135
147
181
Beethoven
Mozart
Vivaldi
7.61×106
4.69×106
3.00×106
-0.74
-5.94
39.1
2.11×104
1.89×104
1.45×104
101
101
102
350
396
305
Enya
5.03×107
-11.8
1.61×104
103
95
Table 2: Means of the variables by artist.
The classical tracks in general have lower LVar than rock tracks. Abba has substantially lower LAve on
average than all other artists, and Vivaldi has substantially higher values on average. The LMax values are
similar on average for all artists. Beatles and Eels have higher LFEner values on average. Classical tracks
have lower LFreq values on average than rock tracks.
3
3.2
Plots
The dotplots in Figure 1 show the distribution of values for each artist. Abba tracks have unusually low
values of LAve. Two Eels tracks have unusually large LVar values. One Beatles track has an unusually low
LFEner value.
LVar
LAve
LMax
Vivaldi
Mozart
Enya
Eels
Beethoven
Beatles
Abba
0.0e+00
5.0e+07
1.0e+08
−100
0
LFEner
100
200
5000
15000
25000
LFreq
Vivaldi
Mozart
Enya
Eels
Beethoven
Beatles
Abba
85
90
95 100 105 110 1150
200
400
600
800
Figure 1: Dotplots of each variable by Artist.
This is a snapshot from the tour, that reveals a
number of features in the data.
Saturday Morning and V6 are two unusual tracks
that are simply outliers.
Several tracks are different to their type of music:
Hey Jude, B4, B8.
There is some obvious clustering. The Abba tracks
are distinguishable from the tracks of other artists,
mostly due to LAve. There is a cluster of rock
tracks, a mixture of Eels and Beatles tracks.
4
4
4.1
Cluster Analysis
Hierarchical
0.0e+00
1.2e+08
0e+00
Saturday Morning
4e+07
0e+00
Saturday Morning
Saturday Morning
All in a Days Work
Yellow Submarine
Love of the Loveless
Wrong About Bobby
Girl
Cant Buy Me Love
Rock Hard Times
I Feel Fine
Help
Ticket to Ride
Penny Lane
Lone Wolf
I Want to Hold Your Hand
Love Me Do
Waterloo
Yesterday
B4
The Good Old Days
Eleanor Rigby
Dancing Queen
Agony
Restraining
Anywhere Is
V6
B8
B3
Mamma Mia
M6
B1
HeyJude
Knowing Me
Take a Chance
M3
V11
Pax Deorum
M5
B5
V10
V8
B2
V4
B6
The Winner
The Memory of Trees
V5
V2
V12
V13
M1
M2
V1
V7
B7
I Have A Dream
SOS
Lay All You
Money
V3
Super Trouper
V9
M4
4e+08
8e+08
Ward
Single
Girl
Knowing Me
Take a Chance
M3
B1
HeyJude
B3
Mamma Mia
M6
V11
Pax Deorum
M5
B5
The Winner
The Memory of Trees
V5
V2
V12
V10
V8
B2
V4
B6
Lay All You
Money
V3
V13
M1
M2
V1
V7
B7
I Have A Dream
SOS
Super Trouper
V9
M4
Restraining
Anywhere Is
Dancing Queen
The Good Old Days
Eleanor Rigby
Agony
V6
B8
Love Me Do
Waterloo
Yesterday
B4
All in a Days Work
Yellow Submarine
Love of the Loveless
Wrong About Bobby
Cant Buy Me Love
Ticket to Ride
Rock Hard Times
I Feel Fine
Help
Penny Lane
Lone Wolf
I Want to Hold Your Hand
2e+07
hclust (*, "ward")
Complete
music.dist
hclust (*, "single")
V13
M1
M2
V1
V7
B7
I Have A Dream
SOS
V11
Pax Deorum
M5
B5
V10
V8
B2
V4
B6
The Winner
The Memory of Trees
V5
V2
V12
Knowing Me
Take a Chance
M3
B1
HeyJude
B3
Mamma Mia
M6
Lay All You
Money
V9
M4
Super Trouper
V3
Love Me Do
Waterloo
Yesterday
B4
Dancing Queen
The Good Old Days
Eleanor Rigby
Restraining
Anywhere Is
Agony
V6
B8
Girl
Cant Buy Me Love
Ticket to Ride
All in a Days Work
Yellow Submarine
Love of the Loveless
Wrong About Bobby
Rock Hard Times
I Feel Fine
Help
Penny Lane
Lone Wolf
I Want to Hold Your Hand
6.0e+07
Wards linkage suggests two clusters are suitable to summarize the data. This would result in one cluster
of 14 purely rock tracks, and a second cluster of 48 mixed tracks. A three cluster solution would break the
large cluster into two, one with 12 tracks (8 rock, 3 classical, 1 new wave) and the other with 36 tracks (10
rock, 24 classical, 2 new wave).
With single linkage individual tracks are sequentially peeled off the pack, illustrating the skewed nature
of the data. Saturday morning, and Girl are singleton clusters. The other 12 tracks from the Wards linkage
first cluster are grouped together by single linkage, too.
Saturday Morning is peeled off into a singleton cluster by complete. linkage. And the other 13 tracks
of from the Wards linkage first cluster are grouped together by complete linkage. The second cluster of 12
tracks (8 rock, 3 classical, 1 new wave) from wards linkage is also grouped together by complete linkage.
music.dist<-dist(d.music[,-c(1:2)])
music.hc1<-hclust(music.dist,method="ward")
music.hc2<-hclust(music.dist,method="single")
music.hc3<-hclust(music.dist,method="complete")
par(mfrow=c(3,1),mar=c(1,2,2,2))
plot(music.hc1,main="Ward",xlab=" ")
text(music.hc1)
5
Comparison of methods, using confusion tables
cl.14<-cutree(music.hc1,4)
cl.24<-cutree(music.hc2,4)
cl.34<-cutree(music.hc3,4)
table(cl.14,cl.24)
table(cl.14,cl.34)
table(cl.34,cl.24)
When the dendrograms are cut into 4 clusters the results are as follows for the different methods.
Ward
1
2
3
4
1
12
36
0
0
Single
2 3
0 0
0 0
12 0
0 1
4
0
0
1
0
Ward
1
2
3
4
1
12
36
0
0
Complete
2 3
0 0
0 0
10 0
0 1
4
0
0
3
0
Complete
1
2
3
4
1
48
0
0
0
Single
2 3
0 0
10 0
0 1
2 0
4
0
0
0
1
It looks like there is a big difference between the methods but its due mostly to the singleton clusters peeled
off in single and complete linkage.
When the dendrograms are cut into 5 clusters the results are as follows for the different methods.
Ward
1
2
3
4
5
1
12
36
0
0
0
Single
2 3
0 0
0 0
4 0
0 1
0 0
4
0
0
0
0
1
5
0
0
0
0
8
Ward
1
2
3
4
5
1
12
0
0
0
0
Complete
2 3
0 0
36 0
0 4
0 0
0 6
4
0
0
0
1
0
5
0
0
0
0
3
Complete
1
2
3
4
5
1
12
36
0
0
0
Single
2 3
0 0
0 0
4 0
0 1
0 0
4
0
0
0
0
1
Wards linkage and complete linkage agree the most. There are two sets of tracks which both methods agree
are cohesive clusters (labeled 1 and 2 by both methods) with 12 and 36 tracks respectively.
6
5
0
0
6
0
2
Because there is a difference in variance between the variables, and we’re using Euclidean distance, we
really should standardize the data before clustering it. These are the results for standardized data.
0
2
4
6
8
0.0
0.5
1.0
1.5
2.0
Saturday Morning
V5
HeyJude
V3
V1
Money
Take a Chance
Super Trouper
Knowing Me
Mamma Mia
Lay All You
Dancing Queen
Waterloo
SOS
I Have A Dream
The Winner
3.0
0
V6
B6
B4
Girl
V2
V7
5 10
20
Saturday Morning
Girl
Rock Hard Times
I Want to Hold Your Hand
All in a Days Work
Yellow Submarine
Love of the Loveless
Wrong About Bobby
Cant Buy Me Love
Ticket to Ride
Lone Wolf
Help
I Feel Fine
Penny Lane
Dancing Queen
Waterloo
Take a Chance
Knowing Me
Mamma Mia
Lay All You
Super Trouper
Money
V5
HeyJude
V2
V7
B4
B1
M6
M3
B3
V10
M2
B6
M1
M5
M4
B5
SOS
I Have A Dream
The Winner
V4
B2
V8
V11
The Memory of Trees
V12
Pax Deorum
V13
V6
Agony
B8
The Good Old Days
Yesterday
Anywhere Is
Love Me Do
Eleanor Rigby
B7
V9
Restraining
V1
V3
30
Ward
Single
V6
V10
M2
V4
B2
V13
Pax Deorum
The Memory of Trees
V12
V8
V11
Rock Hard Times
I Want to Hold Your Hand
Anywhere Is
B7
V9
Restraining
Lone Wolf
Cant Buy Me Love
Ticket to Ride
Help
I Feel Fine
Penny Lane
All in a Days Work
Yellow Submarine
Love of the Loveless
Wrong About Bobby
Love Me Do
Eleanor Rigby
B8
The Good Old Days
Yesterday
Agony
B1
M6
M3
B3
B5
M4
M1
M5
2.5
hclust (*, "ward")
Complete
music.dist
hclust (*, "single")
V2
V7
V5
HeyJude
Dancing Queen
Waterloo
Take a Chance
Mamma Mia
Lay All You
Knowing Me
Super Trouper
Money
SOS
B4
M6
M3
B3
Agony
B1
All in a Days Work
Yellow Submarine
Love of the Loveless
Wrong About Bobby
Love Me Do
Eleanor Rigby
Anywhere Is
B8
The Good Old Days
Yesterday
I Have A Dream
The Winner
V8
V11
V4
B2
The Memory of Trees
V12
B6
M4
B5
M1
M5
V10
M2
V1
V3
Pax Deorum
V13
B7
V9
Restraining
Saturday Morning
Cant Buy Me Love
Ticket to Ride
Lone Wolf
Help
I Feel Fine
Penny Lane
Girl
Rock Hard Times
I Want to Hold Your Hand
Wards linkage suggests five clusters are suitable to summarize the data. Two of the clusters would be
exclusively rock tracks. The remaining clusters would contain a mix of rock and classical and new wave.
Single linkage peels a number of individual tracks off the pack, finding the outliers in the data. V6,
Saturday morning, V2, V7, V5, Hey Jude, V3, V1 are peeled off before larger groups are formed.
Complete linkage also peels off some of the unusual tracks: V6, Saturday morning,V2, V7, V5, Hey Jude.
When the dendrograms are cut into 5 clusters the results are as follows for the different methods.
Ward
1
2
3
4
5
1
8
11
12
14
13
Single
2 3
0 0
0 0
0 1
1 0
0 0
4
0
0
0
1
0
5
0
0
0
0
1
Ward
1
2
3
4
5
Complete
1 2 3
8 0 0
11 0 0
12 0 0
12 2 2
4 0 0
4
0
0
1
0
0
5
0
0
0
0
10
Complete
1
2
3
4
5
1
47
0
2
0
9
Single
2 3
0 0
1 0
0 0
0 1
0 0
4
0
1
0
0
0
There is little agreement at 5 clusters, because single and complete linkage have mostly peeled off singleton
and small clusters, up to this point. There is one group of 10 tracks where Wards and complete agree.
7
5
0
0
0
0
1
We investigate the solution by linking the cluster identities with a tour plot. The 10 tracks where Wards
linkage and complete linkage agree form a tight group in the 5D data space.
8
4.2
k-means
k-means clustering, with k = 2, . . . , 14 is computed on the data. The 5 cluster solution is given below.
Wards
1
2
3
4
5
1
0
0
3
0
14
k-means
2 3
0 0
0 9
1 5
0 0
0 0
4
8
2
1
0
0
5
0
0
3
16
0
The results match for roughly 47
out of 62 cases. The mapping
from Wards to k-means is as follows:
1 → 4 (8 tracks)
2 → 3 (9 tracks)
4 → 5 (16 tracks)
5 → 1 (14 tracks)
Which allows us to rearrange the
confusion table.
Wards
5
3
2
1
4
1
14
3
0
0
0
k-means
2 3
0 0
1 5
0 9
0 0
0 0
4
0
1
2
8
0
5
0
3
0
0
16
The 5-cluster results of k-means agree substantially with Wards linkage hierarchical clustering. The
similar groupings are:
Wards
5
k-means
1
2
3
1
4
4
5
Tracks
All in a Days Work, Saturday Morning, Love of the Loveless, Girl,
Rock Hard Times, Lone Wolf, Wrong About Bobby, I Want to Hold
Your Hand, Cant Buy Me Love, I Feel Fine, Ticket to Ride, Help,
Yellow Submarine, Penny Lane
The Winner, V4, V8, B2, The Memory of Trees, Pax Deorum, V11,
V12, V13
Dancing Queen, Knowing Me, Take a Chance, Mamma Mia, Lay
All You, Super Trouper, Money, Waterloo
V2, V5, V7, V10, M1, M2, M3, M4, M5, M6, B1, B3, B4, B5, B6,
HeyJude
The results are interesting: 8 of the 11 Abba tracks are in one cluster (W1,K4), one cluster contains
purely rock tracks (W5,K1), and one cluster contains all classical tracks except for the unusual Beatles track
Hey Jude (W4,K5).
library(mva)
music.km2<-kmeans(d.music.std[,-c(1,2)],2)
table(cl.15,music.km5$cluster)
9
4.3
SOM
It is important to use standardized data, because SOM is using Euclidean distance. After trying several sizes
of SOM, we settled on a 6 × 6 grid. This is about the largest size map we can fit with 62 data points. We
used the linear initialization, both gaussian and bubble neighborhoods, and differing numbers of iterations.
The favorite model is summarized below.
6
SOM Map
Money
Super Trouper
I Have A Dream
5
Waterloo
Dancing
Mamma
Mia Queen
KnowingTake
Me a Chance
Lay All You
SOS
V4
The
V8 Winner
Pax Deorum
The Memory of Trees
B2V11
y
3
4
V12
V5
2
HeyJude
All in aLove
Days
Work
Me
Do
Wrong About Bobby
RestrainingV13
Yesterday
Anywhere Is
Eleanor Rigby
Yellow Submarine I Feel Fine
B8
B7
V9
M5
M1
The Good Old Days
Love of the Loveless
B6
B5
1
B3 M3
B4
M4 M6
0
V2 V10
Penny
Lane
Ticket
to Ride
Help
Lone
Wolf
Rock Hard
Times
Agony
B1
Cant Buy Me Love
I Want to Hold Your Hand
M2
V7
0
1
Girl
V3
V6
V1
2
3
Saturday Morning
4
5
x
10
6
music.som<-som(d.music.std[,-c(1:2)],6,6,
neigh="bubble",rlen=1000)
xmx<-jitter(music.som$visual$x,factor=3)
xmy<-jitter(music.som$visual$y,factor=3)
par(mfrow=c(1,1),pty="s")
plot(xmx,xmy,type="n",pch=16,xlab="x",
ylab="y",main="SOM Map",
xlim=c(-0.5,6),ylim=c(-0.5,6))
text(xmx,xmy,dimnames(d.music)[[1]])
The rock tracks are mostly in the upper
right, and the classical tracks are mostly
in the lower left. Unusual tracks Girl
and Saturday Morning are at one corner.
The Abba tracks are mostly clustered together. Vivaldi tracks seem to be the most
different classical tracks, on the lower left
fringe.
4.4
Model-based
It isn’t important to standardize the data before model-based clustering because the model accounts for
different variances between variables. We explore the fit for five variance-covariance parametrizations, EII,
VII, EEE, EEV, VVV, for the number of clusters ranging from 1-36. The upper limit matches the number
of clusters used in SOM. The BIC values for the different models are plotted below.
−9000
−11000
BIC
−7000
3 3
3 4
3 4
3 4
5
3 4
4
3 4
3 4
3 4
3 4
3 4
4 3
4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
1
2
1 EII
2 VII
3 EEE
4 EEV
5 VVV
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2
1 1 1 1 1 1
2 2 2 2
1 1
2 2
1
1 1 1 1 1 1 1
2 1
2 1 1
1
0
5
10
15
20
25
30
35
−5400
number of clusters
1
−5500
1
2
1
1
1
1
1
1
2
1
1
1
1
1
2
2
2
2
2
1 EEE
2 EEV
−5600
BIC
1
2
1
2
2
2
2
2
4
6
8
10
12
14
number of clusters
According to BIC the elliptical models are much better than the spherical models (top plot).
Examining only the elliptical models (bottom plot) would suggest that the EEE with 14 clusters is the
best model. The BIC values for EEE are fairly flat from 1-13 clusters, peaks some at 14 and then drops.
For the EEE model, 14 clusters looks best, followed by 10-12, 7, and 1-2. The EEV model has an interesting
range of BIC values for different clusters: 1 cluster is as good as 9 clusters and much better than any other
choice of number.
music.mc<-EMclust(d.music[,-c(1:2)],1:36,c("EII","VII","EEE","EEV","VVV"))
par(pty="m",mfrow=c(2,1))
plot(music.mc)
legend(1,-6300,col=c(1:5),lty=c(1:5),
legend=c("1 EII","2 VII","3 EEE","4 EEV","5 VVV"),bg="white")
music.mc<-EMclust(d.music[,-c(1:2)],1:15,c("EEE","EEV"))
plot(music.mc)
abline(h=seq(-5610,-5400,by=10),col="gray80")
legend(1,-5550,col=c(1:2),lty=c(1:2),
legend=c("1 EEE","2 EEV"),bg="white")
box()
The 14 clusters for the EEE model are (organized some according to similarity of cluster mean):
11
Cluster
Cluster Means
LAve
LMax LFEner
50.2 3.3 × 104
114
-2.8 3.1 × 104
112
-4.3 3.2 × 104
108
Names of tracks
11
12
10
LVar
1.3 × 108
8.3 × 107
4.2 × 107
LFreq
41
246
108
13
5.9 × 107
-3.5
3.1 × 104
111
160
14
2.4 × 107
-20.1
2.9 × 104
107
232
1
1.6 × 107
-24.2
2.7 × 104
106
169
2
4
1.4 × 107
7.2 × 106
216.2
-83.0
3.0 × 104
2.7 × 104
105
102
198
92
3
6.5 × 106
17.5
2.3 × 104
102
274
5
6
9
4.7 × 106
3.1 × 106
2.4 × 106
14.2
1.6
-21.3
1.8 × 104
2.1 × 104
1.3 × 104
86
98
103
209
552
233
7
8
5.3 × 105
5.8 × 105
7.7
-9.2
8.1 × 103
5.7 × 103
99
105
566
222
Saturday Morning
Girl, Cant Buy Me Love
All in a Days Work, Love of the Loveless, Wrong
About Bobby, Yellow Submarine
Rock Hard Times, Lone Wolf, I Want to Hold
Your Hand, I Feel Fine, Ticket to Ride, Help,
Penny Lane
The Good Old Days, Love Me Do, Yesterday,
B4, Waterloo
Dancing Queen, Agony, Eleanor Rigby, B8, Anywhere Is
V6
Knowing Me, Take a Chance, Mamma Mia, Lay
All You, Super Trouper, Money
V1, V3, V9, M3, M6, Restraining, B1, B3, B7,
V13
V5, HeyJude
V7, M4, B5
I Have A Dream, SOS, M1, M2, M5, The Memory of Trees, Pax Deorum, V11
V2, V10, B6
The Winner, V4, V8, B2, V12
The common variance-covariance matrix is



S=


8.5 × 1012
−1.6 × 107
7.2 × 108
1.2 × 106
5.1 × 107
−1.6 × 107
5.9 × 102
−2.3 × 104
2.8 × 100
2.2 × 102
7.2 × 108
−2.3 × 104
5.0 × 106
−2.0 × 103
8.4 × 104
1.2 × 106
2.8 × 100
−2.0 × 103
3.4 × 100
2.4 × 101
5.1 × 107
2.2 × 102
8.4 × 104
2.4 × 101
1.6 × 104






The unusual tracks discovered earlier are in singleton clusters, V6, Saturday Morning. Hey Jude the
unusual Beatles song is grouped with the classical tracks. There are some similar groupings to previous
methods. All in a Days Work, Love of the Loveless, Wrong About Bobby, Yellow Submarine, Rock Hard
Times, Lone Wolf, I Want to Hold Your Hand, I Feel Fine, Ticket to Ride, Help, Penny Lane, The Good Old
Days, Love Me Do, Yesterday are grouped near together. The Abba songs for the most part are grouped
together, although not as neatly as in SOM.
There is some disagreement with SOM. V6 is considered a singleton by model-based clustering, but more
closely aligned with the classical tracks in SOM. Agony is seen to be similar to Dancing Queen, and Eleanor
Rigby by model-based clustering but as closer to the classical tracks by SOM. Perhaps the biggest difference
is that SOM puts the Abba tracks all very close to each other, but model-based clustering breaks them up
over five clusters. Although SOM would group the Abba tracks across 4 nodes.
On closer inspection the model-based clustering doesn’t look so appealing.
4.5
Comparing Methods
The SOM map is a nice way to compare the tracks on a contiuum, rather than discrete clusters, but digesting
the 36 resulting clusters separately could be dizzying. On the map we want to draw bounding lines between
apparent clusters.
There are other methods which can produce similar types of displays as the SOM map: principal component analysis (left plot below) and multidimensional scaling (right plot below).
12
The k-means (EII in model-based) 14 cluster solution for comparison with the EEE model best solution
is:
14
1
3
5
6
7
12
2
4
8
9
10
13
11
V6
V1, V3, V9, B7, V13,
V10, M1, M2, M4, M5, B5, B6
V2, V7
M3, M6, Agony, B1, B3, B4
V5, Hey Jude
V4, V8, B2, The Memory of Trees, Pax Deorum, V11, V12
The Good Old Days, Restraining, Love Me Do, Yesterday, Eleanor Rigbym B8,
Anywhere Is
I Have A Dream, The Winner, SOS
Dancing Queen, Waterloo
Knowing Me, Take a Chance, Mamma Mia, Lay All You, Super Trouper, Money
All in a Days Work, Love of the Loveless, Lone Wolf, Wrong About Bobby, Cant
Buy Me Love, I Feel Fine, Ticket to Ride, Help, Yellow Submarine, Penny Lane
Girl, Rock Hard Times, I Want to Hold Your Hand
Saturday Morning
This looks better in terms of distinguishing between rock and classical track. The Abba tracks are
mostly together, and the other rock tracks that have been found to be similar by other methods are grouped
together. The classical songs are grouped in mostly together, along with the few unusual rock tracks (Agony,
Hey Jude) and the new wave tracks.
Lets examine a smaller number of clusters. Here is the solution for 9 clusters using k-means.
13
6
9
5
7
2
3
1
4
8
V6
V2, V7
V10, M1, M2, M4, M5, B5, B6
V1, V9, M3, M6, Agony, Restraining, B1, B3, B4, B7
V5, HeyJude
The Winner, V3, V4, V8, B2, The Memory of Trees, Pax Deorum, V11, V12, V13
Dancing Queen, Knowing Me, Take a Chance, Mamma Mia, Lay All You, Super
Trouper, I Have A Dream, Money, SOS, Waterloo
All in a Days Work, The Good Old Days, Love of the Loveless, Wrong About Bobby,
Love Me Do, Yesterday, Yellow Submarine, Eleanor Rigby, B8, Anywhere Is
Saturday Morning, Girl, Rock Hard Times, Lone Wolf, I Want to Hold Your Hand,
Cant Buy Me Love, I Feel Fine, Ticket to Ride, Help, Penny Lane
BUT Saturday Morning is not in its own cluster. Here is the solution for 10 clusters:
10
2
5
1
4
8
3
6
7
9
V6
V2, V7, V10, M2
V1, M3, M6, Agony, B1, B3, B4
V5, M4, M5, B5, B6, B7, HeyJude
V3, V4, V8, M1, B2, The Memory of Trees, Pax Deorum, V11, V12, V13
I Have A Dream, The Winner, SOS
Dancing Queen, Knowing Me, Take a Chance, Mamma Mia, Lay All You, Super
Trouper, Money, Waterloo
V9, The Good Old Days, Restraining, Love Me Do, Yesterday, Eleanor Rigby, B8,
Anywhere Is
All in a Days Work, Love of the Loveless, Girl, Rock Hard Times, Lone Wolf, Wrong
About Bobby, I Want to Hold Your Hand, Cant Buy Me Love, I Feel Fine, Ticket
to Ride, Help, Yellow Submarine, Penny Lane
Saturday Morning
14
5
Conclusions
• Here are my final clusters:
Cluster
Cluster Means
LAve
LMax LFEner
216.2 3.0 × 104
105
7.4 1.2 × 104
101
7.5 2.7 × 104
102
2 1.8 × 104
94
12 1.0 × 104
104
Tracks
10
2
5
1
4
LVar
1.4 × 107
1.9 × 106
1.1 × 107
3.0 × 106
1.7 × 106
LFreq
198
690
402
287
168
8
3
2.7 × 106
1.1 × 107
-71
-85
1.3 × 104
2.7 × 104
104
103
243
95
6
1.7 × 107
-3.2
2.5 × 104
106
165
7
5.7 × 107
-3.6
3.1 × 104
110
157
9
1.3 × 108
50.2
3.3 × 104
114
41
V6
V2, V7, V10, M2
V1, M3, M6, Agony, B1, B3, B4
V5, M4, M5, B5, B6, B7, HeyJude
V3, V4, V8, M1, B2, The Memory of Trees, Pax
Deorum, V11, V12, V13
I Have A Dream, The Winner, SOS
Dancing Queen, Knowing Me, Take a Chance,
Mamma Mia, Lay All You, Super Trouper, Money,
Waterloo
V9, The Good Old Days, Restraining, Love Me
Do, Yesterday, Eleanor Rigby, B8, Anywhere Is
All in a Days Work, Love of the Loveless, Girl,
Rock Hard Times, Lone Wolf, Wrong About
Bobby, I Want to Hold Your Hand, Cant Buy Me
Love, I Feel Fine, Ticket to Ride, Help, Yellow
Submarine, Penny Lane
Saturday Morning
• The unusual tracks are:
– V6 because it has a very high LAve value.
– Saturday Morning because it has very high LAve and LVar values.
– Hey Jude and Agony are more similar to classical tracks.
– V9 and B8 are more similar to rock tracks.
– Abba tracks because they have very low LAve.
• New wave tracks are similar to both rock and classical.
• Below is a plot of the cluster means, and the location of the clusters on the SOM map.
15
Cluster means
10
4
2
0
−2
−4
9
4
2
0
−2
−4
8
4
2
0
−2
−4
7
4
2
0
−2
−4
6
4
2
0
−2
−4
5
4
2
0
−2
−4
1
2
3
4
5
6
7
8
9
10
4
4
2
0
−2
−4
3
4
2
0
−2
−4
2
4
2
0
−2
−4
1
LVar
LAve
LMax
Variables
LFEner
LFreq
4
2
0
−2
−4
LVar LAveLMax
LFEner
LFreq
References
Fraley, C. and Raftery, A. E. (2002) Model-based Clustering, Discriminant Analysis, Density Estimation”,
Journal of the American Statistical Association, 97, 611–631, http://www.stat.washington.edu/mclust.
Hastie, T., Tibshirani, R., and Friedman, J. (2001) ”The Elements of Statistical Learning: Data Mining,
Inference, and Prediction”, Springer, New York, ISBN 0 387 95284 5.
Kohonen, T., (2000) Self-organizing Maps (3rd ed), Springer, Berlin, ISBN 3 540 67921 9.
Ripley, B.D. (1996) ”Pattern Recognition and Neural Networks” Cambridge University Press, ISBN 0 521
46086 7.
Venables, W.N. and Ripley, B.D. (2002) Modern Applied Statistics with S, Springer, New York, ISBN 0 387
95457 0.
16
Appendix
d.music<-read.csv("music-plusnew-sub-full.csv",row.names=1)
apply(d.music[,-c(1,2)],2,mean)
apply(d.music[,-c(1,2)],2,sd)
d.music.std<-cbind(d.music[,c(1,2)],apply(d.music[,-c(1,2)],2,f.std.data))
# Summary statistics
apply(d.music[,-c(1,2)],2,mean)
apply(d.music[,-c(1,2)],2,sd)
apply(d.music[d.music[,1]=="Abba",-c(1,2)],2,mean)
apply(d.music[d.music[,1]=="Beatles",-c(1,2)],2,mean)
apply(d.music[d.music[,1]=="Eels",-c(1,2)],2,mean)
apply(d.music[d.music[,1]=="Beethoven",-c(1,2)],2,mean)
apply(d.music[d.music[,1]=="Mozart",-c(1,2)],2,mean)
apply(d.music[d.music[,1]=="Vivaldi",-c(1,2)],2,mean)
apply(d.music[d.music[,1]=="Enya",-c(1,2)],2,mean)
# Plots
library(lattice)
d.music.df<-data.frame(Artist=factor(rep(d.music[,1],5)),
y=as.vector(as.matrix(d.music[,3:7])),
meas=factor(rep(1:5, rep(62,5)), labels=names(d.music[,-c(1,2)])))
postscript("music-dotplot.ps",width=8.0,height=8.0,horizontal=FALSE,
paper="special",family="URWHelvetica")
par(pty="s",mar=c(2,1,1,1))
plt.bg<-trellis.par.get("background")
plt.bg$col<-"grey90"
trellis.par.set("background",plt.bg)
stripplot(Artist~y|meas, data=d.music.df, scales=list(x="free"),
strip=function(...) strip.default(style=1,...),
panel=function(x,y){panel.grid(h=-1,v=5,col="white")
panel.stripplot(x,y,col=1,pch=16)},
xlab="",pch=16,col=1,layout=c(3,2),aspect=1,
as.table=T)
dev.off()
# Hierarchical clustering
music.dist<-dist(d.music[,-c(1:2)])
music.dist<-dist(d.music.std[,-c(1:2)])
music.hc1<-hclust(music.dist,method="ward")
music.hc2<-hclust(music.dist,method="single")
music.hc3<-hclust(music.dist,method="complete")
postscript("music-hclust.ps",width=5.0,height=10.0,horizontal=FALSE,
paper="special",family="URWHelvetica")
par(mfrow=c(3,1),mar=c(1,2,2,2))
plot(music.hc1,main="Ward",xlab=" ")
text(music.hc1)
plot(music.hc2,main="Single",ylab=" ")
text(music.hc2)
plot(music.hc3,main="Complete",ylab=" ")
17
text(music.hc3)
dev.off()
cl.12<-cutree(music.hc1,2)
cl.22<-cutree(music.hc2,2)
cl.32<-cutree(music.hc3,2)
cl.13<-cutree(music.hc1,3)
cl.23<-cutree(music.hc2,3)
cl.33<-cutree(music.hc3,3)
cl.14<-cutree(music.hc1,4)
cl.24<-cutree(music.hc2,4)
cl.34<-cutree(music.hc3,4)
cl.15<-cutree(music.hc1,5)
cl.25<-cutree(music.hc2,5)
cl.35<-cutree(music.hc3,5)
cl.16<-cutree(music.hc1,6)
cl.26<-cutree(music.hc2,6)
cl.36<-cutree(music.hc3,6)
cl.17<-cutree(music.hc1,7)
cl.27<-cutree(music.hc2,7)
cl.37<-cutree(music.hc3,7)
cl.18<-cutree(music.hc1,8)
cl.28<-cutree(music.hc2,8)
cl.38<-cutree(music.hc3,8)
cl.19<-cutree(music.hc1,9)
cl.29<-cutree(music.hc2,9)
cl.39<-cutree(music.hc3,9)
cl.110<-cutree(music.hc1,10)
cl.210<-cutree(music.hc2,10)
cl.310<-cutree(music.hc3,10)
cl.111<-cutree(music.hc1,11)
cl.211<-cutree(music.hc2,11)
cl.311<-cutree(music.hc3,11)
cl.112<-cutree(music.hc1,12)
cl.212<-cutree(music.hc2,12)
cl.312<-cutree(music.hc3,12)
cl.113<-cutree(music.hc1,13)
cl.213<-cutree(music.hc2,13)
cl.313<-cutree(music.hc3,13)
cl.114<-cutree(music.hc1,14)
cl.214<-cutree(music.hc2,14)
cl.314<-cutree(music.hc3,14)
18
table(cl.12,cl.22)
table(cl.12,cl.32)
table(cl.32,cl.22)
table(cl.13,cl.23)
table(cl.13,cl.33)
table(cl.33,cl.23)
table(cl.14,cl.24)
table(cl.14,cl.34)
table(cl.34,cl.24)
table(cl.15,cl.25)
table(cl.15,cl.35)
table(cl.35,cl.25)
table(cl.16,cl.26)
table(cl.16,cl.36)
table(cl.36,cl.26)
table(cl.17,cl.27)
table(cl.17,cl.37)
table(cl.37,cl.27)
table(cl.18,cl.28)
table(cl.18,cl.38)
table(cl.38,cl.28)
table(cl.19,cl.29)
table(cl.19,cl.39)
table(cl.39,cl.29)
table(cl.110,cl.210)
table(cl.110,cl.310)
table(cl.310,cl.210)
table(cl.111,cl.211)
table(cl.111,cl.311)
table(cl.311,cl.211)
table(cl.112,cl.212)
table(cl.112,cl.312)
table(cl.312,cl.212)
table(cl.113,cl.213)
table(cl.113,cl.313)
table(cl.313,cl.213)
table(cl.114,cl.214)
table(cl.114,cl.314)
table(cl.314,cl.214)
for (i in 1:5)
cat(i,",",dimnames(d.music)[[1]][cl.15==i],"\n")
for (i in 1:5)
cat(i,",",dimnames(d.music)[[1]][music.km5$cluster==i],"\n")
dimnames(d.music)[[1]][music.km5$cluster==3&cl.15==2]
library(genegobitree)
library(Rggobi)
ggobi()
setup.gobidend(music.hc1,d.music)
color.click.dn(music.hc1,d.music)
19
library(mva)
music.km2<-kmeans(d.music.std[,-c(1,2)],2)
music.km3<-kmeans(d.music.std[,-c(1,2)],3)
music.km4<-kmeans(d.music.std[,-c(1,2)],4)
music.km5<-kmeans(d.music.std[,-c(1,2)],5)
music.km6<-kmeans(d.music.std[,-c(1,2)],6)
music.km7<-kmeans(d.music.std[,-c(1,2)],7)
music.km8<-kmeans(d.music.std[,-c(1,2)],8)
music.km9<-kmeans(d.music.std[,-c(1,2)],9)
music.km10<-kmeans(d.music.std[,-c(1,2)],10)
music.km11<-kmeans(d.music.std[,-c(1,2)],11)
music.km12<-kmeans(d.music.std[,-c(1,2)],12)
music.km13<-kmeans(d.music.std[,-c(1,2)],13)
music.km14<-kmeans(d.music.std[,-c(1,2)],14)
table(cl.15,music.km5$cluster)
d.music.clust1<-cbind(d.music,cl.12,cl.22,cl.32,cl.13,cl.23,cl.33,
cl.14,cl.24,cl.34,cl.15,cl.25,cl.35,cl.16,cl.26,cl.36,cl.17,cl.27,cl.37,
cl.18,cl.28,cl.38,
music.km2$cluster,music.km3$cluster,music.km4$cluster,music.km5$cluster,
music.km6$cluster,music.km7$cluster,music.km8$cluster)
dimnames(d.music.clust1)[[2]][8]<-"HC-W2"
dimnames(d.music.clust1)[[2]][9]<-"HC-S2"
dimnames(d.music.clust1)[[2]][10]<-"HC-C2"
dimnames(d.music.clust1)[[2]][11]<-"HC-W3"
dimnames(d.music.clust1)[[2]][12]<-"HC-S3"
dimnames(d.music.clust1)[[2]][13]<-"HC-C3"
dimnames(d.music.clust1)[[2]][14]<-"HC-W4"
dimnames(d.music.clust1)[[2]][15]<-"HC-S4"
dimnames(d.music.clust1)[[2]][16]<-"HC-C4"
dimnames(d.music.clust1)[[2]][17]<-"HC-W5"
dimnames(d.music.clust1)[[2]][18]<-"HC-S5"
dimnames(d.music.clust1)[[2]][19]<-"HC-C5"
dimnames(d.music.clust1)[[2]][20]<-"HC-W6"
dimnames(d.music.clust1)[[2]][21]<-"HC-S6"
dimnames(d.music.clust1)[[2]][22]<-"HC-C6"
dimnames(d.music.clust1)[[2]][23]<-"HC-W7"
dimnames(d.music.clust1)[[2]][24]<-"HC-S7"
dimnames(d.music.clust1)[[2]][25]<-"HC-C7"
dimnames(d.music.clust1)[[2]][26]<-"HC-W8"
dimnames(d.music.clust1)[[2]][27]<-"HC-S8"
dimnames(d.music.clust1)[[2]][28]<-"HC-C8"
dimnames(d.music.clust1)[[2]][29]<-"KM-2"
dimnames(d.music.clust1)[[2]][30]<-"KM-3"
dimnames(d.music.clust1)[[2]][31]<-"KM-4"
dimnames(d.music.clust1)[[2]][32]<-"KM-5"
dimnames(d.music.clust1)[[2]][33]<-"KM-6"
dimnames(d.music.clust1)[[2]][34]<-"KM-7"
dimnames(d.music.clust1)[[2]][35]<-"KM-8"
f.writeXML(d.music.clust1,"music-clust1.xml",data.num=1)
for (i in 1:6)
20
cat(i,",",dimnames(d.music)[[1]][music.km6$cluster==i],"\n")
music.som<-som(d.music.std[,-c(1:2)],6,6,rlen=100)
music.som<-som(d.music.std[,-c(1:2)],6,6,rlen=200)
music.som<-som(d.music.std[,-c(1:2)],6,6,rlen=300)
music.som<-som(d.music.std[,-c(1:2)],6,6,rlen=400)
music.som<-som(d.music.std[,-c(1:2)],6,6,rlen=1000)
music.som<-som(d.music.std[,-c(1:2)],6,6,neigh="bubble",rlen=100)
music.som<-som(d.music.std[,-c(1:2)],6,6,neigh="bubble",rlen=200)
music.som<-som(d.music.std[,-c(1:2)],6,6,neigh="bubble",rlen=300)
music.som<-som(d.music.std[,-c(1:2)],6,6,neigh="bubble",rlen=400)
music.som<-som(d.music.std[,-c(1:2)],6,6,neigh="bubble",rlen=1000)
music.som<-som(d.music.std[,-c(1:2)],6,6,init="random",neigh="bubble",rlen=100)
music.som<-som(d.music.std[,-c(1:2)],6,6,init="random",neigh="bubble",
rlen=1000)
xmx<-jitter(music.som$visual$x,factor=3)
xmy<-jitter(music.som$visual$y,factor=3)
par(mfrow=c(1,1),pty="s")
plot(xmx,xmy,type="n",pch=16,xlab="x",ylab="y",main="SOM Map",
xlim=c(-0.5,6),ylim=c(-0.5,6))
text(xmx,xmy,dimnames(d.music)[[1]])
dimnames(music.som$code)<-list(NULL,names(d.music[,-c(1,2)]))
d.music.clust<-cbind(d.music.std,xmx,xmy)
dimnames(d.music.clust)[[2]][8]<-"Map 1"
dimnames(d.music.clust)[[2]][9]<-"Map 2"
d.music.grid<-cbind(rep("0",36),rep("0",36),music.som$code,
music.som$code.sum[,1:2])
dimnames(d.music.grid)[[2]][1]<-"Artist"
dimnames(d.music.grid)[[2]][2]<-"Type"
dimnames(d.music.grid)[[2]][8]<-"Map 1"
dimnames(d.music.grid)[[2]][9]<-"Map 2"
d.music.clust<-rbind(d.music.grid,d.music.clust)
f.writeXML(d.music.clust,
"music-SOM.xml",data.num=2,dat1.id<-c(1:dim(d.music.clust)[1]),
dat2=cbind(c(1:60),c(1:60)),
dat2.source=x33.l[,1],
dat2.destination=x33.l[,2],
dat2.name="SOM",dat2.id=paste(rep("l",60),c(1:60)))
# Favorite model
music.som<-som(d.music[,-c(1:2)],6,6,neigh="bubble",rlen=1000)
music.som<-som(d.music.std[,-c(1:2)],6,6,neigh="bubble",rlen=1000)
xmx<-jitter(music.som$visual$x,factor=3)
xmy<-jitter(music.som$visual$y,factor=3)
postscript("music-som.ps",width=8.0,height=8.0,horizontal=FALSE,
paper="special",family="URWHelvetica")
par(mfrow=c(1,1),pty="s")
plot(xmx,xmy,type="n",pch=16,xlab="x",ylab="y",main="SOM Map",
xlim=c(-0.5,6),ylim=c(-0.5,6))
text(xmx,xmy,dimnames(d.music)[[1]])
dev.off()
# Setting up the net lines
21
n.nodes<-6
x33.l<-NULL
for (i in 1:n.nodes) {
for (j in 1:n.nodes) {
if (j<n.nodes) x33.l<-rbind(x33.l,c((i-1)*n.nodes+j,(i-1)*n.nodes+j+1))
if (i<n.nodes) x33.l<-rbind(x33.l,c((i-1)*n.nodes+j,i*n.nodes+j))
}}
# Model-based
postscript("music-mc1.ps",width=8.0,height=8.0,horizontal=FALSE,
paper="special",family="URWHelvetica")
music.mc<-EMclust(d.music[,-c(1:2)],1:36,c("EII","VII","EEE","EEV","VVV"))
par(pty="m",mfrow=c(2,1))
plot(music.mc)
legend(1,-6300,col=c(1:5),lty=c(1:5),
legend=c("1 EII","2 VII","3 EEE","4 EEV","5 VVV"),bg="white")
music.mc<-EMclust(d.music[,-c(1:2)],1:15,c("EEE","EEV"))
plot(music.mc)
abline(h=seq(-5610,-5400,by=10),col="gray80")
legend(1,-5550,col=c(1:2),lty=c(1:2),
legend=c("1 EEE","2 EEV"),bg="white")
box()
dev.off()
smry<-summary(music.mc,d.music[,-c(1:2)])
cl<-smry$classification
cl.mat<-matrix(0,62,2)
cl.mat[cl==1,1]<-1
cl.mat[cl==2,2]<-1
prm<-mstepEEV(d.music[,-c(1:2)],cl.mat)
d.music[cl==1,1:2]
d.music[cl==2,1:2]
for (i in 1:14)
cat(i,",",dimnames(d.music)[[1]][cl==i],"\n")
smry<-summary(music.mc,d.music[,-c(1:2)])
smry
t(smry$mu)
smry$sigma
mc.clust.dist<-dist(t(smry$mu))
mc.clust.mean<-hclust(mc.clust.dist,method="single")
plot(mc.clust.mean)
music.mc<-EMclust(d.music[,-c(1:2)],11,"EEV")
music.mc2<-EMclust(d.music[,-c(1:2)],2,"EEE")
summary(music.mc2,d.music[,-c(1:2)])
music.mc3<-EMclust(d.music[,-c(1:2)],7,"EEE")
summary(music.mc3,d.music[,-c(1:2)])
mccl<-summary(music.mc3,d.music[,-c(1:2)])$classification
for (i in 1:7)
cat(i,",",dimnames(d.music)[[1]][mccl==i],"\n")
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# Generate the ellipses in 5D
vc<-smry$sigma[,,1]
evc<-eigen(vc)
vc2<-(evc$vectors)%*%diag(sqrt(evc$values))%*%t(evc$vectors)
y1<-f.gen.sphere(500,5)
y1<-y1%*%vc2
y1[,1]<-y1[,1]+smry$mu[1,1]
y1[,2]<-y1[,2]+smry$mu[2,1]
y1[,3]<-y1[,3]+smry$mu[3,1]
y1[,4]<-y1[,4]+smry$mu[4,1]
y1[,5]<-y1[,5]+smry$mu[5,1]
vc<-smry$sigma[,,2]
evc<-eigen(vc)
vc2<-(evc$vectors)%*%diag(sqrt(evc$values))%*%t(evc$vectors)
y2<-f.gen.sphere(500,5)
y2<-y2%*%vc2
y2[,1]<-y2[,1]+smry$mu[1,2]
y2[,2]<-y2[,2]+smry$mu[2,2]
y2[,3]<-y2[,3]+smry$mu[3,2]
y2[,4]<-y2[,4]+smry$mu[4,2]
y2[,5]<-y2[,5]+smry$mu[5,2]
y<-cbind(rep(0,1000),c(rep(1,500),rep(2,500)),
rbind(y1,y2))
y[,1]<-factor(y[,1])
y[,2]<-factor(y[,2])
dimnames(y)<-list(NULL,names(d.music))
d.music.mc<-rbind(d.music,y)
f.writeXML(d.music.mc,"music-mclust.xml",data.num=1)
# Set up a full data set
kmcl6<-music.km6$cluster
dimnames(music.som$code)<-list(NULL,names(d.music[,-c(1,2)]))
d.music.clust<-cbind(d.music,cl6,kmcl6,xmx,xmy)
dimnames(d.music.clust)[[2]][8]<-"HC-W6"
dimnames(d.music.clust)[[2]][9]<-"KM-6"
dimnames(d.music.clust)[[2]][10]<-"Map 1"
dimnames(d.music.clust)[[2]][11]<-"Map 2"
d.music.grid<-cbind(rep("0",36),rep("0",36),music.som$code,rep(0,36),rep(0,36),
music.som$code.sum[,1:2])
dimnames(d.music.grid)[[2]][1]<-"Artist"
dimnames(d.music.grid)[[2]][2]<-"Type"
dimnames(d.music.grid)[[2]][8]<-"HC-W6"
dimnames(d.music.grid)[[2]][9]<-"KM-6"
dimnames(d.music.grid)[[2]][10]<-"Map 1"
dimnames(d.music.grid)[[2]][11]<-"Map 2"
d.music.clust<-rbind(d.music.grid,d.music.clust)
f.writeXML(d.music.clust,
"SOM-music.xml",data.num=2,dat1.id<-c(1:dim(d.music.clust)[1]),
dat2=cbind(c(1:60),c(1:60)),
dat2.source=x33.l[,1],
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dat2.destination=x33.l[,2],
dat2.name="SOM",dat2.id=paste(rep("l",60),c(1:60)))
# Utility functions
f.gen.sphere<-function(n=100,p=5) {
x<-matrix(rnorm(n*p),ncol=p)
xnew<-t(apply(x,1,norm.vec))
xnew
}
norm.vec<-function(x) {
x<-x/norm(x)
x
}
norm<-function(x) { sqrt(sum(x^2))}
# Out put data for ggvis
edges<-NULL
k<-1
for (i in 1:61)
for (j in (i+1):62) {
edges<-rbind(edges,c(i,j,music.dist[k]))
k<-k+1
}
f.writeXML(d.music.std,"music-MDS.xml",data.num=2,dat1.id<-c(1:62),
dat1.name="Music",
dat2=cbind(1:1891,edges[,3]),
dat2.source=edges[,1],dat2.destination=edges[,2],
dat2.name="dist",dat2.id=c(1:1891))
# Comparison of clusters
for (i in 1:14)
cat(i,",",dimnames(d.music)[[1]][music.km14$cluster==i],"\n")
# Check smaller number of clusters
for (i in 1:9)
cat(i,",",dimnames(d.music)[[1]][music.km9$cluster==i],"\n")
for (i in 1:10)
cat(i,",",dimnames(d.music)[[1]][music.km10$cluster==i],"\n")
for (i in 1:11)
cat(i,",",dimnames(d.music)[[1]][music.km11$cluster==i],"\n")
options(digits=2)
for (i in 1:10)
cat(i,",",apply(d.music[music.km10$cluster==i,-c(1,2)],2,mean),"\n")
km10.mn<-NULL
for (i in 1:10)
km10.mn<-rbind(km10.mn,apply(d.music[music.km10$cluster==i,-c(1,2)],2,mean))
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km10.mn.std<-apply(km10.mn,2,f.std.data)
range(km10.mn.std)
postscript("music-means.ps",width=5.0,height=8.0,horizontal=FALSE,
paper="special",family="URWHelvetica")
plot(c(1,5),c(-2.5,2.8),type="n",axes=F,xlab="Variables",ylab="")
rect(0.8,-2.8,5.2,3.1,col="gray80")
abline(v=c(1:5),col="white")
abline(h=seq(-2.5,2.5,by=0.5),col="white")
clrs<-c(1:7,9:11)
for (i in 1:10) {
lines(c(1:5),km10.mn.std[i,],lty=i,col=clrs[i])
points(c(1:5),km10.mn.std[i,],pch=i,col=clrs[i])
}
axis(side=1,at=c(1:5),labels=names(d.music[,-c(1,2)]))
legend(1,-0.8,lty=c(1:10),pch=c(1:10),col=clrs,
legend=c(1:10),bg="gray80")
title("Cluster means")
box()
dev.off()
library(lattice)
d.music.df<-data.frame(cluster=factor(rep(music.km10$cluster,5)),
y=as.vector(as.matrix(d.music.std[,3:7])),
meas=factor(rep(1:5, rep(62,5)), labels=names(d.music[,-c(1,2)])))
postscript("music-means2.ps",width=3.0,height=10.0,horizontal=FALSE,
paper="special",family="URWHelvetica")
plt.bg<-trellis.par.get("background")
plt.bg$col<-"grey90"
trellis.par.set("background",plt.bg)
xyplot(y~meas|cluster,data=d.music.df,xlab="",ylab="",
box.ratio=1,layout=c(1,10),col=1,pch=16,
panel=function(x,y){panel.grid(h=-1,v=5,col="white")
panel.stripplot(x,y,col=1,pch=16)})
dev.off()
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