Dividing the samples into 2 clusters using K means algorithm and euclidean Distance | Big Data | Clustering
Initialize 2 clusters C1 and C2
C1= {1}
C2= {2}
XH =160
XW = 60
|
A
|
B
|
centroid
|
|
|
C1
|
180
|
70
|
(180, 70)
|
|
C2
|
170
|
50
|
(170, 50)
|
For
row 3
C1
[
(160 – 180)2 + (60 – 70)2] ½
= 22.36
C2
[
(160 – 170)2 + (60 – 50)2] ½
= 14.14
C2 has small value. Therefore Row 3 goes into C2
C1= {1}
C2= {2,3}
XH =175
XW = 65
|
A
|
B
|
centroid
|
|
|
C1
|
180
|
70
|
(180, 70)
|
|
C2
|
170+160/2
|
50+60 /2
|
(165, 55)
|
For
row 4
C1
[
(175 – 180)2 + (65 – 70)2] ½
= 7.071
C2
[
(175– 165)2 + (65 – 55)2] ½
= 14.14
C1 has small value. Therefore Row 4 goes into C1
C1= {1,4}
C2= {2,3}
XH =182
XW = 72
|
A
|
B
|
centroid
|
|
|
C1
|
180+175/2
|
70+65/2
|
(177.5, 67.5)
|
|
C2
|
165
|
55
|
(165, 55)
|
For
row 5
C1
[
(182 – 177.5)2 + (72 – 67.5)2] ½
= 6.36
C2
[
(182– 165)2 + (72 – 55)2] ½
= 24.04
C1 has small value. Therefore Row 5 goes into C1
C1= {1,4,5}
C2= {2,3}
XH =185
XW = 75
|
A
|
B
|
centroid
|
|
|
C1
|
177.5+182 /2
|
67.5 +72 /2
|
(179.75, 69.75)
|
|
C2
|
165
|
55
|
(165, 55)
|
For
row 6
C1
[
(185 – 179.75)2 + (75 – 69.75)2] ½
= 7.42
C2
[
(185– 165)2 + (75 – 55)2] ½
= 28.28
C1 has small value. Therefore Row 6 goes into C1
Therefore,
final answer
C1= {1,4,5,6}
C2= {2,3}