tugas clustering.pdf
TRANSCRIPT
Clustering
NAMA KELOMPOK :
RESTI FEBRIANA (135150218113002)
AGUSTIN KARTIKASARI (135150218113006)
IKRAR AMALIA SHOLEKHAH (135150218114014)
SOAL
1. X={2,3,4,10,12}
Jumlah Cluster 2 ?
2. X= {(6,3), (12,4),(18,10),(24,11)}
Jumlah Cluster 2 ?
3.
PEMBAHASAN NO 1
1. X={2,3,4,10,12}
Jumlah Cluster 2 ?
Jawab :
X={2,3,4,10,12}
Dibagi dalam 2 cluster . Dipilih 2 initial centroid yaitu µ1 = 12 , µ2 = 2 dan
menggunakan ukuran city blok distance
- Iterasi 1
X 2 3 4 10 11 12
d1 (x, µ1) 10 9 8 2 1 0
d2 (x, µ2) 0 1 2 8 9 10
Min (d1, d2) C2 C2 C2 C1 C1 C1
Cari centroid baru :
µ1´ = (10 + 11 +12) / 3 = 11
µ2´ = (2 + 3 + 4) / 3 = 3
- Iterasi 1
- PROSES BERHENTI ELEMEN DALAM CLUSTER TETAP
x 2 3 4 10 11 12
d1 (x, µ1´) 9 8 7 1 0 1
d2 (x, µ2´) 1 0 1 7 8 9
Min (d1, d2) C2 C2 C2 C1 C1 C1
PEMBAHASAN NO 1
Ratio perbandingan antara nilai kovarian antarcluster dan di dalam
cluster
Iterasi 1 = (12-2) / (02+12+22+22+12+02)
= 10 / (0+1+4+4+1+0)
= 10 / 10
= 1
Iterasi 2 = (11-3) / (12+02+12+12+02+12)
= 8 / (1+0+1+1+0+1)
= 8 / 4
= 2
PEMBAHASAN NO 1
2. X= {(6,3), (12,4),(18,10),(24,11)}
Jumlah Cluster 2 ?
Jawab :
µ1 = (24,11)
µ2 = (18,10)
- Iterasi 1
PEMBAHASAN NO 2
x (6,3) (12,4) (18,10) (24,11)
d1 (x, µ1) 26 19 7 0
d2 (x, µ2) 19 12 0 7
Min (d1, d2) C2 C2 C2 C1
µ1´ = (24,11)
µ2´ = (6,3) + (12,4) + (18,10) / 3 = (6 + 12 +18) , (3+4+10) /3
= (36,17)/3 = (12, 5.67)
Iterasi 2
PEMBAHASAN NO 2
x (6,3) (12,4) (18,10) (24,11)
d1 (x, µ1´) 20 19 7 0
d2 (x, µ2´) 8.87 1.67 1.67 17.33
Min (d1, d2) C2 C2 C2 C1
Ratio perbandingan antara nilai kovarian antarcluster dan di dalam
cluster
Iterasi 1 = ( (24,11) – (18,10) ) / (𝟏𝟗𝟐+𝟏𝟐𝟐+𝟎𝟐+𝟎𝟐)
= (6+1) / (361+144+0+0)
= 7 / 505
= 0.014
Iterasi 2 = ( (24,11) – (12, 5.67) ) / (𝟖. 𝟖𝟕𝟐+𝟏. 𝟔𝟕𝟐+𝟏. 𝟔𝟕𝟐+𝟎𝟐)
= (12+5.33) / (78.68+2.79+2.79+0)
= 17.33 / 84.26
= 0.146
PEMBAHASAN NO 2
PEMBAHASAN NO 3
3.
Data (1, 1) = |𝟏 − 𝟏|𝟐 + |𝟏 − 𝟏|𝟐 = 𝟎
Data (1, 2) = |𝟒 − 𝟏|𝟐 + |𝟏 − 𝟏|𝟐= |𝟑|𝟐 + |𝟎|𝟐 = 𝟑𝟐 = 𝟗 =3
Data (1, 3) = |𝟏 − 𝟏|𝟐 + |𝟐 − 𝟏|𝟐= |𝟎|𝟐 + |𝟏|𝟐 = 𝟏𝟐 = 𝟏 = 1
Data (1, 4) = |𝟑 − 𝟏|𝟐 + |𝟒 − 𝟏|𝟐= |𝟐|𝟐 + |𝟑|𝟐 = 𝟒 + 𝟗 = 𝟏𝟑
Data (1, 5) = |𝟒 − 𝟏|𝟐 + |𝟒 − 𝟏|𝟐= |𝟒|𝟐 + |𝟑|𝟐 = 𝟏𝟔 + 𝟗 = 𝟐𝟓 = 5
Data (2, 3) = |𝟏 − 𝟒|𝟐 + |𝟐 − 𝟏|𝟐= |𝟑|𝟐 + |𝟏|𝟐 = 𝟗 + 𝟏 = 𝟏𝟎
Data (2, 4) = |𝟑 − 𝟒|𝟐 + |𝟒 − 𝟏|𝟐= |𝟏|𝟐 + |𝟑|𝟐 = 𝟏 + 𝟗 = 𝟏𝟎
Data (2, 5) = |𝟓 − 𝟒|𝟐 + |𝟒 − 𝟏|𝟐= |𝟏|𝟐 + |𝟑|𝟐 = 𝟏 + 𝟗 = 𝟏𝟎
Data (3, 4) = |𝟑 − 𝟏|𝟐 + |𝟒 − 𝟐|𝟐= |𝟐|𝟐 + |𝟐|𝟐 = 𝟒 + 𝟒 = 𝟖
Data (3, 5) = |𝟓 − 𝟏|𝟐 + |𝟒 − 𝟐|𝟐= |𝟒|𝟐 + |𝟐|𝟐 = 𝟏𝟔 + 𝟗 = 𝟐𝟎
Data (4, 5) = |𝟓 − 𝟑|𝟐 + |𝟒 − 𝟒|𝟐= |𝟐|𝟐 + |𝟎|𝟐 = 𝟒 = 2
Tabel Jarak
X 1 2 3 4 5
1 0 3 1 13 5
2 3 0 10 10 10
3 1 10 0 8 20
4 13 10 5 0 2
5 5 10 20 2 0
Single Linkage
Data terkecil = 1
Data terpilih 1,3
Menghitung jarak antar kelompok 1 & 3 dengan kelompok lain
𝐷𝑚𝑖𝑛(1,3)2 = (3, 10) = 3
𝐷𝑚𝑖𝑛(1,3)4 = ( 130, 8) = 8
𝐷𝑚𝑖𝑛(1,3)5 = (5, 20) = 20
x (1,3) 2 4 5
(1,3) 0 3 8 20
2 3 0 10 10
4 8 10 0 2
5 20 10 2 0
