a.
| I | B | C | D | E |
|---|---|---|---|---|
| / | < | |||
| A | 9 | 1 | 4 | 6 |
| B | 4 | 1 | 4 | |
| C | 4 | 4 | ||
| D | 4 |
A merged with C
| B | D | E | |
|---|---|---|---|
| / | < | ||
| AC | 13/2 | 5/2 | 5 |
| B | 1 | 4 | |
| D | 4 |
B merged with D
| BD | E | |
|---|---|---|
| / | ||
| AC | 9 | 5 |
| BD | 4 |
BD merged with E
| BDE | |
|---|---|
| / | |
| AC | 7 |
Reformat table to check with python
| I | A | B | C | D | E |
|---|---|---|---|---|---|
| / | < | ||||
| A | 0 | 9 | 1 | 4 | 6 |
| B | 9 | 0 | 4 | 1 | 4 |
| C | 1 | 4 | 0 | 4 | 4 |
| D | 4 | 1 | 4 | 0 | 4 |
| E | 6 | 4 | 4 | 4 | 0 |
import scipy.cluster.hierarchy as shc
from scipy.spatial.distance import squareform, pdist
import pandas as pd
import matplotlib.pyplot as plt
from scipy.spatial.distance import squareform, pdist
D = pd.DataFrame(D)
D.columns = ["I", "A", "B", "C", "D", "E"]
D = D.set_index("I")
D = D.drop("/")
fig, ax = plt.subplots()
dend = shc.dendrogram(shc.linkage(squareform(D), method='average'), labels=D.index, ax=ax)
plt.show()
b.
- \(d(ij)k = \alpha_id_{ik} + \alpha_jd_{jk} + \beta d_{ij} + \gamma|d_{ik} - d_{jk}\)
- \(d(ij)k = \frac{n_i}{n_i + n_j} d_{ik} + \frac{n_j}{n_i + n_j} d_{jk} + 0 \)
- \(\sum_{x \in C_i \cup C_j}a = \sum_{x \in C_i}a + \sum_{x \in C_j}a\)
- \(\delta(C_i\cup C_j, C_k) = \frac{\left[\sum \limits_{x \in C_i} \sum \limits_{y \in C_k} d(x, y)\right] + \left[\sum \limits_{x \in C_j} \sum \limits_{y \in C_k} d(x, y)\right]}{(n_i + n_j)n_k} \)
- \(d_{ik} = \frac{\sum_{x \in C_i}\sum_{y \in C_k}}{n_in_k} \)
- \( \sum _{x \in C_i} \sum _{x \in C_k} d(x, y) = n_id_{ik}/n_k\)
- \(\delta(C_i \cup C_j, C_k) = \frac{n_id_{ik} + n_jd_{jk}}{(n_i + n_j)}\)
- \(d(ij)k = \delta(C_i \cup C_j, C_k) = \frac{n_i}{n_i + n_j} d_{ik} + \frac{n_j}{n_i + n_j} d_{jk}\)
2.
a.
Neighbors are within \(\epsilon\) of each point
D = [1, 3, 4, 5, 8, 9, 10, 11, 20]
N = []
e = 2
for point in D:
n = []
for p in range(point-2, point+e+1):
if p in D and p is not point:
n.append(p)
N.append(n)
D = list(zip(D, N))
D
b.
A point is a core point if it has at least min points
min_points = 3
C = []
for point, neighbors in D:
if len(neighbors) >= min_points:
C.append(point)
C
c.
A point is a border point if it is within \(\epsilon\) of a core point (a neighbor) and is not also a core point
B = []
for point, neighbors in D:
if point in C:
B += [n for n in neighbors if n not in C]
B = set(B)
B
d.
A point is a noise point if is not a border point or a core point.
# * operator unpacks list
[i for i in list(zip(*D))[0] if i not in B and i not in C]
e.
clusters = C
def reach(c, D):
N = []
for i in range(len(D)):
if D[i] <= c+e and D[i] >= c-e:
N.append(i)
if N == []:
return []
left = min(N)
right = max(N)
return reach(D[left], D[0:left]) + [D[i] for i in N] + reach(D[right], D[right+1:-1])
P = list(zip(*D))[0]
#reach(C[0], P)
[reach(c, P) for c in C]
As you can see, there can be some extra handling to avoid duplicate clusters when core points are in the same cluster. 20 is a noise point, so should not be added to any of the clusters.
3.
a.
D = {"point":["A", "B", "C", "D", "E", "F", "G", "H"],"x":[1, 1, 1, 1, 3, 3, 3, 3], "y":[1, 2, 4, 5, 1, 2, 4, 5]}
import matplotlib.pyplot as plt
import pandas as pd
D = pd.DataFrame(D)
D = D.set_index("point")
fig, ax = plt.subplots()
ax.scatter(D["x"], D["y"] )
for i, x, y in zip(D.index.values, D["x"], D["y"]):
if x > 2:
ax.annotate(i, (x, y), textcoords="offset points", xytext=(-10, -3), ha="center")
else:
ax.annotate(i, (x, y), textcoords="offset points", xytext=(10, -3), ha="center")
plt.show()
b.
let's recompute the distance matrix for fun!
import scipy.cluster.hierarchy as shc
from scipy.spatial.distance import squareform, pdist
dist = pd.DataFrame(squareform(pdist(D[['x', 'y']]), 'euclidean'),
columns=D.index.values,
index=D.index.values)
dist
A B C D E F G \
A 0.000000 1.000000 3.000000 4.000000 2.000000 2.236068 3.605551
B 1.000000 0.000000 2.000000 3.000000 2.236068 2.000000 2.828427
C 3.000000 2.000000 0.000000 1.000000 3.605551 2.828427 2.000000
D 4.000000 3.000000 1.000000 0.000000 4.472136 3.605551 2.236068
E 2.000000 2.236068 3.605551 4.472136 0.000000 1.000000 3.000000
F 2.236068 2.000000 2.828427 3.605551 1.000000 0.000000 2.000000
G 3.605551 2.828427 2.000000 2.236068 3.000000 2.000000 0.000000
H 4.472136 3.605551 2.236068 2.000000 4.000000 3.000000 1.000000
H
A 4.472136
B 3.605551
C 2.236068
D 2.000000
E 4.000000
F 3.000000
G 1.000000
H 0.000000
This kind of clustering works by finding the closest euclidian points, merging them into a cluster, then defining the distance to the cluster as the closest point in the cluster. We'll use scipy to avoid the tedium.
fig, ax = plt.subplots()
dend = shc.dendrogram(shc.linkage(D[['x', 'y']], method='single'), labels=D.index, ax=ax)
plt.show()
c.
This kind of clustering works by finding the closest euclidian points, merging them into a cluster, then defining the distance to the cluster as the farthest point point in the cluster. We'll use scipy again to avoid the tedium.
fig, ax = plt.subplots()
dend = shc.dendrogram(shc.linkage(D[['x', 'y']], method='complete'), labels=D.index, ax=ax)
plt.show()