2
$\begingroup$

I'm trying to run the DBSCAN algorithm on this .csv. In the first part of my program I load it and plot the data inside it to check its distribution. This is the first part of the code:

import csv
import sys
import os
from os.path import join
from sklearn.cluster import DBSCAN
import matplotlib.pyplot as plt
import numpy as np

def load_data(file_path, file_name):
   with open(join(file_path, file_name)) as csv_file:
       data_file = csv.reader(csv_file,delimiter=',')
       temp1 = next(data_file)
       n_samples = int(temp1[0])
       print("n_samples=")
       print(n_samples)
       n_features = int(temp1[1])
       temp2 = next(data_file)
       feature_names = np.array(temp2[:n_features])

       data_list = [iter for iter in data_file]
               
       data = np.asarray(data_list, dtype=np.float64)                  
       
   return(data,feature_names,n_samples,n_features)

# --- Main program ---

file_path="Datasets/"    
file_name3="CURE-complete.csv"    
data3,feature_names3,n_samples3,n_features3 = load_data(file_path, file_name3)
fig = plt.figure(figsize=(8,8))
ax = fig.add_subplot(111)
fig.subplots_adjust(top=1)
ax.set_title('Dataset n. 3 of data points')
ax.set_xlabel(feature_names3[0])
ax.set_ylabel(feature_names3[1])
plt.plot(data3[:,0], data3[:,1], '.', markersize=1.2, markeredgecolor = 'blue')
plt.show()

This is how the data are in represented: enter image description here

Since I need to know how many clusters should I split the dataset into, I used the Means algorithm which returned 4 (this is the elbow of the diagram, the best value):

sse = {}
for k in range(1, 11):
    kmeans = KMeans(n_clusters=k, random_state=0).fit(data3)
    sse[k] = kmeans.inertia_
    
kn = KneeLocator(
    list(sse.keys()),
    list(sse.values()),
    curve='convex',
    direction='decreasing',
    interp_method='polynomial',
)

plt.figure()
plt.plot(list(sse.keys()), list(sse.values()))
plt.vlines(kn.knee, plt.ylim()[0], plt.ylim()[1], linestyles='dashed')
plt.xlabel("Number of clusters")
plt.ylabel("SSE")
plt.show()

This is the diagram: enter image description here

Then, I run the DBSCAN algorithm with the following parameter: min_sample = 10 (fixed value, I need to use it) and eps = 2. This is the code:

np.random.seed(5)
dbscan2 = DBSCAN(eps=2, min_samples=10).fit(data3)
fig = plt.figure(figsize=(20,10))

ax = fig.add_subplot(111)
fig.subplots_adjust(top=1)
ax.set_title('Clustered points in dataset n. 3')

ax.set_xlabel('x')
ax.set_ylabel('y')

# set the list of colors to be selected when plotting the different clusters
color=['b','g','r','c','m','y','k','w']

# number of clusters given by the diagram
k=4

#plot the dataset
for clu in range(k):
    # collect the sequence of cooordinates of the points in each given cluster (determined by clu)
    data_list_x = [data3[i,0] for i in range(n_samples3) if dbscan2.labels_[i]==clu]
    data_list_y = [data3[i,1] for i in range(n_samples3) if dbscan2.labels_[i]==clu]
    plt.scatter(data_list_x, data_list_y, s=10, edgecolors='none', c=color[clu], alpha=0.5)

plt.show()

This is the result: enter image description here

Is my process incorrect? I do not understand why the plots looks so different. I would expect a digram like to the first one but with different color for each cluster.

$\endgroup$

1 Answer 1

1
+50
$\begingroup$

KMeans and DBSCAN are two different types of Clustering techniques.

The elbow method you used to get the best cluster count should be used in K-Means only.
You used that value i.e. K=4 to assign colors to the scatterplot, while the parameter is not used in DBSCAN fit method.

Actually that is not a valid parm for DBSCAN
You will have to control "esp" to control the number of Clusters

Fit with esp=6 resulted in 112 Clusters. You only need these few lines of code

dbscan2 = DBSCAN(eps=6, min_samples=10).fit(data3)
fig = plt.figure(figsize=(12,7))

# max(dbscan2.labels_) # This is the number of Cluster
plt.scatter(data3[:,0], data3[:,1], s=10, edgecolors='none', c=dbscan2.labels_, alpha=0.5, cmap='hsv')

enter image description here

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.