Copyright (c) 2015-2017 Sebastian Raschka
Python Machine Learning - Code Examples¶
Chapter 11 - Working with Unlabeled Data – Clustering Analysis¶
Note that the optional watermark extension is a small IPython notebook plugin that I developed to make the code reproducible. You can just skip the following line(s).
In [1]:
%load_ext watermark
%watermark -a 'Sebastian Raschka' -u -d -v -p numpy,pandas,matplotlib,scipy,sklearn
Sebastian Raschka last updated: 2017-01-03 CPython 3.5.2 IPython 5.1.0 numpy 1.11.3 pandas 0.19.1 matplotlib 1.5.3 scipy 0.18.1 sklearn 0.18.1
The use of watermark is optional. You can install this IPython extension via "pip install watermark". For more information, please see: https://github.com/rasbt/watermark.
Overview¶
In [2]:
from IPython.display import Image
%matplotlib inline
Grouping objects by similarity using k-means¶
In [3]:
from sklearn.datasets import make_blobs
X, y = make_blobs(n_samples=150,
n_features=2,
centers=3,
cluster_std=0.5,
shuffle=True,
random_state=0)
In [4]:
import matplotlib.pyplot as plt
plt.scatter(X[:, 0], X[:, 1], c='white', marker='o', s=50)
plt.grid()
plt.tight_layout()
#plt.savefig('./figures/spheres.png', dpi=300)
plt.show()
In [5]:
from sklearn.cluster import KMeans
km = KMeans(n_clusters=3,
init='random',
n_init=10,
max_iter=300,
tol=1e-04,
random_state=0)
y_km = km.fit_predict(X)
plt.scatter(X[y_km == 0, 0],
X[y_km == 0, 1],
s=50,
c='lightgreen',
marker='s',
label='cluster 1')
plt.scatter(X[y_km == 1, 0],
X[y_km == 1, 1],
s=50,
c='orange',
marker='o',
label='cluster 2')
plt.scatter(X[y_km == 2, 0],
X[y_km == 2, 1],
s=50,
c='lightblue',
marker='v',
label='cluster 3')
plt.scatter(km.cluster_centers_[:, 0],
km.cluster_centers_[:, 1],
s=250,
marker='*',
c='red',
label='centroids')
plt.legend()
plt.grid()
plt.tight_layout()
#plt.savefig('./figures/centroids.png', dpi=300)
plt.show()
K-means++¶
...
Hard versus soft clustering¶
...
Using the elbow method to find the optimal number of clusters¶
In [6]:
print('Distortion: %.2f' % km.inertia_)
Distortion: 72.48
In [7]:
distortions = []
for i in range(1, 11):
km = KMeans(n_clusters=i,
init='k-means++',
n_init=10,
max_iter=300,
random_state=0)
km.fit(X)
distortions.append(km.inertia_)
plt.plot(range(1, 11), distortions, marker='o')
plt.xlabel('Number of clusters')
plt.ylabel('Distortion')
plt.tight_layout()
#plt.savefig('./figures/elbow.png', dpi=300)
plt.show()
Quantifying the quality of clustering via silhouette plots¶
In [8]:
import numpy as np
from matplotlib import cm
from sklearn.metrics import silhouette_samples
km = KMeans(n_clusters=3,
init='k-means++',
n_init=10,
max_iter=300,
tol=1e-04,
random_state=0)
y_km = km.fit_predict(X)
cluster_labels = np.unique(y_km)
n_clusters = cluster_labels.shape[0]
silhouette_vals = silhouette_samples(X, y_km, metric='euclidean')
y_ax_lower, y_ax_upper = 0, 0
yticks = []
for i, c in enumerate(cluster_labels):
c_silhouette_vals = silhouette_vals[y_km == c]
c_silhouette_vals.sort()
y_ax_upper += len(c_silhouette_vals)
color = cm.jet(float(i) / n_clusters)
plt.barh(range(y_ax_lower, y_ax_upper), c_silhouette_vals, height=1.0,
edgecolor='none', color=color)
yticks.append((y_ax_lower + y_ax_upper) / 2.)
y_ax_lower += len(c_silhouette_vals)
silhouette_avg = np.mean(silhouette_vals)
plt.axvline(silhouette_avg, color="red", linestyle="--")
plt.yticks(yticks, cluster_labels + 1)
plt.ylabel('Cluster')
plt.xlabel('Silhouette coefficient')
plt.tight_layout()
# plt.savefig('./figures/silhouette.png', dpi=300)
plt.show()
Comparison to "bad" clustering:
In [9]:
km = KMeans(n_clusters=2,
init='k-means++',
n_init=10,
max_iter=300,
tol=1e-04,
random_state=0)
y_km = km.fit_predict(X)
plt.scatter(X[y_km == 0, 0],
X[y_km == 0, 1],
s=50,
c='lightgreen',
marker='s',
label='cluster 1')
plt.scatter(X[y_km == 1, 0],
X[y_km == 1, 1],
s=50,
c='orange',
marker='o',
label='cluster 2')
plt.scatter(km.cluster_centers_[:, 0], km.cluster_centers_[:, 1],
s=250, marker='*', c='red', label='centroids')
plt.legend()
plt.grid()
plt.tight_layout()
#plt.savefig('./figures/centroids_bad.png', dpi=300)
plt.show()
In [10]:
cluster_labels = np.unique(y_km)
n_clusters = cluster_labels.shape[0]
silhouette_vals = silhouette_samples(X, y_km, metric='euclidean')
y_ax_lower, y_ax_upper = 0, 0
yticks = []
for i, c in enumerate(cluster_labels):
c_silhouette_vals = silhouette_vals[y_km == c]
c_silhouette_vals.sort()
y_ax_upper += len(c_silhouette_vals)
color = cm.jet(float(i) / n_clusters)
plt.barh(range(y_ax_lower, y_ax_upper), c_silhouette_vals, height=1.0,
edgecolor='none', color=color)
yticks.append((y_ax_lower + y_ax_upper) / 2.)
y_ax_lower += len(c_silhouette_vals)
silhouette_avg = np.mean(silhouette_vals)
plt.axvline(silhouette_avg, color="red", linestyle="--")
plt.yticks(yticks, cluster_labels + 1)
plt.ylabel('Cluster')
plt.xlabel('Silhouette coefficient')
plt.tight_layout()
# plt.savefig('./figures/silhouette_bad.png', dpi=300)
plt.show()
Organizing clusters as a hierarchical tree¶
In [11]:
Image(filename='./images/11_05.png', width=400)
Out[11]:
In [12]:
import pandas as pd
import numpy as np
np.random.seed(123)
variables = ['X', 'Y', 'Z']
labels = ['ID_0', 'ID_1', 'ID_2', 'ID_3', 'ID_4']
X = np.random.random_sample([5, 3])*10
df = pd.DataFrame(X, columns=variables, index=labels)
df
Out[12]:
| X | Y | Z | |
|---|---|---|---|
| ID_0 | 6.964692 | 2.861393 | 2.268515 |
| ID_1 | 5.513148 | 7.194690 | 4.231065 |
| ID_2 | 9.807642 | 6.848297 | 4.809319 |
| ID_3 | 3.921175 | 3.431780 | 7.290497 |
| ID_4 | 4.385722 | 0.596779 | 3.980443 |
Performing hierarchical clustering on a distance matrix¶
In [13]:
from scipy.spatial.distance import pdist, squareform
row_dist = pd.DataFrame(squareform(pdist(df, metric='euclidean')),
columns=labels,
index=labels)
row_dist
Out[13]:
| ID_0 | ID_1 | ID_2 | ID_3 | ID_4 | |
|---|---|---|---|---|---|
| ID_0 | 0.000000 | 4.973534 | 5.516653 | 5.899885 | 3.835396 |
| ID_1 | 4.973534 | 0.000000 | 4.347073 | 5.104311 | 6.698233 |
| ID_2 | 5.516653 | 4.347073 | 0.000000 | 7.244262 | 8.316594 |
| ID_3 | 5.899885 | 5.104311 | 7.244262 | 0.000000 | 4.382864 |
| ID_4 | 3.835396 | 6.698233 | 8.316594 | 4.382864 | 0.000000 |
We can either pass a condensed distance matrix (upper triangular) from the pdist function, or we can pass the "original" data array and define the metric='euclidean' argument in linkage. However, we should not pass the squareform distance matrix, which would yield different distance values although the overall clustering could be the same.
In [14]:
# 1. incorrect approach: Squareform distance matrix
from scipy.cluster.hierarchy import linkage
row_clusters = linkage(row_dist, method='complete', metric='euclidean')
pd.DataFrame(row_clusters,
columns=['row label 1', 'row label 2',
'distance', 'no. of items in clust.'],
index=['cluster %d' % (i + 1)
for i in range(row_clusters.shape[0])])
Out[14]:
| row label 1 | row label 2 | distance | no. of items in clust. | |
|---|---|---|---|---|
| cluster 1 | 0.0 | 4.0 | 6.521973 | 2.0 |
| cluster 2 | 1.0 | 2.0 | 6.729603 | 2.0 |
| cluster 3 | 3.0 | 5.0 | 8.539247 | 3.0 |
| cluster 4 | 6.0 | 7.0 | 12.444824 | 5.0 |
In [15]:
# 2. correct approach: Condensed distance matrix
row_clusters = linkage(pdist(df, metric='euclidean'), method='complete')
pd.DataFrame(row_clusters,
columns=['row label 1', 'row label 2',
'distance', 'no. of items in clust.'],
index=['cluster %d' % (i + 1)
for i in range(row_clusters.shape[0])])
Out[15]:
| row label 1 | row label 2 | distance | no. of items in clust. | |
|---|---|---|---|---|
| cluster 1 | 0.0 | 4.0 | 3.835396 | 2.0 |
| cluster 2 | 1.0 | 2.0 | 4.347073 | 2.0 |
| cluster 3 | 3.0 | 5.0 | 5.899885 | 3.0 |
| cluster 4 | 6.0 | 7.0 | 8.316594 | 5.0 |
In [16]:
# 3. correct approach: Input sample matrix
row_clusters = linkage(df.values, method='complete', metric='euclidean')
pd.DataFrame(row_clusters,
columns=['row label 1', 'row label 2',
'distance', 'no. of items in clust.'],
index=['cluster %d' % (i + 1)
for i in range(row_clusters.shape[0])])
Out[16]:
| row label 1 | row label 2 | distance | no. of items in clust. | |
|---|---|---|---|---|
| cluster 1 | 0.0 | 4.0 | 3.835396 | 2.0 |
| cluster 2 | 1.0 | 2.0 | 4.347073 | 2.0 |
| cluster 3 | 3.0 | 5.0 | 5.899885 | 3.0 |
| cluster 4 | 6.0 | 7.0 | 8.316594 | 5.0 |
In [17]:
from scipy.cluster.hierarchy import dendrogram
# make dendrogram black (part 1/2)
# from scipy.cluster.hierarchy import set_link_color_palette
# set_link_color_palette(['black'])
row_dendr = dendrogram(row_clusters,
labels=labels,
# make dendrogram black (part 2/2)
# color_threshold=np.inf
)
plt.tight_layout()
plt.ylabel('Euclidean distance')
#plt.savefig('./figures/dendrogram.png', dpi=300,
# bbox_inches='tight')
plt.show()
Attaching dendrograms to a heat map¶
In [18]:
# plot row dendrogram
fig = plt.figure(figsize=(8, 8), facecolor='white')
axd = fig.add_axes([0.09, 0.1, 0.2, 0.6])
# note: for matplotlib < v1.5.1, please use orientation='right'
row_dendr = dendrogram(row_clusters, orientation='left')
# reorder data with respect to clustering
df_rowclust = df.iloc[row_dendr['leaves'][::-1]]
axd.set_xticks([])
axd.set_yticks([])
# remove axes spines from dendrogram
for i in axd.spines.values():
i.set_visible(False)
# plot heatmap
axm = fig.add_axes([0.23, 0.1, 0.6, 0.6]) # x-pos, y-pos, width, height
cax = axm.matshow(df_rowclust, interpolation='nearest', cmap='hot_r')
fig.colorbar(cax)
axm.set_xticklabels([''] + list(df_rowclust.columns))
axm.set_yticklabels([''] + list(df_rowclust.index))
# plt.savefig('./figures/heatmap.png', dpi=300)
plt.show()
Applying agglomerative clustering via scikit-learn¶
In [19]:
from sklearn.cluster import AgglomerativeClustering
ac = AgglomerativeClustering(n_clusters=2,
affinity='euclidean',
linkage='complete')
labels = ac.fit_predict(X)
print('Cluster labels: %s' % labels)
Cluster labels: [0 1 1 0 0]
Locating regions of high density via DBSCAN¶
In [20]:
Image(filename='./images/11_11.png', width=500)
Out[20]:
In [21]:
from sklearn.datasets import make_moons
X, y = make_moons(n_samples=200, noise=0.05, random_state=0)
plt.scatter(X[:, 0], X[:, 1])
plt.tight_layout()
# plt.savefig('./figures/moons.png', dpi=300)
plt.show()
K-means and hierarchical clustering:
In [22]:
f, (ax1, ax2) = plt.subplots(1, 2, figsize=(8, 3))
km = KMeans(n_clusters=2, random_state=0)
y_km = km.fit_predict(X)
ax1.scatter(X[y_km == 0, 0], X[y_km == 0, 1],
c='lightblue', marker='o', s=40, label='cluster 1')
ax1.scatter(X[y_km == 1, 0], X[y_km == 1, 1],
c='red', marker='s', s=40, label='cluster 2')
ax1.set_title('K-means clustering')
ac = AgglomerativeClustering(n_clusters=2,
affinity='euclidean',
linkage='complete')
y_ac = ac.fit_predict(X)
ax2.scatter(X[y_ac == 0, 0], X[y_ac == 0, 1], c='lightblue',
marker='o', s=40, label='cluster 1')
ax2.scatter(X[y_ac == 1, 0], X[y_ac == 1, 1], c='red',
marker='s', s=40, label='cluster 2')
ax2.set_title('Agglomerative clustering')
plt.legend()
plt.tight_layout()
#plt.savefig('./figures/kmeans_and_ac.png', dpi=300)
plt.show()
Density-based clustering:
In [23]:
from sklearn.cluster import DBSCAN
db = DBSCAN(eps=0.2, min_samples=5, metric='euclidean')
y_db = db.fit_predict(X)
plt.scatter(X[y_db == 0, 0], X[y_db == 0, 1],
c='lightblue', marker='o', s=40,
label='cluster 1')
plt.scatter(X[y_db == 1, 0], X[y_db == 1, 1],
c='red', marker='s', s=40,
label='cluster 2')
plt.legend()
plt.tight_layout()
#plt.savefig('./figures/moons_dbscan.png', dpi=300)
plt.show()
Summary¶
...