Notebook version: 1.1 (Nov 17, 2017)
Author: Jesús Cid Sueiro (jcid@tsc.uc3m.es)
Jerónimo Arenas García (jarenas@tsc.uc3m.es)
Changes: v.1.0 - First complete version.
v.1.1 - Python 3 version
In [1]:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from scipy import stats
# use seaborn plotting defaults
import seaborn as sns; sns.set()
from sklearn.cluster import KMeans
from sklearn.datasets.samples_generator import make_blobs, make_circles
from sklearn.utils import shuffle
from sklearn.metrics.pairwise import rbf_kernel
from sklearn.cluster import SpectralClustering
# For the graph representation
import networkx as nx
The first one consists of 4 compact clusters generated from a Gaussian distribution. This is the kind of dataset that are best suited to centroid-based clustering algorithms like $K$-means. If the goal of the clustering algorithm is to minimize the intra-cluster distances and find a representative prototype or centroid for each cluster, $K$-means may be a good option.
In [2]:
N = 300
nc = 4
Xs, ys = make_blobs(n_samples=N, centers=nc,
random_state=6, cluster_std=0.60, shuffle = False)
X, y = shuffle(Xs, ys, random_state=0)
plt.scatter(X[:, 0], X[:, 1], s=30);
plt.axis('equal')
plt.show()
Note that we have computed two data matrices:
Note that both matrices contain the same data (rows) but in different order. The sorted matrix will be useful later for illustration purposes, but keep in mind that, in a real clustering application, vector ${\bf y}$ is unknown (learning is not supervised), and only a data matrix with an arbitrary ordering (like ${\bf X}$) will be available.
The second dataset contains two concentric rings. One could expect from a clustering algorithm to identify two different clusters, one per each ring of points. If this is the case, $K$-means or any other algorithm focused on minimizing distances to some cluster centroids is not a good choice.
In [3]:
X2s, y2s = make_circles(n_samples=N, factor=.5, noise=.05, shuffle=False)
X2, y2 = shuffle(X2s, y2s, random_state=0)
plt.scatter(X2[:, 0], X2[:, 1], s=30)
plt.axis('equal')
plt.show()
Note, again, that we have computed both the sorted (${\bf X}_{2s}$) and the shuffled (${\bf X}_2$) versions of the dataset in the code above.
In [4]:
# <SOL>
est = KMeans(n_clusters=4)
clusters = est.fit_predict(X)
plt.scatter(X[:, 0], X[:, 1], c=clusters, s=30, cmap='rainbow')
plt.axis('equal')
clusters = est.fit_predict(X2)
plt.figure()
plt.scatter(X2[:, 0], X2[:, 1], c=clusters, s=30, cmap='rainbow')
plt.axis('equal')
plt.show()
# </SOL>
Spectral clustering algorithms are focused on connectivity: clusters are determined by maximizing some measure of intra-cluster connectivity and maximizing some form of inter-cluster connectivity.
To implement a spectral clustering algorithm we must specify a similarity measure between data points. In this session, we will use the rbf kernel, that computes the similarity between ${\bf x}$ and ${\bf y}$ as:
$$\kappa({\bf x},{\bf y}) = \exp(-\gamma \|{\bf x}-{\bf y}\|^2)$$Other similarity functions can be used, like the kernel functions implemented in Scikit-learn (see the metrics module).
For a dataset ${\cal S} = \{{\bf x}_0,\ldots,{\bf x}_{N-1}\}$, the $N\times N$ affinity matrix ${\bf K}$ contains the similarity measure between each pair of samples. Thus, its components are
$$K_{ij} = \kappa\left({\bf x}_i, {\bf x}_j\right)$$The following fragment of code illustrates all pairs of distances between any two points in the dataset.
In [5]:
gamma = 0.5
K = rbf_kernel(X, X, gamma=gamma)
In [6]:
plt.imshow(K, cmap='hot')
plt.colorbar()
plt.title('RBF Affinity Matrix for gamma = ' + str(gamma))
plt.grid('off')
plt.show()
Despite the apparent randomness of the affinity matrix, it contains some hidden structure, that we can uncover by visualizing the affinity matrix computed with the sorted data matrix, ${\bf X}_s$.
In [7]:
Ks = rbf_kernel(Xs, Xs, gamma=gamma)
plt.imshow(Ks, cmap='hot')
plt.colorbar()
plt.title('RBF Affinity Matrix for gamma = ' + str(gamma))
plt.grid('off')
plt.show()
Note that, despite their completely different appearance, both affinity matrices contain the same values, but with a different order of rows and columns.
For this dataset, the sorted affinity matrix is almost block diagonal. Note, also, that the block-wise form of this matrix depends on parameter $\gamma$.
Out from the diagonal block, similarities are close to zero. We can enforze a block diagonal structure be setting to zero the small similarity values.
For instance, by thresholding ${\bf K}_s$ with threshold $t$, we get the truncated (and sorted) affinity matrix $$ \overline{K}_{s,ij} = K_{s,ij} \cdot \text{u}(K_{s,ij} - t) $$
(where $\text{u}()$ is the step function) which is block diagonal.
In [8]:
t = 0.001
# Kt = <FILL IN> # Truncated affinity matrix
Kt = K*(K>t) # Truncated affinity matrix
# Kst = <FILL IN> # Truncated and sorted affinity matrix
Kst = Ks*(Ks>t) # Truncated and sorted affinity matrix
Any similarity matrix defines a weighted graph in such a way that the weight of the edge linking ${\bf x}_i$ and ${\bf x}_j$ is $K_{ij}$.
If $K$ is a full matrix, the graph is fully connected (there is and edge connecting every pair of nodes). But we can get a more interesting sparse graph by setting to zero the edges with a small weights.
For instance, let us visualize the graph for the truncated affinity matrix $\overline{\bf K}$ with threshold $t$. You can also check the effect of increasing or decreasing $t$.
In [9]:
G = nx.from_numpy_matrix(Kt)
graphplot = nx.draw(G, X, node_size=40, width=0.5,)
plt.axis('equal')
plt.show()
Note that, for this dataset, the graph connects edges from the same cluster only. Therefore, the number of diagonal blocks in $\overline{\bf K}_s$ is equal to the number of connected components in the graph.
Note, also, the graph does not depend on the sample ordering in the data matrix: the graphs for any matrix ${\bf K}$ and its sorted version ${\bf K}_s$ are the same.
The Laplacian matrix of a given affinity matrix ${\bf K}$ is given by $${\bf L} = {\bf D} - {\bf K}$$ where ${\bf D}$ is the diagonal degree matrix given by $$D_{ii}=\sum^{n}_{j} K_{ij}$$
If ${\bf L}$ is a block diagonal with blocks ${\bf L}_0, {\bf L}_1, \ldots, {\bf L}_{c-1}$, then it has at least $c$ orthogonal eigenvectors with zero eigenvalue: indeed, each block ${\bf L}_i$ is the Laplacian matrix of the graph containing the samples in the $i$ connected component, therefore, according to property P2,
$${\bf L}_i \cdot {\bf 1}_{N_i} = {\bf 0}_{N_i}$$where $N_i$ is the number of samples in the $i$-th connected component. Therefore, if
$${\bf v}_i = \left(\begin{array}{l} {\bf 0}_{N_0} \\ \vdots \\ {\bf 0}_{N_{i-1}} \\ {\bf 1}_{N_i} \\ {\bf 0}_{N_{i+1}} \\ \vdots \\ {\bf 0}_{N_{c-1}} \end{array} \right) $$then $${\bf L} \cdot {\bf v}_{i} = {\bf 0}_{N}$$
We can compute the Laplacian matrix for the given dataset and visualize the eigenvalues:
In [10]:
Dst = np.diag(np.sum(Kst, axis=1))
Lst = Dst - Kst
# Next, we compute the eigenvalues of the matrix
w = np.linalg.eigvalsh(Lst)
plt.figure()
plt.plot(w, marker='.');
plt.title('Eigenvalues of the matrix')
plt.show()
Verify that ${\bf 1}_N$ is an eigenvector with zero eigenvalues. To do so, compute ${\bf L}_{st} \cdot {\bf 1}_N$ and verify that its euclidean norm is close to zero (it may be not exactly zero due to finite precission errors).
Verify that vectors ${\bf v}_i$ defined above (that you can compute using vi = (ys==i)) also have zero eigenvalue.
In [11]:
# <SOL>
print(np.linalg.norm(Lst.dot(np.ones((N,1)))))
for i in range(nc):
vi = (ys==i)
print(np.linalg.norm(Lst.dot(vi)))
# </SOL>
Verify that the spectral properties of the Laplacian matrix computed from ${\bf K}_{st}$ still apply using the unsorted matrix, ${\bf K}_t$: compute ${\bf L}_{t} \cdot {\bf v}'_{i}$, where ${\bf v}'_i$ is a binary vector with components equal to 1 at the positions corresponding to samples in cluster $i$ (that you can compute using vi = (y==i))), and verify that its euclidean norm is close to zero.
In [12]:
# <SOL>
Dt = np.diag(np.sum(Kt, axis=1))
Lt = Dt - Kt
print(np.linalg.norm(Lt.dot(np.ones((N,1)))))
for i in range(nc):
vi = (y==i)
print(np.linalg.norm(Lt.dot(vi)))
# </SOL>
Note that the position of 1's in eigenvectors ${\bf v}_i$ points out the samples in the $i$-th connected component. This suggest the following tentative clustering algorithm:
This is the grounding idea of some spectral clustering algorithms. In this precise form, this algorithm does not usually work, for several reasons that we will discuss next, but with some modifications it becomes a powerfull method.
One of the reasons why the algorithm above may not work is that vectors ${\bf v}'_0, \ldots,{\bf v}'_{c-1}$ are not the only zero eigenvectors or ${\bf L}_t$: any linear combination of them is also a zero eigenvector. Eigenvector computation algorithms may return a different set of orthogonal eigenvectors.
However, one can expect that eigenvector should have similar component in the positions corresponding to samples in the same connected component.
In [13]:
wst, vst = np.linalg.eigh(Lst)
for n in range(nc):
plt.plot(vst[:,n], '.-')
In [14]:
plt.imshow(vst[:,:nc], aspect='auto')
plt.grid(False)
plt.title('Display of first 4 eigenvectors of Ks')
Out[14]:
In [15]:
# <SOL>
D = np.diag(np.sum(K, axis=1))
L = D - K
w, v = np.linalg.eigh(L)
for n in range(nc):
plt.plot(v[:,n], '.-')
# </SOL>
Note that, despite the eigenvector components can not be used as a straighforward cluster indicator, they are strongly informative of the clustering structure.
Therefore we can define vectors ${\bf z}_n = (v_{n0}, \ldots, v_{n,c-1})$ and apply a centroid based algorithm (like $K$-means) to identify all points with similar eigenvector components. The corresponding samples in ${\bf X}$ become the final clusters of the spectral clustering algorithm.
One possible way to identify the cluster structure is to apply a $K$-means algorithm over the eigenvector coordinates. The steps of the spectral clustering algorithm become the following
Summarizing, the steps of the spectral clustering algorithm for a data matrix ${\bf X}$ are the following:
In [16]:
# <SOL>
g = 20
t = 0.1
K2 = rbf_kernel(X2, X2, gamma=g)
K2t = K2*(K2>t)
G2 = nx.from_numpy_matrix(K2t)
graphplot = nx.draw(G2, X2, node_size=40, width=0.5)
plt.axis('equal')
plt.show()
# </SOL>
In [17]:
# <SOL>
D2t = np.diag(np.sum(K2t, axis=1))
L2t = D2t - K2t
w2t, v2t = np.linalg.eigh(L2t)
Z2t = v2t[:,0:2]
plt.scatter(Z2t[:,0], Z2t[:,1], s=20)
plt.show()
# </SOL>
In [18]:
est = KMeans(n_clusters=2)
clusters = est.fit_predict(Z2t)
In [19]:
plt.scatter(X2[:, 0], X2[:, 1], c=clusters, s=50, cmap='rainbow')
plt.axis('equal')
plt.show()
The spectral clustering algorithm in Scikit-learn requires the number of clusters to be specified. It works well for a small number of clusters but is not advised when using many clusters and/or data.
Finally, we are going to run spectral clustering on both datasets. Spend a few minutes figuring out the meaning of parameters of the Spectral Clustering implementation of Scikit-learn:
http://scikit-learn.org/stable/modules/generated/sklearn.cluster.SpectralClustering.html
Note that there is not equivalent parameter to our threshold $t$, which has been useful for the graph representations. However, playing with $\gamma$ should be enough to get a good clustering.
The following piece of code executes the algorithm with an 'rbf' kernel. You can manually adjust the number of clusters and the parameter of the kernel to study the behavior of the algorithm. When you are done, you can also:
In [20]:
n_clusters = 4
gamma = .1 # Warning do not exceed gamma=100
SpClus = SpectralClustering(n_clusters=n_clusters,affinity='rbf',
gamma=gamma)
SpClus.fit(X)
plt.scatter(X[:, 0], X[:, 1], c=SpClus.labels_.astype(np.int), s=50,
cmap='rainbow')
plt.axis('equal')
plt.show()
In [21]:
nc = 2
gamma = 50 #Warning do not exceed gamma=300
SpClus = SpectralClustering(n_clusters=nc, affinity='rbf', gamma=gamma)
SpClus.fit(X2)
plt.scatter(X2[:, 0], X2[:, 1], c=SpClus.labels_.astype(np.int), s=50,
cmap='rainbow')
plt.axis('equal')
plt.show()
In [22]:
nc = 5
SpClus = SpectralClustering(n_clusters=nc, affinity='nearest_neighbors')
SpClus.fit(X2)
plt.scatter(X2[:, 0], X2[:, 1], c=SpClus.labels_.astype(np.int), s=50,
cmap='rainbow')
plt.axis('equal')
plt.show()
Bottom-up approach:
In practice, this creates a hierarchical tree, that can be visualized with a dendogram. We can cut the tree at different levels, in each case obtaining a different number of clusters.
We merge the two closest clusters, where the distance between clusters is defined as:
Hierarchical clustering may lead to clusters of very different sizes. Complete linkage is the worst strategy, while Ward gives the most regular sizes. However, the affinity (or distance used in clustering) cannot be varied with Ward, thus for non Euclidean metrics, average linkage is a good alternative.
There are at least three different implementations of the algorithm:
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