Chapter 9 – Unsupervised Learning

This notebook contains all the sample code in chapter 9.

# Setup¶

First, let's import a few common modules, ensure MatplotLib plots figures inline and prepare a function to save the figures. We also check that Python 3.5 or later is installed (although Python 2.x may work, it is deprecated so we strongly recommend you use Python 3 instead), as well as Scikit-Learn ≥0.20.

In [1]:
# Python ≥3.5 is required
import sys
assert sys.version_info >= (3, 5)

# Scikit-Learn ≥0.20 is required
import sklearn
assert sklearn.__version__ >= "0.20"

# Common imports
import numpy as np
import os

# to make this notebook's output stable across runs
np.random.seed(42)

# To plot pretty figures
%matplotlib inline
import matplotlib as mpl
import matplotlib.pyplot as plt
mpl.rc('axes', labelsize=14)
mpl.rc('xtick', labelsize=12)
mpl.rc('ytick', labelsize=12)

# Where to save the figures
PROJECT_ROOT_DIR = "."
CHAPTER_ID = "unsupervised_learning"
IMAGES_PATH = os.path.join(PROJECT_ROOT_DIR, "images", CHAPTER_ID)
os.makedirs(IMAGES_PATH, exist_ok=True)

def save_fig(fig_id, tight_layout=True, fig_extension="png", resolution=300):
path = os.path.join(IMAGES_PATH, fig_id + "." + fig_extension)
print("Saving figure", fig_id)
if tight_layout:
plt.tight_layout()
plt.savefig(path, format=fig_extension, dpi=resolution)


# Clustering¶

## Introduction – Classification vs Clustering¶

In [2]:
from sklearn.datasets import load_iris

In [3]:
data = load_iris()
X = data.data
y = data.target
data.target_names

Out[3]:
array(['setosa', 'versicolor', 'virginica'], dtype='<U10')
In [4]:
plt.figure(figsize=(9, 3.5))

plt.subplot(121)
plt.plot(X[y==0, 2], X[y==0, 3], "yo", label="Iris setosa")
plt.plot(X[y==1, 2], X[y==1, 3], "bs", label="Iris versicolor")
plt.plot(X[y==2, 2], X[y==2, 3], "g^", label="Iris virginica")
plt.xlabel("Petal length", fontsize=14)
plt.ylabel("Petal width", fontsize=14)
plt.legend(fontsize=12)

plt.subplot(122)
plt.scatter(X[:, 2], X[:, 3], c="k", marker=".")
plt.xlabel("Petal length", fontsize=14)
plt.tick_params(labelleft=False)

save_fig("classification_vs_clustering_plot")
plt.show()

Saving figure classification_vs_clustering_plot


A Gaussian mixture model (explained below) can actually separate these clusters pretty well (using all 4 features: petal length & width, and sepal length & width).

In [5]:
from sklearn.mixture import GaussianMixture

In [6]:
y_pred = GaussianMixture(n_components=3, random_state=42).fit(X).predict(X)


Let's map each cluster to a class. Instead of hard coding the mapping (as is done in the book, for simplicity), we will pick the most common class for each cluster (using the scipy.stats.mode() function):

In [7]:
from scipy import stats

mapping = {}
for class_id in np.unique(y):
mode, _ = stats.mode(y_pred[y==class_id])
mapping[mode[0]] = class_id

mapping

Out[7]:
{1: 0, 2: 1, 0: 2}
In [8]:
y_pred = np.array([mapping[cluster_id] for cluster_id in y_pred])

In [9]:
plt.plot(X[y_pred==0, 2], X[y_pred==0, 3], "yo", label="Cluster 1")
plt.plot(X[y_pred==1, 2], X[y_pred==1, 3], "bs", label="Cluster 2")
plt.plot(X[y_pred==2, 2], X[y_pred==2, 3], "g^", label="Cluster 3")
plt.xlabel("Petal length", fontsize=14)
plt.ylabel("Petal width", fontsize=14)
plt.legend(loc="upper left", fontsize=12)
plt.show()

In [10]:
np.sum(y_pred==y)

Out[10]:
145
In [11]:
np.sum(y_pred==y) / len(y_pred)

Out[11]:
0.9666666666666667

Note: the results in this notebook may differ slightly from the book. This is because algorithms can sometimes be tweaked a bit between Scikit-Learn versions.

## K-Means¶

Let's start by generating some blobs:

In [12]:
from sklearn.datasets import make_blobs

In [13]:
blob_centers = np.array(
[[ 0.2,  2.3],
[-1.5 ,  2.3],
[-2.8,  1.8],
[-2.8,  2.8],
[-2.8,  1.3]])
blob_std = np.array([0.4, 0.3, 0.1, 0.1, 0.1])

In [14]:
X, y = make_blobs(n_samples=2000, centers=blob_centers,
cluster_std=blob_std, random_state=7)


Now let's plot them:

In [15]:
def plot_clusters(X, y=None):
plt.scatter(X[:, 0], X[:, 1], c=y, s=1)
plt.xlabel("$x_1$", fontsize=14)
plt.ylabel("$x_2$", fontsize=14, rotation=0)

In [16]:
plt.figure(figsize=(8, 4))
plot_clusters(X)
save_fig("blobs_plot")
plt.show()

Saving figure blobs_plot


### Fit and Predict¶

Let's train a K-Means clusterer on this dataset. It will try to find each blob's center and assign each instance to the closest blob:

In [17]:
from sklearn.cluster import KMeans

In [18]:
k = 5
kmeans = KMeans(n_clusters=k, random_state=42)
y_pred = kmeans.fit_predict(X)


Each instance was assigned to one of the 5 clusters:

In [19]:
y_pred

Out[19]:
array([4, 1, 0, ..., 3, 0, 1], dtype=int32)
In [20]:
y_pred is kmeans.labels_

Out[20]:
True

And the following 5 centroids (i.e., cluster centers) were estimated:

In [21]:
kmeans.cluster_centers_

Out[21]:
array([[ 0.20876306,  2.25551336],
[-2.80389616,  1.80117999],
[-1.46679593,  2.28585348],
[-2.79290307,  2.79641063],
[-2.80037642,  1.30082566]])

Note that the KMeans instance preserves the labels of the instances it was trained on. Somewhat confusingly, in this context, the label of an instance is the index of the cluster that instance gets assigned to:

In [22]:
kmeans.labels_

Out[22]:
array([4, 1, 0, ..., 3, 0, 1], dtype=int32)

Of course, we can predict the labels of new instances:

In [23]:
X_new = np.array([[0, 2], [3, 2], [-3, 3], [-3, 2.5]])
kmeans.predict(X_new)

Out[23]:
array([0, 0, 3, 3], dtype=int32)

### Decision Boundaries¶

Let's plot the model's decision boundaries. This gives us a Voronoi diagram:

In [24]:
def plot_data(X):
plt.plot(X[:, 0], X[:, 1], 'k.', markersize=2)

def plot_centroids(centroids, weights=None, circle_color='w', cross_color='k'):
if weights is not None:
centroids = centroids[weights > weights.max() / 10]
plt.scatter(centroids[:, 0], centroids[:, 1],
marker='o', s=35, linewidths=8,
color=circle_color, zorder=10, alpha=0.9)
plt.scatter(centroids[:, 0], centroids[:, 1],
marker='x', s=2, linewidths=12,
color=cross_color, zorder=11, alpha=1)

def plot_decision_boundaries(clusterer, X, resolution=1000, show_centroids=True,
show_xlabels=True, show_ylabels=True):
mins = X.min(axis=0) - 0.1
maxs = X.max(axis=0) + 0.1
xx, yy = np.meshgrid(np.linspace(mins[0], maxs[0], resolution),
np.linspace(mins[1], maxs[1], resolution))
Z = clusterer.predict(np.c_[xx.ravel(), yy.ravel()])
Z = Z.reshape(xx.shape)

plt.contourf(Z, extent=(mins[0], maxs[0], mins[1], maxs[1]),
cmap="Pastel2")
plt.contour(Z, extent=(mins[0], maxs[0], mins[1], maxs[1]),
linewidths=1, colors='k')
plot_data(X)
if show_centroids:
plot_centroids(clusterer.cluster_centers_)

if show_xlabels:
plt.xlabel("$x_1$", fontsize=14)
else:
plt.tick_params(labelbottom=False)
if show_ylabels:
plt.ylabel("$x_2$", fontsize=14, rotation=0)
else:
plt.tick_params(labelleft=False)

In [25]:
plt.figure(figsize=(8, 4))
plot_decision_boundaries(kmeans, X)
save_fig("voronoi_plot")
plt.show()

Saving figure voronoi_plot


Not bad! Some of the instances near the edges were probably assigned to the wrong cluster, but overall it looks pretty good.

### Hard Clustering vs Soft Clustering¶

Rather than arbitrarily choosing the closest cluster for each instance, which is called hard clustering, it might be better measure the distance of each instance to all 5 centroids. This is what the transform() method does:

In [26]:
kmeans.transform(X_new)

Out[26]:
array([[2.88633901, 0.32995317, 2.9042344 , 1.49439034, 2.81093633],
[5.84236351, 2.80290755, 5.84739223, 4.4759332 , 5.80730058],
[1.71086031, 3.29399768, 0.29040966, 1.69136631, 1.21475352],
[1.21567622, 3.21806371, 0.36159148, 1.54808703, 0.72581411]])

You can verify that this is indeed the Euclidian distance between each instance and each centroid:

In [27]:
np.linalg.norm(np.tile(X_new, (1, k)).reshape(-1, k, 2) - kmeans.cluster_centers_, axis=2)

Out[27]:
array([[2.88633901, 0.32995317, 2.9042344 , 1.49439034, 2.81093633],
[5.84236351, 2.80290755, 5.84739223, 4.4759332 , 5.80730058],
[1.71086031, 3.29399768, 0.29040966, 1.69136631, 1.21475352],
[1.21567622, 3.21806371, 0.36159148, 1.54808703, 0.72581411]])

### K-Means Algorithm¶

The K-Means algorithm is one of the fastest clustering algorithms, and also one of the simplest:

• First initialize $k$ centroids randomly: $k$ distinct instances are chosen randomly from the dataset and the centroids are placed at their locations.
• Repeat until convergence (i.e., until the centroids stop moving):
• Assign each instance to the closest centroid.
• Update the centroids to be the mean of the instances that are assigned to them.

The KMeans class applies an optimized algorithm by default. To get the original K-Means algorithm (for educational purposes only), you must set init="random", n_init=1and algorithm="full". These hyperparameters will be explained below.

Let's run the K-Means algorithm for 1, 2 and 3 iterations, to see how the centroids move around:

In [28]:
kmeans_iter1 = KMeans(n_clusters=5, init="random", n_init=1,
algorithm="full", max_iter=1, random_state=0)
kmeans_iter2 = KMeans(n_clusters=5, init="random", n_init=1,
algorithm="full", max_iter=2, random_state=0)
kmeans_iter3 = KMeans(n_clusters=5, init="random", n_init=1,
algorithm="full", max_iter=3, random_state=0)
kmeans_iter1.fit(X)
kmeans_iter2.fit(X)
kmeans_iter3.fit(X)

Out[28]:
KMeans(algorithm='full', init='random', max_iter=3, n_clusters=5, n_init=1,
random_state=0)

And let's plot this:

In [29]:
plt.figure(figsize=(10, 8))

plt.subplot(321)
plot_data(X)
plot_centroids(kmeans_iter1.cluster_centers_, circle_color='r', cross_color='w')
plt.ylabel("$x_2$", fontsize=14, rotation=0)
plt.tick_params(labelbottom=False)
plt.title("Update the centroids (initially randomly)", fontsize=14)

plt.subplot(322)
plot_decision_boundaries(kmeans_iter1, X, show_xlabels=False, show_ylabels=False)
plt.title("Label the instances", fontsize=14)

plt.subplot(323)
plot_decision_boundaries(kmeans_iter1, X, show_centroids=False, show_xlabels=False)
plot_centroids(kmeans_iter2.cluster_centers_)

plt.subplot(324)
plot_decision_boundaries(kmeans_iter2, X, show_xlabels=False, show_ylabels=False)

plt.subplot(325)
plot_decision_boundaries(kmeans_iter2, X, show_centroids=False)
plot_centroids(kmeans_iter3.cluster_centers_)

plt.subplot(326)
plot_decision_boundaries(kmeans_iter3, X, show_ylabels=False)

save_fig("kmeans_algorithm_plot")
plt.show()

Saving figure kmeans_algorithm_plot


### K-Means Variability¶

In the original K-Means algorithm, the centroids are just initialized randomly, and the algorithm simply runs a single iteration to gradually improve the centroids, as we saw above.

However, one major problem with this approach is that if you run K-Means multiple times (or with different random seeds), it can converge to very different solutions, as you can see below:

In [30]:
def plot_clusterer_comparison(clusterer1, clusterer2, X, title1=None, title2=None):
clusterer1.fit(X)
clusterer2.fit(X)

plt.figure(figsize=(10, 3.2))

plt.subplot(121)
plot_decision_boundaries(clusterer1, X)
if title1:
plt.title(title1, fontsize=14)

plt.subplot(122)
plot_decision_boundaries(clusterer2, X, show_ylabels=False)
if title2:
plt.title(title2, fontsize=14)

In [31]:
kmeans_rnd_init1 = KMeans(n_clusters=5, init="random", n_init=1,
algorithm="full", random_state=2)
kmeans_rnd_init2 = KMeans(n_clusters=5, init="random", n_init=1,
algorithm="full", random_state=5)

plot_clusterer_comparison(kmeans_rnd_init1, kmeans_rnd_init2, X,
"Solution 1", "Solution 2 (with a different random init)")

save_fig("kmeans_variability_plot")
plt.show()

Saving figure kmeans_variability_plot


### Inertia¶

To select the best model, we will need a way to evaluate a K-Mean model's performance. Unfortunately, clustering is an unsupervised task, so we do not have the targets. But at least we can measure the distance between each instance and its centroid. This is the idea behind the inertia metric:

In [32]:
kmeans.inertia_

Out[32]:
211.5985372581683

As you can easily verify, inertia is the sum of the squared distances between each training instance and its closest centroid:

In [33]:
X_dist = kmeans.transform(X)
np.sum(X_dist[np.arange(len(X_dist)), kmeans.labels_]**2)

Out[33]:
211.5985372581684

The score() method returns the negative inertia. Why negative? Well, it is because a predictor's score() method must always respect the "greater is better" rule.

In [34]:
kmeans.score(X)

Out[34]:
-211.5985372581683

### Multiple Initializations¶

So one approach to solve the variability issue is to simply run the K-Means algorithm multiple times with different random initializations, and select the solution that minimizes the inertia. For example, here are the inertias of the two "bad" models shown in the previous figure:

In [35]:
kmeans_rnd_init1.inertia_

Out[35]:
219.8438540223319
In [36]:
kmeans_rnd_init2.inertia_

Out[36]:
236.95563196978728

As you can see, they have a higher inertia than the first "good" model we trained, which means they are probably worse.

When you set the n_init hyperparameter, Scikit-Learn runs the original algorithm n_init times, and selects the solution that minimizes the inertia. By default, Scikit-Learn sets n_init=10.

In [37]:
kmeans_rnd_10_inits = KMeans(n_clusters=5, init="random", n_init=10,
algorithm="full", random_state=2)
kmeans_rnd_10_inits.fit(X)

Out[37]:
KMeans(algorithm='full', init='random', n_clusters=5, random_state=2)

As you can see, we end up with the initial model, which is certainly the optimal K-Means solution (at least in terms of inertia, and assuming $k=5$).

In [38]:
plt.figure(figsize=(8, 4))
plot_decision_boundaries(kmeans_rnd_10_inits, X)
plt.show()


### K-Means++¶

Instead of initializing the centroids entirely randomly, it is preferable to initialize them using the following algorithm, proposed in a 2006 paper by David Arthur and Sergei Vassilvitskii:

• Take one centroid $c_1$, chosen uniformly at random from the dataset.
• Take a new center $c_i$, choosing an instance $\mathbf{x}_i$ with probability: $D(\mathbf{x}_i)^2$ / $\sum\limits_{j=1}^{m}{D(\mathbf{x}_j)}^2$ where $D(\mathbf{x}_i)$ is the distance between the instance $\mathbf{x}_i$ and the closest centroid that was already chosen. This probability distribution ensures that instances that are further away from already chosen centroids are much more likely be selected as centroids.
• Repeat the previous step until all $k$ centroids have been chosen.

The rest of the K-Means++ algorithm is just regular K-Means. With this initialization, the K-Means algorithm is much less likely to converge to a suboptimal solution, so it is possible to reduce n_init considerably. Most of the time, this largely compensates for the additional complexity of the initialization process.

To set the initialization to K-Means++, simply set init="k-means++" (this is actually the default):

In [39]:
KMeans()

Out[39]:
KMeans()
In [40]:
good_init = np.array([[-3, 3], [-3, 2], [-3, 1], [-1, 2], [0, 2]])
kmeans = KMeans(n_clusters=5, init=good_init, n_init=1, random_state=42)
kmeans.fit(X)
kmeans.inertia_

Out[40]:
211.62337889822362

### Accelerated K-Means¶

The K-Means algorithm can be significantly accelerated by avoiding many unnecessary distance calculations: this is achieved by exploiting the triangle inequality (given three points A, B and C, the distance AC is always such that AC ≤ AB + BC) and by keeping track of lower and upper bounds for distances between instances and centroids (see this 2003 paper by Charles Elkan for more details).

To use Elkan's variant of K-Means, just set algorithm="elkan". Note that it does not support sparse data, so by default, Scikit-Learn uses "elkan" for dense data, and "full" (the regular K-Means algorithm) for sparse data.

In [41]:
%timeit -n 50 KMeans(algorithm="elkan", random_state=42).fit(X)

90.4 ms ± 612 µs per loop (mean ± std. dev. of 7 runs, 50 loops each)

In [42]:
%timeit -n 50 KMeans(algorithm="full", random_state=42).fit(X)

91.6 ms ± 171 µs per loop (mean ± std. dev. of 7 runs, 50 loops each)


There's no big difference in this case, as the dataset is fairly small.

### Mini-Batch K-Means¶

Scikit-Learn also implements a variant of the K-Means algorithm that supports mini-batches (see this paper):

In [43]:
from sklearn.cluster import MiniBatchKMeans

In [44]:
minibatch_kmeans = MiniBatchKMeans(n_clusters=5, random_state=42)
minibatch_kmeans.fit(X)

Out[44]:
MiniBatchKMeans(n_clusters=5, random_state=42)
In [45]:
minibatch_kmeans.inertia_

Out[45]:
211.93186531476786

If the dataset does not fit in memory, the simplest option is to use the memmap class, just like we did for incremental PCA in the previous chapter. First let's load MNIST:

In [46]:
import urllib.request
from sklearn.datasets import fetch_openml

mnist = fetch_openml('mnist_784', version=1)
mnist.target = mnist.target.astype(np.int64)

In [47]:
from sklearn.model_selection import train_test_split

X_train, X_test, y_train, y_test = train_test_split(
mnist["data"], mnist["target"], random_state=42)


Next, let's write it to a memmap:

In [48]:
filename = "my_mnist.data"
X_mm = np.memmap(filename, dtype='float32', mode='write', shape=X_train.shape)
X_mm[:] = X_train

In [49]:
minibatch_kmeans = MiniBatchKMeans(n_clusters=10, batch_size=10, random_state=42)
minibatch_kmeans.fit(X_mm)

Out[49]:
MiniBatchKMeans(batch_size=10, n_clusters=10, random_state=42)

If your data is so large that you cannot use memmap, things get more complicated. Let's start by writing a function to load the next batch (in real life, you would load the data from disk):

In [50]:
def load_next_batch(batch_size):
return X[np.random.choice(len(X), batch_size, replace=False)]


Now we can train the model by feeding it one batch at a time. We also need to implement multiple initializations and keep the model with the lowest inertia:

In [51]:
np.random.seed(42)

In [52]:
k = 5
n_init = 10
n_iterations = 100
batch_size = 100
init_size = 500  # more data for K-Means++ initialization
evaluate_on_last_n_iters = 10

best_kmeans = None

for init in range(n_init):
minibatch_kmeans = MiniBatchKMeans(n_clusters=k, init_size=init_size)
minibatch_kmeans.partial_fit(X_init)

minibatch_kmeans.sum_inertia_ = 0
for iteration in range(n_iterations):
minibatch_kmeans.partial_fit(X_batch)
if iteration >= n_iterations - evaluate_on_last_n_iters:
minibatch_kmeans.sum_inertia_ += minibatch_kmeans.inertia_

if (best_kmeans is None or
minibatch_kmeans.sum_inertia_ < best_kmeans.sum_inertia_):
best_kmeans = minibatch_kmeans

In [53]:
best_kmeans.score(X)

Out[53]:
-211.70999744411446

Mini-batch K-Means is much faster than regular K-Means:

In [54]:
%timeit KMeans(n_clusters=5, random_state=42).fit(X)

45.7 ms ± 245 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [55]:
%timeit MiniBatchKMeans(n_clusters=5, random_state=42).fit(X)

10.2 ms ± 107 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


That's much faster! However, its performance is often lower (higher inertia), and it keeps degrading as k increases. Let's plot the inertia ratio and the training time ratio between Mini-batch K-Means and regular K-Means:

In [56]:
from timeit import timeit

In [57]:
times = np.empty((100, 2))
inertias = np.empty((100, 2))
for k in range(1, 101):
kmeans_ = KMeans(n_clusters=k, random_state=42)
minibatch_kmeans = MiniBatchKMeans(n_clusters=k, random_state=42)
print("\r{}/{}".format(k, 100), end="")
times[k-1, 0] = timeit("kmeans_.fit(X)", number=10, globals=globals())
times[k-1, 1]  = timeit("minibatch_kmeans.fit(X)", number=10, globals=globals())
inertias[k-1, 0] = kmeans_.inertia_
inertias[k-1, 1] = minibatch_kmeans.inertia_

100/100
In [58]:
plt.figure(figsize=(10,4))

plt.subplot(121)
plt.plot(range(1, 101), inertias[:, 0], "r--", label="K-Means")
plt.plot(range(1, 101), inertias[:, 1], "b.-", label="Mini-batch K-Means")
plt.xlabel("$k$", fontsize=16)
plt.title("Inertia", fontsize=14)
plt.legend(fontsize=14)
plt.axis([1, 100, 0, 100])

plt.subplot(122)
plt.plot(range(1, 101), times[:, 0], "r--", label="K-Means")
plt.plot(range(1, 101), times[:, 1], "b.-", label="Mini-batch K-Means")
plt.xlabel("$k$", fontsize=16)
plt.title("Training time (seconds)", fontsize=14)
plt.axis([1, 100, 0, 6])

save_fig("minibatch_kmeans_vs_kmeans")
plt.show()

Saving figure minibatch_kmeans_vs_kmeans


### Finding the optimal number of clusters¶

What if the number of clusters was set to a lower or greater value than 5?

In [59]:
kmeans_k3 = KMeans(n_clusters=3, random_state=42)
kmeans_k8 = KMeans(n_clusters=8, random_state=42)

plot_clusterer_comparison(kmeans_k3, kmeans_k8, X, "$k=3$", "$k=8$")
plt.show()

Saving figure bad_n_clusters_plot


Ouch, these two models don't look great. What about their inertias?

In [60]:
kmeans_k3.inertia_

Out[60]:
653.2223267580945
In [61]:
kmeans_k8.inertia_

Out[61]:
118.44108623570081

No, we cannot simply take the value of $k$ that minimizes the inertia, since it keeps getting lower as we increase $k$. Indeed, the more clusters there are, the closer each instance will be to its closest centroid, and therefore the lower the inertia will be. However, we can plot the inertia as a function of $k$ and analyze the resulting curve:

In [62]:
kmeans_per_k = [KMeans(n_clusters=k, random_state=42).fit(X)
for k in range(1, 10)]
inertias = [model.inertia_ for model in kmeans_per_k]

In [63]:
plt.figure(figsize=(8, 3.5))
plt.plot(range(1, 10), inertias, "bo-")
plt.xlabel("$k$", fontsize=14)
plt.ylabel("Inertia", fontsize=14)
plt.annotate('Elbow',
xy=(4, inertias[3]),
xytext=(0.55, 0.55),
textcoords='figure fraction',
fontsize=16,
arrowprops=dict(facecolor='black', shrink=0.1)
)
plt.axis([1, 8.5, 0, 1300])
save_fig("inertia_vs_k_plot")
plt.show()

Saving figure inertia_vs_k_plot


As you can see, there is an elbow at $k=4$, which means that less clusters than that would be bad, and more clusters would not help much and might cut clusters in half. So $k=4$ is a pretty good choice. Of course in this example it is not perfect since it means that the two blobs in the lower left will be considered as just a single cluster, but it's a pretty good clustering nonetheless.

In [64]:
plot_decision_boundaries(kmeans_per_k[4-1], X)
plt.show()


Another approach is to look at the silhouette score, which is the mean silhouette coefficient over all the instances. An instance's silhouette coefficient is equal to $(b - a)/\max(a, b)$ where $a$ is the mean distance to the other instances in the same cluster (it is the mean intra-cluster distance), and $b$ is the mean nearest-cluster distance, that is the mean distance to the instances of the next closest cluster (defined as the one that minimizes $b$, excluding the instance's own cluster). The silhouette coefficient can vary between -1 and +1: a coefficient close to +1 means that the instance is well inside its own cluster and far from other clusters, while a coefficient close to 0 means that it is close to a cluster boundary, and finally a coefficient close to -1 means that the instance may have been assigned to the wrong cluster.

Let's plot the silhouette score as a function of $k$:

In [65]:
from sklearn.metrics import silhouette_score

In [66]:
silhouette_score(X, kmeans.labels_)

Out[66]:
0.655517642572828
In [67]:
silhouette_scores = [silhouette_score(X, model.labels_)
for model in kmeans_per_k[1:]]

In [68]:
plt.figure(figsize=(8, 3))
plt.plot(range(2, 10), silhouette_scores, "bo-")
plt.xlabel("$k$", fontsize=14)
plt.ylabel("Silhouette score", fontsize=14)
plt.axis([1.8, 8.5, 0.55, 0.7])
save_fig("silhouette_score_vs_k_plot")
plt.show()

Saving figure silhouette_score_vs_k_plot


As you can see, this visualization is much richer than the previous one: in particular, although it confirms that $k=4$ is a very good choice, but it also underlines the fact that $k=5$ is quite good as well.

An even more informative visualization is given when you plot every instance's silhouette coefficient, sorted by the cluster they are assigned to and by the value of the coefficient. This is called a silhouette diagram:

In [69]:
from sklearn.metrics import silhouette_samples
from matplotlib.ticker import FixedLocator, FixedFormatter

plt.figure(figsize=(11, 9))

for k in (3, 4, 5, 6):
plt.subplot(2, 2, k - 2)

y_pred = kmeans_per_k[k - 1].labels_
silhouette_coefficients = silhouette_samples(X, y_pred)

ticks = []
for i in range(k):
coeffs = silhouette_coefficients[y_pred == i]
coeffs.sort()

color = mpl.cm.Spectral(i / k)
plt.fill_betweenx(np.arange(pos, pos + len(coeffs)), 0, coeffs,
facecolor=color, edgecolor=color, alpha=0.7)
ticks.append(pos + len(coeffs) // 2)

plt.gca().yaxis.set_major_locator(FixedLocator(ticks))
plt.gca().yaxis.set_major_formatter(FixedFormatter(range(k)))
if k in (3, 5):
plt.ylabel("Cluster")

if k in (5, 6):
plt.gca().set_xticks([-0.1, 0, 0.2, 0.4, 0.6, 0.8, 1])
plt.xlabel("Silhouette Coefficient")
else:
plt.tick_params(labelbottom=False)

plt.axvline(x=silhouette_scores[k - 2], color="red", linestyle="--")
plt.title("$k={}$".format(k), fontsize=16)

save_fig("silhouette_analysis_plot")
plt.show()

Saving figure silhouette_analysis_plot


As you can see, $k=5$ looks like the best option here, as all clusters are roughly the same size, and they all cross the dashed line, which represents the mean silhouette score.

### Limits of K-Means¶

In [70]:
X1, y1 = make_blobs(n_samples=1000, centers=((4, -4), (0, 0)), random_state=42)
X1 = X1.dot(np.array([[0.374, 0.95], [0.732, 0.598]]))
X2, y2 = make_blobs(n_samples=250, centers=1, random_state=42)
X2 = X2 + [6, -8]
X = np.r_[X1, X2]
y = np.r_[y1, y2]

In [71]:
plot_clusters(X)

In [72]:
kmeans_good = KMeans(n_clusters=3, init=np.array([[-1.5, 2.5], [0.5, 0], [4, 0]]), n_init=1, random_state=42)
kmeans_good.fit(X)

Out[72]:
KMeans(n_clusters=3, random_state=42)
In [73]:
plt.figure(figsize=(10, 3.2))

plt.subplot(121)
plot_decision_boundaries(kmeans_good, X)
plt.title("Inertia = {:.1f}".format(kmeans_good.inertia_), fontsize=14)

plt.subplot(122)

plt.show()

Saving figure bad_kmeans_plot


### Using clustering for image segmentation¶

In [74]:
# Download the ladybug image
images_path = os.path.join(PROJECT_ROOT_DIR, "images", "unsupervised_learning")
os.makedirs(images_path, exist_ok=True)
urllib.request.urlretrieve(url, os.path.join(images_path, filename))

Downloading ladybug.png

Out[74]:
('./images/unsupervised_learning/ladybug.png',
<http.client.HTTPMessage at 0x7fa4386b0090>)
In [75]:
from matplotlib.image import imread
image.shape

Out[75]:
(533, 800, 3)
In [76]:
X = image.reshape(-1, 3)
kmeans = KMeans(n_clusters=8, random_state=42).fit(X)
segmented_img = kmeans.cluster_centers_[kmeans.labels_]
segmented_img = segmented_img.reshape(image.shape)

In [77]:
segmented_imgs = []
n_colors = (10, 8, 6, 4, 2)
for n_clusters in n_colors:
kmeans = KMeans(n_clusters=n_clusters, random_state=42).fit(X)
segmented_img = kmeans.cluster_centers_[kmeans.labels_]
segmented_imgs.append(segmented_img.reshape(image.shape))

In [78]:
plt.figure(figsize=(10,5))

plt.subplot(231)
plt.imshow(image)
plt.title("Original image")
plt.axis('off')

for idx, n_clusters in enumerate(n_colors):
plt.subplot(232 + idx)
plt.imshow(segmented_imgs[idx])
plt.title("{} colors".format(n_clusters))
plt.axis('off')

save_fig('image_segmentation_diagram', tight_layout=False)
plt.show()

Saving figure image_segmentation_diagram