# The $k$-Means Algorithm¶

Again, we start by generating some artificial data:

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plt.jet() # set the color map. When your colors are lost, re-run this.
import sklearn.datasets as datasets
X, Y = datasets.make_blobs(centers=4, cluster_std=0.5, random_state=0)


As always, we first plot the data to get a feeling of what we're dealing with:

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plt.scatter(X[:,0], X[:,1]); The data looks like it may contain four different "types" of data point.

In fact, this is how it was created above.

We can plot this information as well, using color:

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plt.scatter(X[:,0], X[:,1], c=Y); Normally, you do not know the information in Y, however.

You could try to recover it from the data alone.

This is what the kMeans algorithm does.

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from sklearn.cluster import KMeans
kmeans = KMeans(4, random_state=8)
Y_hat = kmeans.fit(X).labels_


Now the label assignments should be quite similar to Y, up to a different ordering of the colors:

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plt.scatter(X[:,0], X[:,1], c=Y_hat); Often, you're not so much interested in the assignments to the means.

You'll want to have a closer look at the means $\mu$.

The means in $\mu$ can be seen as representatives of their respective cluster.

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plt.scatter(X[:,0], X[:,1], c=Y_hat, alpha=0.4)
mu = kmeans.cluster_centers_
plt.scatter(mu[:,0], mu[:,1], s=100, c=np.unique(Y_hat))
print mu

[[-1.47935679  3.11716896]
[-1.26811733  7.76378266]
[ 1.99186903  0.96561071]
[ 0.92578447  4.32475792]] ## $k$-Means on Images¶

In this final example, we use the $k$-Means algorithm on the classical MNIST dataset.

The MNIST dataset contains images of hand-written digits.

Let's first fetch the dataset from the internet (which may take a while, note the asterisk [*]):

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from sklearn.datasets import fetch_mldata
from sklearn.cluster import KMeans
from sklearn.utils import shuffle
X_digits, _,_, Y_digits = fetch_mldata("MNIST Original").values() # fetch dataset from internet
X_digits, Y_digits = shuffle(X_digits,Y_digits) # shuffle dataset (which is ordered!)
X_digits = X_digits[-5000:]       # take only the last instances, to shorten runtime of KMeans


Let's have a look at some of the instances in the dataset we just loaded:

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plt.rc("image", cmap="binary") # use black/white palette for plotting
for i in xrange(10):
plt.subplot(2,5,i+1)
plt.imshow(X_digits[i].reshape(28,28))
plt.xticks(())
plt.yticks(())
plt.tight_layout() Warning: This takes quite a few seconds, so be patient until the asterisk [*] disappears!

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kmeans = KMeans(20)
mu_digits = kmeans.fit(X_digits).cluster_centers_


Let's have a closer look at the means. Even though there are 10 digits, some of them are over/under-represented. Do you understand why?

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plt.figure(figsize=(16,6))
for i in xrange(2*(mu_digits.shape/2)): # loop over all means
plt.subplot(2,mu_digits.shape/2,i+1)
plt.imshow(mu_digits[i].reshape(28,28))
plt.xticks(())
plt.yticks(())
plt.tight_layout() ## Playing with $k$-Means¶

This website allows you to play with the $k$-Means algorithm on another artificial dataset, the "Mickey Mouse" dataset.

You "step" through the algorithm by pressing the buttons in the correct order.

Try to get a feeling of how the algorithm proceeds.

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from IPython.display import HTML
HTML('<iframe src=http://icperformance.com/wp-content/demos/kmeansmouse.html width=500 height=450></iframe>')

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# Playing with this Notebook¶

Try to see what happens when you

• Increase the standard deviation of the clusters in this notebook
• Choose a "wrong" number of clusters by:

1. changing the number of clusters generated
2. changing the number of clusters used by KMeans
• What happens to result of the $k$-Means algorithm when you have multiplied one axis of the matrix $X$ with a large value?

For example, the 0-th axis with 100:

X[:,0] *= 100

Why does the result change?

• Combine the $k$-Means algorithm with the PCA algorithm

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