## Required Packages

import sys
import sklearn
import numpy as np
import matplotlib.pyplot as plt

%matplotlib inline


## Version Check

print('Python: {}'.format(sys.version))
print('Scikit-learn: {}'.format(sklearn.__version__))
print('NumPy: {}'.format(np.__version__))

Python: 3.7.6 (default, Jan  8 2020, 20:23:39) [MSC v.1916 64 bit (AMD64)]
Scikit-learn: 0.22.1
NumPy: 1.18.1


For the convenience, we will load the MNIST dataset from tensorflow Keras Library. Or you can download it directly from here.

import tensorflow as tf
from tensorflow.keras.datasets import mnist

(X_train, y_train), (X_test, y_test) = mnist.load_data()

# Print shape of dataset
print("Training: {}".format(X_train.shape))
print("Test: {}".format(X_test.shape))

Training: (60000, 28, 28)
Test: (10000, 28, 28)


As you can see, the original dataset contains 28x28x1 pixel image. Let's print it out, and what it looks like.

fig, axs = plt.subplots(3, 3, figsize = (12, 12))
plt.gray()

# loop through subplots and add mnist images
for i, ax in enumerate(axs.flat):
ax.imshow(X_train[i])
ax.axis('off')
ax.set_title('Number {}'.format(y_train[i]))

# display the figure
plt.show()


## Preprocessing

### Reshape

Images stored as NumPy arrays are 2-dimensional arrays. However, the K-means clustering algorithm provided by scikit-learn ingests 1-dimensional arrays; as a result, we will need to reshape each image. (in other words, we need to flatten the data)

Clustering algorithms almost always use 1-dimensional data. For example, if you were clustering a set of X, Y coordinates, each point would be passed to the clustering algorithm as a 1-dimensional array with a length of two (example: [2,4] or [-1, 4]). If you were using 3-dimensional data, the array would have a length of 3 (example: [2, 4, 1] or [-1, 4, 5]).

MNIST contains images that are 28 by 28 pixels; as a result, they will have a length of 784 once we reshape them into a 1-dimensional array.

X_train = X_train.reshape(len(X_train), -1)
print(X_train.shape)

(60000, 784)


### Normalization

Also, One approach to help training is normalization. In order to do this, we need to convert each pixel value into 0 to 1 range. The maximum value of pixel in grayscale is 255, so it can normalize it by dividing 255. Of course, its overall shape is same as before.

X_train = X_train.astype(np.float32) / 255.


## Applying K-means Clustering

Since the size of the MNIST dataset is quite large, we will use the mini-batch implementation of k-means clustering (MiniBatchKMeans) provided by scikit-learn. This will dramatically reduce the amount of time it takes to fit the algorithm to the data.

Here, we just choose the n_clusters argument to the n_digits(the size of unique labels, in our case, 10), and set the default parameters in MiniBatchKMeans.

And as you know that, K-means clustering is one of the unsupervised learning. That means it doesn't require any label to train.

from sklearn.cluster import MiniBatchKMeans

n_digits = len(np.unique(y_train))
print(n_digits)

10

kmeans = MiniBatchKMeans(n_clusters=n_digits)
kmeans.fit(X_train)

MiniBatchKMeans(batch_size=100, compute_labels=True, init='k-means++',
init_size=None, max_iter=100, max_no_improvement=10,
n_clusters=10, n_init=3, random_state=None,
reassignment_ratio=0.01, tol=0.0, verbose=0)

We can find the labels of each input that is generated from K means model.

kmeans.labels_

array([7, 8, 3, ..., 7, 9, 7])

But these are not real label of each image, since the output of the kmeans.labels_ is just group id for clustering. For example, 6 in kmeans.labels_ has similar features with another 6 in kmeans.labels_. There is no more meaning from the label.

To match it with real label, we can tackle the follow things:

• Combine each images in the same group
• Check Frequency distribution of actual labels (using np.bincount)
• Find the Maximum frequent label (through np.argmax), and set the label.
def infer_cluster_labels(kmeans, actual_labels):
"""
Associates most probable label with each cluster in KMeans model
returns: dictionary of clusters assigned to each label
"""

inferred_labels = {}

# Loop through the clusters
for i in range(kmeans.n_clusters):

# find index of points in cluster
labels = []
index = np.where(kmeans.labels_ == i)

# append actual labels for each point in cluster
labels.append(actual_labels[index])

# determine most common label
if len(labels[0]) == 1:
counts = np.bincount(labels[0])
else:
counts = np.bincount(np.squeeze(labels))

# assign the cluster to a value in the inferred_labels dictionary
if np.argmax(counts) in inferred_labels:
# append the new number to the existing array at this slot
inferred_labels[np.argmax(counts)].append(i)
else:
# create a new array in this slot
inferred_labels[np.argmax(counts)] = [i]

return inferred_labels

def infer_data_labels(X_labels, cluster_labels):
"""
Determines label for each array, depending on the cluster it has been assigned to.
returns: predicted labels for each array
"""

# empty array of len(X)
predicted_labels = np.zeros(len(X_labels)).astype(np.uint8)

for i, cluster in enumerate(X_labels):
for key, value in cluster_labels.items():
if cluster in value:
predicted_labels[i] = key

return predicted_labels

cluster_labels = infer_cluster_labels(kmeans, y_train)
X_clusters = kmeans.predict(X_train)
predicted_labels = infer_data_labels(X_clusters, cluster_labels)
print(predicted_labels[:20])
print(y_train[:20])

[8 0 4 1 7 2 1 8 1 7 3 1 3 6 1 7 2 8 6 7]
[5 0 4 1 9 2 1 3 1 4 3 5 3 6 1 7 2 8 6 9]


As a result, some predicted label is mismatched, but most of case, the k-means model can correctly cluster of each group.

## Evaluating Clustering Algorithm

With the functions defined above, we can now determine the accuracy of our algorithms. Since we are using this clustering algorithm for classification, accuracy is ultimately the most important metric; however, there are other metrics out there that can be applied directly to the clusters themselves, regardless of the associated labels. Two of these metrics that we will use are inertia and homogeneity. (See the detailed description of homogeneity_score)

Furthermore, earlier we made the assumption that K = 10 was the appropriate number of clusters; however, this might not be the case. Let's fit the K-means clustering algorithm with several different values of K, than evaluate the performance using our metrics.

from sklearn.metrics import homogeneity_score

def calc_metrics(estimator, data, labels):
print('Number of Clusters: {}'.format(estimator.n_clusters))
# Inertia
inertia = estimator.inertia_
print("Inertia: {}".format(inertia))
# Homogeneity Score
homogeneity = homogeneity_score(labels, estimator.labels_)
print("Homogeneity score: {}".format(homogeneity))
return inertia, homogeneity

from sklearn.metrics import accuracy_score

clusters = [10, 16, 36, 64, 144, 256]
iner_list = []
homo_list = []
acc_list = []

for n_clusters in clusters:
estimator = MiniBatchKMeans(n_clusters=n_clusters)
estimator.fit(X_train)

inertia, homo = calc_metrics(estimator, X_train, y_train)
iner_list.append(inertia)
homo_list.append(homo)

# Determine predicted labels
cluster_labels = infer_cluster_labels(estimator, y_train)
prediction = infer_data_labels(estimator.labels_, cluster_labels)

acc = accuracy_score(y_train, prediction)
acc_list.append(acc)
print('Accuracy: {}\n'.format(acc))

Number of Clusters: 10
Inertia: 2383375.0
Homogeneity score: 0.46576292303121536
Accuracy: 0.56465

Number of Clusters: 16
Inertia: 2208197.5
Homogeneity score: 0.5531322770474518
Accuracy: 0.65095

Number of Clusters: 36
Inertia: 1961340.875
Homogeneity score: 0.6783212163972349
Accuracy: 0.767

Number of Clusters: 64
Inertia: 1822361.625
Homogeneity score: 0.727585914263205
Accuracy: 0.7895166666666666

Number of Clusters: 144
Inertia: 1635514.25
Homogeneity score: 0.8048996371912126
Accuracy: 0.8673833333333333

Number of Clusters: 256
Inertia: 1519708.25
Homogeneity score: 0.8428113183818001
Accuracy: 0.9000333333333334


fig, ax = plt.subplots(1, 2, figsize=(16, 10))
ax[0].plot(clusters, iner_list, label='inertia', marker='o')
ax[1].plot(clusters, homo_list, label='homogeneity', marker='o')
ax[1].plot(clusters, acc_list, label='accuracy', marker='^')
ax[0].legend(loc='best')
ax[1].legend(loc='best')
ax[0].grid('on')
ax[1].grid('on')
ax[0].set_title('Inertia of each clusters')
ax[1].set_title('Homogeneity and Accuracy of each clusters')
plt.show()


As a result, we found out that when the K value is increased, the accuracy and homogeneity is also increased. We can also check the performance on test dataset.

X_test = X_test.reshape(len(X_test), -1)
X_test = X_test.astype(np.float32) / 255.

kmeans = MiniBatchKMeans(n_clusters=256)
kmeans.fit(X_test)

cluster_labels = infer_cluster_labels(kmeans, y_test)

test_clusters = kmeans.predict(X_test)
prediction = infer_data_labels(kmeans.predict(X_test), cluster_labels)
print('Accuracy: {}'.format(accuracy_score(y_test, prediction)))

Accuracy: 0.8877


There we have MiniBatchKmeans Clustering model with almost 90% accuracy. One definite way to check the model performance is to visualize the real image.

For the convenience, we decrease the n_clusters to 36.

kmeans = MiniBatchKMeans(n_clusters = 36)
kmeans.fit(X_test)

# record centroid values
centroids = kmeans.cluster_centers_

# reshape centroids into images
images = centroids.reshape(36, 28, 28)
images *= 255
images = images.astype(np.uint8)

# determine cluster labels
cluster_labels = infer_cluster_labels(kmeans, y_test)
prediction = infer_data_labels(kmeans.predict(X_test), cluster_labels)

# create figure with subplots using matplotlib.pyplot
fig, axs = plt.subplots(6, 6, figsize = (20, 20))
plt.gray()

# loop through subplots and add centroid images
for i, ax in enumerate(axs.flat):

# determine inferred label using cluster_labels dictionary
for key, value in cluster_labels.items():
if i in value:
ax.set_title('Inferred Label: {}'.format(key), color='blue')