Cluster points after KMeans clustering (scikit study)

Question

I did clustering using Kmeans using sklearn. While it has a method for printing centroids, I find it rather bizzare, that scikit-learn has no way to print out the cluster points of each cluster (or what I have not seen so far). Is there an easy way to get the cluster points of each cluster?

I currently have this rather complex code, where V is a dataset:

def getClusterPoints(V, labels): clusters = {} for l in range(0, max(labels)+1): data_points = [] indices = [i for i, x in enumerate(labels) if x == l] for idx in indices: data_points.append(V[idx]) clusters[l] = data_points return clusters 

Suggestions / links are greatly appreciated.

Thanks! PD.

Answer 1

for example

 import numpy as np from sklearn.cluster import KMeans from sklearn import datasets iris = datasets.load_iris() X = iris.data y = iris.target estimator = KMeans(n_clusters=3) estimator.fit(X) 

You can get clusters of each point on

 estimator.labels_ 

Of:

 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1], dtype=int32) 

Then we get point indices for each cluster

 {i: np.where(estimator.labels_ == i)[0] for i in range(estimator.n_clusters)} 

Of:

 {0: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49]), 1: array([ 50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 101, 106, 113, 114, 119, 121, 123, 126, 127, 133, 138, 142, 146, 149]), 2: array([ 52, 77, 100, 102, 103, 104, 105, 107, 108, 109, 110, 111, 112, 115, 116, 117, 118, 120, 122, 124, 125, 128, 129, 130, 131, 132, 134, 135, 136, 137, 139, 140, 141, 143, 144, 145, 147, 148])} 

Edit

If you want to use an array of points in X as values, not an array of indices:

 {i: X[np.where(estimator.labels_ == i)] for i in range(estimator.n_clusters)} 

Answer 2

If you read the documentation , you can see that kmeans has the labels_ attribute. This attribute provides clusters.

See the full example below:

 import matplotlib.pyplot as plt from sklearn.cluster import MiniBatchKMeans, KMeans from sklearn.metrics.pairwise import pairwise_distances_argmin from sklearn.datasets.samples_generator import make_blobs import numpy as np ############################################################################## # Generate sample data np.random.seed(0) batch_size = 45 centers = [[1, 1], [-1, -1], [1, -1]] n_clusters = len(centers) X, labels_true = make_blobs(n_samples=3000, centers=centers, cluster_std=0.7) ############################################################################## # Compute clustering with Means k_means = KMeans(init='k-means++', n_clusters=3, n_init=10) k_means.fit(X) ############################################################################## # Plot the results for i in set(k_means.labels_): index = k_means.labels_ == i plt.plot(X[index,0], X[index,1], 'o') plt.show()