Python sklearn-K means how to get points in each cluster

Question

I am using the sklearn.cluster KMeans package. Once I finish clustering, if I need to know which values ​​were grouped together, how can I do this?

Say I had 100 data points, and KMeans gave me 5 clusters. Now I want to know which data points are in cluster 5. How can I do this.

Is there a function to give a cluster identifier and it will list all the data points in that cluster

Thanks.

Answer 1

I had a similar requirement and I use pandas to create a new data frame with a dataset index and column labels.

data = pd.read_csv('filename') km = KMeans(n_clusters=5).fit(data) cluster_map = pd.DataFrame() cluster_map['data_index'] = data.index.values cluster_map['cluster'] = km.labels_ 

Once a DataFrame is available, it’s easy enough to filter, For example, to filter out all data points in cluster 3

 cluster_map[cluster_map.cluster == 3] 

Answer 2

If you have a large dataset and need to retrieve clusters on demand, you will see some speedup with numpy.where . Here is an example of an iris dataset:

 from sklearn.cluster import KMeans from sklearn import datasets import numpy as np centers = [[1, 1], [-1, -1], [1, -1]] iris = datasets.load_iris() X = iris.data y = iris.target km = KMeans(n_clusters=3) km.fit(X) 

Define a function to retrieve the cluster_id indices you specified. (Here are two functions, for comparison, they both return the same value):

 def ClusterIndicesNumpy(clustNum, labels_array): #numpy return np.where(labels_array == clustNum)[0] def ClusterIndicesComp(clustNum, labels_array): #list comprehension return np.array([i for i, x in enumerate(labels_array) if x == clustNum]) 

Let's say you need all the samples in cluster 2 :

 ClusterIndicesNumpy(2, km.labels_) 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]) 

Numpy wins the test:

 %timeit ClusterIndicesNumpy(2,km.labels_) 100000 loops, best of 3: 4 µs per loop %timeit ClusterIndicesComp(2,km.labels_) 1000 loops, best of 3: 479 µs per loop 

Now you can extract all the data from your cluster 2 as follows:

 X[ClusterIndicesNumpy(2,km.labels_)] array([[ 6.9, 3.1, 4.9, 1.5], [ 6.7, 3. , 5. , 1.7], [ 6.3, 3.3, 6. , 2.5], ... #truncated 

Double check the first three indexes from the truncated array above:

 print X[52], km.labels_[52] print X[77], km.labels_[77] print X[100], km.labels_[100] [ 6.9 3.1 4.9 1.5] 2 [ 6.7 3. 5. 1.7] 2 [ 6.3 3.3 6. 2.5] 2 

Answer 3

You can see the labels_ attribute

for instance

 km = KMeans(2) km.fit([[1,2,3],[2,3,4],[5,6,7]]) print km.labels_ output: array([1, 1, 0], dtype=int32) 

As you can see, the first and second point is cluster 1 , the last point in cluster 0 .

Answer 4

To get the identifiers of points / samples / observations that are inside each cluster, do the following:

An example of using iris data and a good pythonic way:

 import numpy as np from sklearn.cluster import KMeans from sklearn import datasets np.random.seed(0) # Use Iris data iris = datasets.load_iris() X = iris.data y = iris.target # KMeans with 3 clusters clf = KMeans(n_clusters=3) clf.fit(X,y) #Coordinates of cluster centers with shape [n_clusters, n_features] clf.cluster_centers_ #Labels of each point clf.labels_ # Nice Pythonic way to get the indices of the points for each corresponding cluster mydict = {i: np.where(clf.labels_ == i)[0] for i in range(clf.n_clusters)} # Transform this dictionary into list (if you need a list as result) dictlist = [] for key, value in mydict.iteritems(): temp = [key,value] dictlist.append(temp) 

RESULTS

 #dict format {0: 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]), 1: 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]), 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])} # list format [[0, 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])], [1, 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])], [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])]] 

Answer 5

You can just store tags in an array. Convert an array to a data frame. Then combine the data that you used to create the K tools with the new data frame with clusters.

Display a data frame. Now you should see a row with the corresponding cluster. If you want to list all the data with a specific cluster, use something like data.loc [data ['cluster_label_name'] == 2], assuming your cluster 2 for now.