CODEWITHSUNDEEP
pythonpandasscikit-learncluster-analysishierarchical-clustering
myradio
myradio

Reputation: 1775

Get the clusters from agglomerative for a given truncation

From the example of scikit-learn using scipy, (only changing the truncate_mode from 'level' to 'lastp'),

import numpy as np

from matplotlib import pyplot as plt
from scipy.cluster.hierarchy import dendrogram
from sklearn.datasets import load_iris
from sklearn.cluster import AgglomerativeClustering


def plot_dendrogram(model, **kwargs):
    # Create linkage matrix and then plot the dendrogram

    # create the counts of samples under each node
    counts = np.zeros(model.children_.shape[0])
    n_samples = len(model.labels_)
    for i, merge in enumerate(model.children_):
        current_count = 0
        for child_idx in merge:
            if child_idx < n_samples:
                current_count += 1  # leaf node
            else:
                current_count += counts[child_idx - n_samples]
        counts[i] = current_count

    linkage_matrix = np.column_stack(
        [model.children_, model.distances_, counts]
    ).astype(float)

    # Plot the corresponding dendrogram
    dendrogram(linkage_matrix, **kwargs)


iris = load_iris()
X = iris.data

# setting distance_threshold=0 ensures we compute the full tree.
model = AgglomerativeClustering(distance_threshold=0, n_clusters=None)

model = model.fit(X)
plt.title("Hierarchical Clustering Dendrogram")
# plot the top three levels of the dendrogram
plot_dendrogram(model, truncate_mode="lastp", p=10)
plt.xlabel("Number of points in node (or index of point if no parenthesis).")
plt.show()

I get,

dendogram example

Now, how can I get the elements that belong to each of the clusters for that specific truncation? I would like to know the indices of 21, 17, 12 etc elements that constitute each of those clusters.

Upvotes: 2

Views: 641

Answers (1)

Jykr
Jykr

Reputation: 1

This is bit hacky but you can get the full dendrogram with custom labels then obtain it from the return value of dendrogram.

# Fit model with desired number of clusters
model = AgglomerativeClustering(10, compute_full_tree=True, compute_distances=True).fit(X)

# Return dendrogram output
def plot_dendrogram(model, **kwargs):
    # Create linkage matrix and then plot the dendrogram

    # create the counts of samples under each node
    counts = np.zeros(model.children_.shape[0])
    n_samples = len(model.labels_)
    for i, merge in enumerate(model.children_):
        current_count = 0
        for child_idx in merge:
            if child_idx < n_samples:
                current_count += 1  # leaf node
            else:
                current_count += counts[child_idx - n_samples]
        counts[i] = current_count

    linkage_matrix = np.column_stack(
        [model.children_, model.distances_, counts]
    ).astype(float)

    # Plot the corresponding dendrogram
    d = dendrogram(linkage_matrix, no_labels=False, **kwargs)
    return d

# Plot dendrogram
d = plot_dendrogram(clustering, leaf_rotation=90, truncate_mode='lastp', p=10)

# Obtain full dendrogram output without plotting, provide clustering labels
d2 = plot_dendrogram(clustering, leaf_rotation=90, no_plot=True, labels=clustering.labels_)

# d2 includes sorted cluster labels
leaf_labels = np.array(d2['ivl'])
agg_labels = np.unique(leaf_labels, return_index=True)[1]
labels = [leaf_labels[idx] for idx in sorted(agg_labels)]

print(labels)
>>> [10, 4, 6, 1, 8, 11, 3, 7, 2, 5, 9, 0]

Upvotes: 0

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