We Keep Coding

How to find the index of the point closest to K means cluster centers using sklearn?

I have used python's sklearn package for K-means clustering. So far I am able to get the coordinates of the cluster centers using the following code.
import numpy as np
from sklearn.cluster import KMeans
p50 = np.load('tsnep400.npy')
kmeans = KMeans(n_clusters=50).fit(p50)
np.savetxt('kmeans_50clusters_centers_tsnep400', kmeans.cluster_centers_, fmt='%1.3f')
np.savetxt('kmeans_50clusters_tsnep400.dat', kmeans.labels_, fmt='%1.1d')
centroids = {i: np.where(kmeans.labels_ == i)[0] for i in range(kmeans.n_clusters)}
np.save('kmeans_50clusters_memebers_tsnep400.npy',centroids)
How do I find the index of the point closest to cluster centers?

According to the scikit-learn documentation, the attribute .labels_ contains the labels of each point, by their index. Thus, you can use this to group each of your points into a cluster and then calculate the distance to each cluster center. You can use the following code for this:
from scipy.spatial.distance import euclidean
# Loop over all clusters and find index of closest point to the cluster center and append to closest_pt_idx list.
closest_pt_idx = []
for iclust in range(kmeans.n_clusters):
# get all points assigned to each cluster:
cluster_pts = p50[kmeans.labels_ == iclust]
# get all indices of points assigned to this cluster:
cluster_pts_indices = np.where(kmeans.labels_ == iclust)[0]
cluster_cen = kmeans.cluster_centers_[iclust]
min_idx = np.argmin([euclidean(p50[idx], cluster_cen) for idx in cluster_pts_indices])
# Testing:
print('closest point to cluster center: ', cluster_pts[min_idx])
print('closest index of point to cluster center: ', cluster_pts_indices[min_idx])
print(' ', p50[cluster_pts_indices[min_idx]])
closest_pt_idx.append(cluster_pts_indices[min_idx])

How to get the optimal number of clusters using hierarchical cluster analysis automatically in python?

I want to use hierarchical cluster analysis to get the optimal number (K) of clusters automatically, then apply this K to K-means clustering in python.
After studying many article, I know some methods tell us that we can plot the graph to determine K, but have any methods can output a real number automatically in python?

The hierarchical clustering method is based on dendrogram to determine the optimal number of clusters. Plot the dendrogram using a code similar to the following:
# General imports
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
# Special imports
from scipy.cluster.hierarchy import dendrogram, linkage
# Load data, fill in appropriately
X = []
# How to cluster the data, single is minimal distance between clusters
linked = linkage(X, 'single')
# Plot dendrogram
plt.figure(figsize=(10, 7))
dendrogram(linked,
orientation='top',
labels=labelList,
distance_sort='descending',
show_leaf_counts=True)
plt.show()
In the dendrogram locate the largest vertical difference between nodes, and in the middle pass an horizontal line. The number of vertical lines intersecting it is the optimal number of clusters (when affinity is calculated using the method set in linkage).
See example here: https://stackabuse.com/hierarchical-clustering-with-python-and-scikit-learn/
How to automatically read a dendrogram and extract that number is something I would also like to know.
Added in edit:
There is a way to do so using SK Learn package. See the following example:
#==========================================================================
# Hierarchical Clustering - Automatic determination of number of clusters
#==========================================================================
# General imports
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from os import path
# Special imports
from scipy.cluster.hierarchy import dendrogram, linkage
import scipy.cluster.hierarchy as shc
from sklearn.cluster import AgglomerativeClustering
# %matplotlib inline
print("============================================================")
print(" Hierarchical Clustering demo - num of clusters ")
print("============================================================")
print(" ")
folder = path.dirname(path.realpath(__file__)) # set current folder
# Load data
customer_data = pd.read_csv( path.join(folder, "hierarchical-clustering-with-python-and-scikit-learn-shopping-data.csv"))
# print(customer_data.shape)
print("In this data there should be 5 clusters...")
# Retain only the last two columns
data = customer_data.iloc[:, 3:5].values
# # Plot dendrogram using SciPy
# plt.figure(figsize=(10, 7))
# plt.title("Customer Dendograms")
# dend = shc.dendrogram(shc.linkage(data, method='ward'))
# plt.show()
# Initialize hiererchial clustering method, in order for the algorithm to determine the number of clusters
# put n_clusters=None, compute_full_tree = True,
# best distance threshold value for this dataset is distance_threshold = 200
cluster = AgglomerativeClustering(n_clusters=None, affinity='euclidean', linkage='ward', compute_full_tree=True, distance_threshold=200)
# Cluster the data
cluster.fit_predict(data)
print(f"Number of clusters = {1+np.amax(cluster.labels_)}")
# Display the clustering, assigning cluster label to every datapoint
print("Classifying the points into clusters:")
print(cluster.labels_)
# Display the clustering graphically in a plot
plt.scatter(data[:,0],data[:,1], c=cluster.labels_, cmap='rainbow')
plt.title(f"SK Learn estimated number of clusters = {1+np.amax(cluster.labels_)}")
plt.show()
print(" ")
The data was taken from here: https://stackabuse.s3.amazonaws.com/files/hierarchical-clustering-with-python-and-scikit-learn-shopping-data.csv

I found a solution I am using in my code. It involves color_list that counts amount of numbers of "connections". If one wants to extract the number of "leaves" (clusters) just decrease the number by 1:
https://www.youtube.com/watch?v=4DInt3H2UNE

Get the cluster size in sklearn in python

I am using sklearn DBSCAN to cluster my data as follows.
#Apply DBSCAN (sims == my data as list of lists)
db1 = DBSCAN(min_samples=1, metric='precomputed').fit(sims)
db1_labels = db1.labels_
db1n_clusters_ = len(set(db1_labels)) - (1 if -1 in db1_labels else 0)
#Returns the number of clusters (E.g., 10 clusters)
print('Estimated number of clusters: %d' % db1n_clusters_)
Now I want to get the top 3 clusters sorted from the size (number of data points in each cluster). Please let me know how to obtain the cluster size in sklearn?

Another option would be to use numpy.unique:
db1_labels = db1.labels_
labels, counts = np.unique(db1_labels[db1_labels>=0], return_counts=True)
print labels[np.argsort(-counts)[:3]]

Well you can Bincount Function in Numpy to get the frequencies of labels. For example, we will use the example for DBSCAN using scikit-learn:
#Store the labels
labels = db.labels_
#Then get the frequency count of the non-negative labels
counts = np.bincount(labels[labels>=0])
print counts
#Output : [243 244 245]
Then to get the top 3 values use argsort in numpy. In our example since there are only 3 clusters, I will extract the top 2 values :
top_labels = np.argsort(-counts)[:2]
print top_labels
#Output : [2 1]
#To get their respective frequencies
print counts[top_labels]

Gap Statistic Method

import sys
import numpy as np
import scipy.io as sio
import matplotlib.pyplot as plt
from sklearn.cluster import KMeans
from sklearn.svm import SVC
filename = sys.argv[1]
datafile = sio.loadmat(filename)
data = datafile['bow']
sizedata=[len(data), len(data[0])]
gap=[]
SD=[]
for knum in xrange(10,20):
print knum
#Clustering original Data
kmeanspp = KMeans(n_clusters=knum,init = 'k-means++',max_iter = 100,n_jobs = 1)
kmeanspp.fit(data)
dispersion = kmeanspp.inertia_
#Clustering Reference Data
nrefs = 10
refDisp = np.zeros(nrefs)
for nref in xrange(nrefs):
refdata = np.random.random_sample((sizedata[0],sizedata[1]))
refkmeans = KMeans(n_clusters=knum,init='k-means++',max_iter=100,n_jobs=1)
refkmeans.fit(refdata)
refdisp = refkmeans.inertia_
refDisp[nref]=np.log(refdisp)
mean_log_refdisp = np.mean(refDisp)
gap.append(mean_log_refdisp-np.log(dispersion))
#Calculating standard deviaiton
sd = (sum([(r-m)**2 for r,m in zip(refDisp,[mean_log_refdisp]*nrefs)])/nrefs)**0.5
SD.append(sd)
SD = [sd*((1+(1/nrefs))**0.5) for sd in SD]
#determining optimal k
opt_k = None
diff = []
for i in xrange(len(gap)-1):
diff = (SD[i+1]-(gap[i+1]-gap[i]))
if diff>0:
opt_k = i+10
break
print diff
plt.plot(np.linspace(10,19,10,True),gap)
plt.show()
Here I am trying to implement the Gap Statistic method for determining the optimal number of clusters. But the problem is that every time I run the code I get a different value for k.
What is the solution to the problem?
How can the value of optimal k differ for the same data?
I have stored the data in a .mat file beforehand and I am passing it as an argument via terminal
I am looking for the smallest value of k for which Gap(k)>= Gap(k+1)-s(k+1) where s(k+1) = sd(k+1)*square_root(1+(1/B)) where sd is the standard deviation of the reference distribution and B is the number of copies of Monte Carlo sample
Otherwise stated, I am searching for the value of k for which
s(k+1)-Gap(k+1)+Gap(k)>=0

Couple of problems with your simulation:
1- sd = (sum([(r-m)**2 for r,m in zip(refDisp,[mean_log_refdisp]*nrefs)])/nrefs)**0.5
Why did you multiply the second component of zip by nrefs that is not needed according to the original paper.
2-
if diff>0:
opt_k = i+10
break
if diff>0 you want diff>=0 since equality can happen a
About why you get different number of clusters each time, as people said it is monte carlo simulation so there can be randomness and also it depends on what you are clustering and your dataset. I suggest you to test your algorithms against Silhouette and Elbow to get a better idea about number of clusters.

One option is to run your function several times and then average the gap statistics and the s values, and find the smallest k where the average s(k+1)-Gap(k+1)+Gap(k) is greater than
This will take longer but give a more reliable result.

sklearn agglomerative clustering linkage matrix

I'm trying to draw a complete-link scipy.cluster.hierarchy.dendrogram, and I found that scipy.cluster.hierarchy.linkage is slower than sklearn.AgglomerativeClustering.
However, sklearn.AgglomerativeClustering doesn't return the distance between clusters and the number of original observations, which scipy.cluster.hierarchy.dendrogram needs. Is there a way to take them?

It's possible, but it isn't pretty. It requires (at a minimum) a small rewrite of AgglomerativeClustering.fit (source). The difficulty is that the method requires a number of imports, so it ends up getting a bit nasty looking. To add in this feature:
Insert the following line after line 748:
kwargs['return_distance'] = True
Replace line 752 with:
self.children_, self.n_components_, self.n_leaves_, parents, self.distance = \
This will give you a new attribute, distance, that you can easily call.
A couple things to note:
When doing this, I ran into this issue about the check_array function on line 711. This can be fixed by using check_arrays (from sklearn.utils.validation import check_arrays). You can modify that line to become X = check_arrays(X)[0]. This appears to be a bug (I still have this issue on the most recent version of scikit-learn).
Depending on which version of sklearn.cluster.hierarchical.linkage_tree you have, you may also need to modify it to be the one provided in the source.
To make things easier for everyone, here is the full code that you will need to use:
from heapq import heapify, heappop, heappush, heappushpop
import warnings
import sys
import numpy as np
from scipy import sparse
from sklearn.base import BaseEstimator, ClusterMixin
from sklearn.externals.joblib import Memory
from sklearn.externals import six
from sklearn.utils.validation import check_arrays
from sklearn.utils.sparsetools import connected_components
from sklearn.cluster import _hierarchical
from sklearn.cluster.hierarchical import ward_tree
from sklearn.cluster._feature_agglomeration import AgglomerationTransform
from sklearn.utils.fast_dict import IntFloatDict
def _fix_connectivity(X, connectivity, n_components=None,
affinity="euclidean"):
"""
Fixes the connectivity matrix
- copies it
- makes it symmetric
- converts it to LIL if necessary
- completes it if necessary
"""
n_samples = X.shape[0]
if (connectivity.shape[0] != n_samples or
connectivity.shape[1] != n_samples):
raise ValueError('Wrong shape for connectivity matrix: %s '
'when X is %s' % (connectivity.shape, X.shape))
# Make the connectivity matrix symmetric:
connectivity = connectivity + connectivity.T
# Convert connectivity matrix to LIL
if not sparse.isspmatrix_lil(connectivity):
if not sparse.isspmatrix(connectivity):
connectivity = sparse.lil_matrix(connectivity)
else:
connectivity = connectivity.tolil()
# Compute the number of nodes
n_components, labels = connected_components(connectivity)
if n_components > 1:
warnings.warn("the number of connected components of the "
"connectivity matrix is %d > 1. Completing it to avoid "
"stopping the tree early." % n_components,
stacklevel=2)
# XXX: Can we do without completing the matrix?
for i in xrange(n_components):
idx_i = np.where(labels == i)[0]
Xi = X[idx_i]
for j in xrange(i):
idx_j = np.where(labels == j)[0]
Xj = X[idx_j]
D = pairwise_distances(Xi, Xj, metric=affinity)
ii, jj = np.where(D == np.min(D))
ii = ii[0]
jj = jj[0]
connectivity[idx_i[ii], idx_j[jj]] = True
connectivity[idx_j[jj], idx_i[ii]] = True
return connectivity, n_components
# average and complete linkage
def linkage_tree(X, connectivity=None, n_components=None,
n_clusters=None, linkage='complete', affinity="euclidean",
return_distance=False):
"""Linkage agglomerative clustering based on a Feature matrix.
The inertia matrix uses a Heapq-based representation.
This is the structured version, that takes into account some topological
structure between samples.
Parameters
----------
X : array, shape (n_samples, n_features)
feature matrix representing n_samples samples to be clustered
connectivity : sparse matrix (optional).
connectivity matrix. Defines for each sample the neighboring samples
following a given structure of the data. The matrix is assumed to
be symmetric and only the upper triangular half is used.
Default is None, i.e, the Ward algorithm is unstructured.
n_components : int (optional)
Number of connected components. If None the number of connected
components is estimated from the connectivity matrix.
NOTE: This parameter is now directly determined directly
from the connectivity matrix and will be removed in 0.18
n_clusters : int (optional)
Stop early the construction of the tree at n_clusters. This is
useful to decrease computation time if the number of clusters is
not small compared to the number of samples. In this case, the
complete tree is not computed, thus the 'children' output is of
limited use, and the 'parents' output should rather be used.
This option is valid only when specifying a connectivity matrix.
linkage : {"average", "complete"}, optional, default: "complete"
Which linkage critera to use. The linkage criterion determines which
distance to use between sets of observation.
- average uses the average of the distances of each observation of
the two sets
- complete or maximum linkage uses the maximum distances between
all observations of the two sets.
affinity : string or callable, optional, default: "euclidean".
which metric to use. Can be "euclidean", "manhattan", or any
distance know to paired distance (see metric.pairwise)
return_distance : bool, default False
whether or not to return the distances between the clusters.
Returns
-------
children : 2D array, shape (n_nodes-1, 2)
The children of each non-leaf node. Values less than `n_samples`
correspond to leaves of the tree which are the original samples.
A node `i` greater than or equal to `n_samples` is a non-leaf
node and has children `children_[i - n_samples]`. Alternatively
at the i-th iteration, children[i][0] and children[i][1]
are merged to form node `n_samples + i`
n_components : int
The number of connected components in the graph.
n_leaves : int
The number of leaves in the tree.
parents : 1D array, shape (n_nodes, ) or None
The parent of each node. Only returned when a connectivity matrix
is specified, elsewhere 'None' is returned.
distances : ndarray, shape (n_nodes-1,)
Returned when return_distance is set to True.
distances[i] refers to the distance between children[i][0] and
children[i][1] when they are merged.
See also
--------
ward_tree : hierarchical clustering with ward linkage
"""
X = np.asarray(X)
if X.ndim == 1:
X = np.reshape(X, (-1, 1))
n_samples, n_features = X.shape
linkage_choices = {'complete': _hierarchical.max_merge,
'average': _hierarchical.average_merge,
}
try:
join_func = linkage_choices[linkage]
except KeyError:
raise ValueError(
'Unknown linkage option, linkage should be one '
'of %s, but %s was given' % (linkage_choices.keys(), linkage))
if connectivity is None:
from scipy.cluster import hierarchy # imports PIL
if n_clusters is not None:
warnings.warn('Partial build of the tree is implemented '
'only for structured clustering (i.e. with '
'explicit connectivity). The algorithm '
'will build the full tree and only '
'retain the lower branches required '
'for the specified number of clusters',
stacklevel=2)
if affinity == 'precomputed':
# for the linkage function of hierarchy to work on precomputed
# data, provide as first argument an ndarray of the shape returned
# by pdist: it is a flat array containing the upper triangular of
# the distance matrix.
i, j = np.triu_indices(X.shape[0], k=1)
X = X[i, j]
elif affinity == 'l2':
# Translate to something understood by scipy
affinity = 'euclidean'
elif affinity in ('l1', 'manhattan'):
affinity = 'cityblock'
elif callable(affinity):
X = affinity(X)
i, j = np.triu_indices(X.shape[0], k=1)
X = X[i, j]
out = hierarchy.linkage(X, method=linkage, metric=affinity)
children_ = out[:, :2].astype(np.int)
if return_distance:
distances = out[:, 2]
return children_, 1, n_samples, None, distances
return children_, 1, n_samples, None
if n_components is not None:
warnings.warn(
"n_components is now directly calculated from the connectivity "
"matrix and will be removed in 0.18",
DeprecationWarning)
connectivity, n_components = _fix_connectivity(X, connectivity)
connectivity = connectivity.tocoo()
# Put the diagonal to zero
diag_mask = (connectivity.row != connectivity.col)
connectivity.row = connectivity.row[diag_mask]
connectivity.col = connectivity.col[diag_mask]
connectivity.data = connectivity.data[diag_mask]
del diag_mask
if affinity == 'precomputed':
distances = X[connectivity.row, connectivity.col]
else:
# FIXME We compute all the distances, while we could have only computed
# the "interesting" distances
distances = paired_distances(X[connectivity.row],
X[connectivity.col],
metric=affinity)
connectivity.data = distances
if n_clusters is None:
n_nodes = 2 * n_samples - 1
else:
assert n_clusters <= n_samples
n_nodes = 2 * n_samples - n_clusters
if return_distance:
distances = np.empty(n_nodes - n_samples)
# create inertia heap and connection matrix
A = np.empty(n_nodes, dtype=object)
inertia = list()
# LIL seems to the best format to access the rows quickly,
# without the numpy overhead of slicing CSR indices and data.
connectivity = connectivity.tolil()
# We are storing the graph in a list of IntFloatDict
for ind, (data, row) in enumerate(zip(connectivity.data,
connectivity.rows)):
A[ind] = IntFloatDict(np.asarray(row, dtype=np.intp),
np.asarray(data, dtype=np.float64))
# We keep only the upper triangular for the heap
# Generator expressions are faster than arrays on the following
inertia.extend(_hierarchical.WeightedEdge(d, ind, r)
for r, d in zip(row, data) if r < ind)
del connectivity
heapify(inertia)
# prepare the main fields
parent = np.arange(n_nodes, dtype=np.intp)
used_node = np.ones(n_nodes, dtype=np.intp)
children = []
# recursive merge loop
for k in xrange(n_samples, n_nodes):
# identify the merge
while True:
edge = heappop(inertia)
if used_node[edge.a] and used_node[edge.b]:
break
i = edge.a
j = edge.b
if return_distance:
# store distances
distances[k - n_samples] = edge.weight
parent[i] = parent[j] = k
children.append((i, j))
# Keep track of the number of elements per cluster
n_i = used_node[i]
n_j = used_node[j]
used_node[k] = n_i + n_j
used_node[i] = used_node[j] = False
# update the structure matrix A and the inertia matrix
# a clever 'min', or 'max' operation between A[i] and A[j]
coord_col = join_func(A[i], A[j], used_node, n_i, n_j)
for l, d in coord_col:
A[l].append(k, d)
# Here we use the information from coord_col (containing the
# distances) to update the heap
heappush(inertia, _hierarchical.WeightedEdge(d, k, l))
A[k] = coord_col
# Clear A[i] and A[j] to save memory
A[i] = A[j] = 0
# Separate leaves in children (empty lists up to now)
n_leaves = n_samples
# # return numpy array for efficient caching
children = np.array(children)[:, ::-1]
if return_distance:
return children, n_components, n_leaves, parent, distances
return children, n_components, n_leaves, parent
# Matching names to tree-building strategies
def _complete_linkage(*args, **kwargs):
kwargs['linkage'] = 'complete'
return linkage_tree(*args, **kwargs)
def _average_linkage(*args, **kwargs):
kwargs['linkage'] = 'average'
return linkage_tree(*args, **kwargs)
_TREE_BUILDERS = dict(
ward=ward_tree,
complete=_complete_linkage,
average=_average_linkage,
)
def _hc_cut(n_clusters, children, n_leaves):
"""Function cutting the ward tree for a given number of clusters.
Parameters
----------
n_clusters : int or ndarray
The number of clusters to form.
children : list of pairs. Length of n_nodes
The children of each non-leaf node. Values less than `n_samples` refer
to leaves of the tree. A greater value `i` indicates a node with
children `children[i - n_samples]`.
n_leaves : int
Number of leaves of the tree.
Returns
-------
labels : array [n_samples]
cluster labels for each point
"""
if n_clusters > n_leaves:
raise ValueError('Cannot extract more clusters than samples: '
'%s clusters where given for a tree with %s leaves.'
% (n_clusters, n_leaves))
# In this function, we store nodes as a heap to avoid recomputing
# the max of the nodes: the first element is always the smallest
# We use negated indices as heaps work on smallest elements, and we
# are interested in largest elements
# children[-1] is the root of the tree
nodes = [-(max(children[-1]) + 1)]
for i in xrange(n_clusters - 1):
# As we have a heap, nodes[0] is the smallest element
these_children = children[-nodes[0] - n_leaves]
# Insert the 2 children and remove the largest node
heappush(nodes, -these_children[0])
heappushpop(nodes, -these_children[1])
label = np.zeros(n_leaves, dtype=np.intp)
for i, node in enumerate(nodes):
label[_hierarchical._hc_get_descendent(-node, children, n_leaves)] = i
return label
class AgglomerativeClustering(BaseEstimator, ClusterMixin):
"""
Agglomerative Clustering
Recursively merges the pair of clusters that minimally increases
a given linkage distance.
Parameters
----------
n_clusters : int, default=2
The number of clusters to find.
connectivity : array-like or callable, optional
Connectivity matrix. Defines for each sample the neighboring
samples following a given structure of the data.
This can be a connectivity matrix itself or a callable that transforms
the data into a connectivity matrix, such as derived from
kneighbors_graph. Default is None, i.e, the
hierarchical clustering algorithm is unstructured.
affinity : string or callable, default: "euclidean"
Metric used to compute the linkage. Can be "euclidean", "l1", "l2",
"manhattan", "cosine", or 'precomputed'.
If linkage is "ward", only "euclidean" is accepted.
memory : Instance of joblib.Memory or string (optional)
Used to cache the output of the computation of the tree.
By default, no caching is done. If a string is given, it is the
path to the caching directory.
n_components : int (optional)
Number of connected components. If None the number of connected
components is estimated from the connectivity matrix.
NOTE: This parameter is now directly determined from the connectivity
matrix and will be removed in 0.18
compute_full_tree : bool or 'auto' (optional)
Stop early the construction of the tree at n_clusters. This is
useful to decrease computation time if the number of clusters is
not small compared to the number of samples. This option is
useful only when specifying a connectivity matrix. Note also that
when varying the number of clusters and using caching, it may
be advantageous to compute the full tree.
linkage : {"ward", "complete", "average"}, optional, default: "ward"
Which linkage criterion to use. The linkage criterion determines which
distance to use between sets of observation. The algorithm will merge
the pairs of cluster that minimize this criterion.
- ward minimizes the variance of the clusters being merged.
- average uses the average of the distances of each observation of
the two sets.
- complete or maximum linkage uses the maximum distances between
all observations of the two sets.
pooling_func : callable, default=np.mean
This combines the values of agglomerated features into a single
value, and should accept an array of shape [M, N] and the keyword
argument ``axis=1``, and reduce it to an array of size [M].
Attributes
----------
labels_ : array [n_samples]
cluster labels for each point
n_leaves_ : int
Number of leaves in the hierarchical tree.
n_components_ : int
The estimated number of connected components in the graph.
children_ : array-like, shape (n_nodes-1, 2)
The children of each non-leaf node. Values less than `n_samples`
correspond to leaves of the tree which are the original samples.
A node `i` greater than or equal to `n_samples` is a non-leaf
node and has children `children_[i - n_samples]`. Alternatively
at the i-th iteration, children[i][0] and children[i][1]
are merged to form node `n_samples + i`
"""
def __init__(self, n_clusters=2, affinity="euclidean",
memory=Memory(cachedir=None, verbose=0),
connectivity=None, n_components=None,
compute_full_tree='auto', linkage='ward',
pooling_func=np.mean):
self.n_clusters = n_clusters
self.memory = memory
self.n_components = n_components
self.connectivity = connectivity
self.compute_full_tree = compute_full_tree
self.linkage = linkage
self.affinity = affinity
self.pooling_func = pooling_func
def fit(self, X, y=None):
"""Fit the hierarchical clustering on the data
Parameters
----------
X : array-like, shape = [n_samples, n_features]
The samples a.k.a. observations.
Returns
-------
self
"""
X = check_arrays(X)[0]
memory = self.memory
if isinstance(memory, six.string_types):
memory = Memory(cachedir=memory, verbose=0)
if self.linkage == "ward" and self.affinity != "euclidean":
raise ValueError("%s was provided as affinity. Ward can only "
"work with euclidean distances." %
(self.affinity, ))
if self.linkage not in _TREE_BUILDERS:
raise ValueError("Unknown linkage type %s."
"Valid options are %s" % (self.linkage,
_TREE_BUILDERS.keys()))
tree_builder = _TREE_BUILDERS[self.linkage]
connectivity = self.connectivity
if self.connectivity is not None:
if callable(self.connectivity):
connectivity = self.connectivity(X)
connectivity = check_arrays(
connectivity, accept_sparse=['csr', 'coo', 'lil'])
n_samples = len(X)
compute_full_tree = self.compute_full_tree
if self.connectivity is None:
compute_full_tree = True
if compute_full_tree == 'auto':
# Early stopping is likely to give a speed up only for
# a large number of clusters. The actual threshold
# implemented here is heuristic
compute_full_tree = self.n_clusters < max(100, .02 * n_samples)
n_clusters = self.n_clusters
if compute_full_tree:
n_clusters = None
# Construct the tree
kwargs = {}
kwargs['return_distance'] = True
if self.linkage != 'ward':
kwargs['linkage'] = self.linkage
kwargs['affinity'] = self.affinity
self.children_, self.n_components_, self.n_leaves_, parents, \
self.distance = memory.cache(tree_builder)(X, connectivity,
n_components=self.n_components,
n_clusters=n_clusters,
**kwargs)
# Cut the tree
if compute_full_tree:
self.labels_ = _hc_cut(self.n_clusters, self.children_,
self.n_leaves_)
else:
labels = _hierarchical.hc_get_heads(parents, copy=False)
# copy to avoid holding a reference on the original array
labels = np.copy(labels[:n_samples])
# Reasign cluster numbers
self.labels_ = np.searchsorted(np.unique(labels), labels)
return self
Below is a simple example showing how to use the modified AgglomerativeClustering class:
import numpy as np
import AgglomerativeClustering # Make sure to use the new one!!!
d = np.array(
[
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
)
clustering = AgglomerativeClustering(n_clusters=2, compute_full_tree=True,
affinity='euclidean', linkage='complete')
clustering.fit(d)
print clustering.distance
That example has the following output:
[ 5.19615242 10.39230485]
This can then be compared to a scipy.cluster.hierarchy.linkage implementation:
import numpy as np
from scipy.cluster.hierarchy import linkage
d = np.array(
[
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
)
print linkage(d, 'complete')
Output:
[[ 1. 2. 5.19615242 2. ]
[ 0. 3. 10.39230485 3. ]]
Just for kicks I decided to follow up on your statement about performance:
import AgglomerativeClustering
from scipy.cluster.hierarchy import linkage
import numpy as np
import time
l = 1000; iters = 50
d = [np.random.random(100) for _ in xrange(1000)]
t = time.time()
for _ in xrange(iters):
clustering = AgglomerativeClustering(n_clusters=l-1,
affinity='euclidean', linkage='complete')
clustering.fit(d)
scikit_time = (time.time() - t) / iters
print 'scikit-learn Time: {0}s'.format(scikit_time)
t = time.time()
for _ in xrange(iters):
linkage(d, 'complete')
scipy_time = (time.time() - t) / iters
print 'SciPy Time: {0}s'.format(scipy_time)
print 'scikit-learn Speedup: {0}'.format(scipy_time / scikit_time)
This gave me the following results:
scikit-learn Time: 0.566560001373s
SciPy Time: 0.497740001678s
scikit-learn Speedup: 0.878530077083
According to this, the implementation from Scikit-Learn takes 0.88x the execution time of the SciPy implementation, i.e. SciPy's implementation is 1.14x faster. It should be noted that:
I modified the original scikit-learn implementation
I only did a small number of iterations
I only tested a small number of test cases (both cluster size as well as number of items per dimension should be tested)
I ran SciPy second, so it is had the advantage of obtaining more cache hits on the source data
The two methods don't exactly do the same thing.
With all of that in mind, you should really evaluate which method performs better for your specific application. There are also functional reasons to go with one implementation over the other.

I made a scipt to do it without modifying sklearn and without recursive functions. Before using note that:
Merge distance can sometimes decrease with respect to the children
merge distance. I added three ways to handle those cases: Take the
max, do nothing or increase with the l2 norm. The l2 norm logic has not been verified yet. Please check yourself what suits you best.
Import the packages:
from sklearn.cluster import AgglomerativeClustering
import numpy as np
import matplotlib.pyplot as plt
from scipy.cluster.hierarchy import dendrogram
Function to compute weights and distances:
def get_distances(X,model,mode='l2'):
distances = []
weights = []
children=model.children_
dims = (X.shape[1],1)
distCache = {}
weightCache = {}
for childs in children:
c1 = X[childs[0]].reshape(dims)
c2 = X[childs[1]].reshape(dims)
c1Dist = 0
c1W = 1
c2Dist = 0
c2W = 1
if childs[0] in distCache.keys():
c1Dist = distCache[childs[0]]
c1W = weightCache[childs[0]]
if childs[1] in distCache.keys():
c2Dist = distCache[childs[1]]
c2W = weightCache[childs[1]]
d = np.linalg.norm(c1-c2)
cc = ((c1W*c1)+(c2W*c2))/(c1W+c2W)
X = np.vstack((X,cc.T))
newChild_id = X.shape[0]-1
# How to deal with a higher level cluster merge with lower distance:
if mode=='l2': # Increase the higher level cluster size suing an l2 norm
added_dist = (c1Dist**2+c2Dist**2)**0.5
dNew = (d**2 + added_dist**2)**0.5
elif mode == 'max': # If the previrous clusters had higher distance, use that one
dNew = max(d,c1Dist,c2Dist)
elif mode == 'actual': # Plot the actual distance.
dNew = d
wNew = (c1W + c2W)
distCache[newChild_id] = dNew
weightCache[newChild_id] = wNew
distances.append(dNew)
weights.append( wNew)
return distances, weights
Make sample data of 2 clusters with 2 subclusters:
# Make 4 distributions, two of which form a bigger cluster
X1_1 = np.random.randn(25,2)+[8,1.5]
X1_2 = np.random.randn(25,2)+[8,-1.5]
X2_1 = np.random.randn(25,2)-[8,3]
X2_2 = np.random.randn(25,2)-[8,-3]
# Merge the four distributions
X = np.vstack([X1_1,X1_2,X2_1,X2_2])
# Plot the clusters
colors = ['r']*25 + ['b']*25 + ['g']*25 + ['y']*25
plt.scatter(X[:,0],X[:,1],c=colors)
Sample data:
Fit the clustering model
model = AgglomerativeClustering(n_clusters=2,linkage="ward")
model.fit(X)
Call the function to find the distances, and pass it to the dendogram
distance, weight = get_distances(X,model)
linkage_matrix = np.column_stack([model.children_, distance, weight]).astype(float)
plt.figure(figsize=(20,10))
dendrogram(linkage_matrix)
plt.show()
Ouput dendogram:

Update: I recommend this solution - https://stackoverflow.com/a/47769506/1333621, if you found my attempt useful please examine Arjun's solution and re-examine your vote
You will need to generate a "linkage matrix" from children_ array
where every row in the linkage matrix has the format [idx1, idx2, distance, sample_count].
This is not meant to be a paste-and-run solution, I'm not keeping track of what I needed to import - but it should be pretty clear anyway.
Here is one way to generate the required structure Z and visualize the result
X is your n_samples x n_features input data
cluster
agg_cluster = sklearn.cluster.AgglomerativeClustering(n_clusters=n)
agg_labels = agg_cluster.fit_predict(X)
some empty data structures
Z = []
# should really call this cluster dict
node_dict = {}
n_samples = len(X)
write a recursive function to gather all leaf nodes associated with a given cluster, compute distances, and centroid positions
def get_all_children(k, verbose=False):
i,j = agg_cluster.children_[k]
if k in node_dict:
return node_dict[k]['children']
if i < leaf_count:
left = [i]
else:
# read the AgglomerativeClustering doc. to see why I select i-n_samples
left = get_all_children(i-n_samples)
if j < leaf_count:
right = [j]
else:
right = get_all_children(j-n_samples)
if verbose:
print k,i,j,left, right
left_pos = np.mean(map(lambda ii: X[ii], left),axis=0)
right_pos = np.mean(map(lambda ii: X[ii], right),axis=0)
# this assumes that agg_cluster used euclidean distances
dist = metrics.pairwise_distances([left_pos,right_pos],metric='euclidean')[0,1]
all_children = [x for y in [left,right] for x in y]
pos = np.mean(map(lambda ii: X[ii], all_children),axis=0)
# store the results to speed up any additional or recursive evaluations
node_dict[k] = {'top_child':[i,j],'children':all_children, 'pos':pos,'dist':dist, 'node_i':k + n_samples}
return all_children
#return node_di|ct
populate node_dict and generate Z - with distance and n_samples per node
for k,x in enumerate(agg_cluster.children_):
get_all_children(k,verbose=False)
# Every row in the linkage matrix has the format [idx1, idx2, distance, sample_count].
Z = [[v['top_child'][0],v['top_child'][1],v['dist'],len(v['children'])] for k,v in node_dict.iteritems()]
# create a version with log scaled distances for easier visualization
Z_log =[[v['top_child'][0],v['top_child'][1],np.log(1.0+v['dist']),len(v['children'])] for k,v in node_dict.iteritems()]
plot it using scipy dendrogram
from scipy.cluster import hierarchy
plt.figure()
dn = hierarchy.dendrogram(Z_log,p=4,truncate_mode='level')
plt.show()
be disappointed by how opaque this visualization is and wish you could interactively drill down into larger clusters and examine directional (not scalar) distances between centroids :( - maybe a bokeh solution exists?
references
http://docs.scipy.org/doc/scipy/reference/generated/scipy.cluster.hierarchy.dendrogram.html
https://joernhees.de/blog/2015/08/26/scipy-hierarchical-clustering-and-dendrogram-tutorial/#Selecting-a-Distance-Cut-Off-aka-Determining-the-Number-of-Clusters

I think the official example of sklearn on the AgglomerativeClustering would be helpful.
Plot Hierarchical Clustering Dendrogram:
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='level', p=3)
plt.xlabel("Number of points in node (or index of point if no parenthesis).")
plt.show()
NB This solution relies on distances_ variable which only is set when calling AgglomerativeClustering with the distance_threshold parameter.

I ran into the same problem when setting n_clusters.
I think the problem is that if you set n_clusters, the distances don't get evaluated.
If you set n_clusters = None and set a distance_threshold, then it works with the code provided on sklearn.
I understand that this will probably not help in your situation but I hope a fix is underway.