Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
328 views
in Technique[技术] by (71.8m points)

python - 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?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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:

  1. Insert the following line after line 748:

    kwargs['return_distance'] = True

  2. 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:

  1. 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).

  2. 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, 

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...