Is it possible to specify your own distance function using scikit-learn K-Means Clustering?_问答_开发者

Is it possible to specify your own distance function using scikit-learn K-Means Clustering?

Here's a small kmeans that uses any of the 20-odd distances in scipy.spatial.distance, or a user function.
Comments would be welcome (this has had only one user so far, not enough); in particular, what are your N, dim, k, metric ?

#!/usr/bin/env python
# kmeans.py using any of the 20-odd metrics in scipy.spatial.distance
# kmeanssample 2 pass, first sample sqrt(N)

from __future__ import division
import random
import numpy as np
from scipy.spatial.distance import cdist  # $scipy/spatial/distance.py
    # http://docs.scipy.org/doc/scipy/reference/spatial.html
from scipy.sparse import issparse  # $scipy/sparse/csr.py

__date__ = "2011-11-17 Nov denis"
    # X sparse, any cdist metric: real app ?
    # centres get dense rapidly, metrics in high dim hit distance whiteout
    # vs unsupervised / semi-supervised svm

#...............................................................................
def kmeans( X, centres, delta=.001, maxiter=10, metric="euclidean", p=2, verbose=1 ):
    """ centres, Xtocentre, distances = kmeans( X, initial centres ... )
    in:
        X N x dim  may be sparse
        centres k x dim: initial centres, e.g. random.sample( X, k )
        delta: relative error, iterate until the average distance to centres
            is within delta of the previous average distance
        maxiter
        metric: any of the 20-odd in scipy.spatial.distance
            "chebyshev" = max, "cityblock" = L1, "minkowski" with p=
            or a function( Xvec, centrevec ), e.g. Lqmetric below
        p: for minkowski metric -- local mod cdist for 0 < p < 1 too
        verbose: 0 silent, 2 prints running distances
    out:
        centres, k x dim
        Xtocentre: each X -> its nearest centre, ints N -> k
        distances, N
    see also: kmeanssample below, class Kmeans below.
    """
    if not issparse(X):
        X = np.asanyarray(X)  # ?
    centres = centres.todense() if issparse(centres) \
        else centres.copy()
    N, dim = X.shape
    k, cdim = centres.shape
    if dim != cdim:
        raise ValueError( "kmeans: X %s and centres %s must have the same number of columns" % (
            X.shape, centres.shape ))
    if verbose:
        print "kmeans: X %s  centres %s  delta=%.2g  maxiter=%d  metric=%s" % (
            X.shape, centres.shape, delta, maxiter, metric)
    allx = np.arange(N)
    prevdist = 0
    for jiter in range( 1, maxiter+1 ):
        D = cdist_sparse( X, centres, metric=metric, p=p )  # |X| x |centres|
        xtoc = D.argmin(axis=1)  # X -> nearest centre
        distances = D[allx,xtoc]
        avdist = distances.mean()  # median ?
        if verbose >= 2:
            print "kmeans: av |X - nearest centre| = %.4g" % avdist
        if (1 - delta) * prevdist <= avdist <= prevdist \
        or jiter == maxiter:
            break
        prevdist = avdist
        for jc in range(k):  # (1 pass in C)
            c = np.where( xtoc == jc )[0]
            if len(c) > 0:
                centres[jc] = X[c].mean( axis=0 )
    if verbose:
        print "kmeans: %d iterations  cluster sizes:" % jiter, np.bincount(xtoc)
    if verbose >= 2:
        r50 = np.zeros(k)
        r90 = np.zeros(k)
        for j in range(k):
            dist = distances[ xtoc == j ]
            if len(dist) > 0:
                r50[j], r90[j] = np.percentile( dist, (50, 90) )
        print "kmeans: cluster 50 % radius", r50.astype(int)
        print "kmeans: cluster 90 % radius", r90.astype(int)
            # scale L1 / dim, L2 / sqrt(dim) ?
    return centres, xtoc, distances

#...............................................................................
def kmeanssample( X, k, nsample=0, **kwargs ):
    """ 2-pass kmeans, fast for large N:
        1) kmeans a random sample of nsample ~ sqrt(N) from X
        2) full kmeans, starting from those centres
    """
        # merge w kmeans ? mttiw
        # v large N: sample N^1/2, N^1/2 of that
        # seed like sklearn ?
    N, dim = X.shape
    if nsample == 0:
        nsample = max( 2*np.sqrt(N), 10*k )
    Xsample = randomsample( X, int(nsample) )
    pass1centres = randomsample( X, int(k) )
    samplecentres = kmeans( Xsample, pass1centres, **kwargs )[0]
    return kmeans( X, samplecentres, **kwargs )

def cdist_sparse( X, Y, **kwargs ):
    """ -> |X| x |Y| cdist array, any cdist metric
        X or Y may be sparse -- best csr
    """
        # todense row at a time, v slow if both v sparse
    sxy = 2*issparse(X) + issparse(Y)
    if sxy == 0:
        return cdist( X, Y, **kwargs )
    d = np.empty( (X.shape[0], Y.shape[0]), np.float64 )
    if sxy == 2:
        for j, x in enumerate(X):
            d[j] = cdist( x.todense(), Y, **kwargs ) [0]
    elif sxy == 1:
        for k, y in enumerate(Y):
            d[:,k] = cdist( X, y.todense(), **kwargs ) [0]
    else:
        for j, x in enumerate(X):
            for k, y in enumerate(Y):
                d[j,k] = cdist( x.todense(), y.todense(), **kwargs ) [0]
    return d

def randomsample( X, n ):
    """ random.sample of the rows of X
        X may be sparse -- best csr
    """
    sampleix = random.sample( xrange( X.shape[0] ), int(n) )
    return X[sampleix]

def nearestcentres( X, centres, metric="euclidean", p=2 ):
    """ each X -> nearest centre, any metric
            euclidean2 (~ withinss) is more sensitive to outliers,
            cityblock (manhattan, L1) less sensitive
    """
    D = cdist( X, centres, metric=metric, p=p )  # |X| x |centres|
    return D.argmin(axis=1)

def Lqmetric( x, y=None, q=.5 ):
    # yes a metric, may increase weight of near matches; see ...
    return (np.abs(x - y) ** q) .mean() if y is not None \
        else (np.abs(x) ** q) .mean()

#...............................................................................
class Kmeans:
    """ km = Kmeans( X, k= or centres=, ... )
        in: either initial centres= for kmeans
            or k= [nsample=] for kmeanssample
        out: km.centres, km.Xtocentre, km.distances
        iterator:
            for jcentre, J in km:
                clustercentre = centres[jcentre]
                J indexes e.g. X[J], classes[J]
    """
    def __init__( self, X, k=0, centres=None, nsample=0, **kwargs ):
        self.X = X
        if centres is None:
            self.centres, self.Xtocentre, self.distances = kmeanssample(
                X, k=k, nsample=nsample, **kwargs )
        else:
            self.centres, self.Xtocentre, self.distances = kmeans(
                X, centres, **kwargs )

    def __iter__(self):
        for jc in range(len(self.centres)):
            yield jc, (self.Xtocentre == jc)

#...............................................................................
if __name__ == "__main__":
    import random
    import sys
    from time import time

    N = 10000
    dim = 10
    ncluster = 10
    kmsample = 100  # 0: random centres, > 0: kmeanssample
    kmdelta = .001
    kmiter = 10
    metric = "cityblock"  # "chebyshev" = max, "cityblock" L1,  Lqmetric
    seed = 1

    exec( "\n".join( sys.argv[1:] ))  # run this.py N= ...
    np.set_printoptions( 1, threshold=200, edgeitems=5, suppress=True )
    np.random.seed(seed)
    random.seed(seed)

    print "N %d  dim %d  ncluster %d  kmsample %d  metric %s" % (
        N, dim, ncluster, kmsample, metric)
    X = np.random.exponential( size=(N,dim) )
        # cf scikits-learn datasets/
    t0 = time()
    if kmsample > 0:
        centres, xtoc, dist = kmeanssample( X, ncluster, nsample=kmsample,
            delta=kmdelta, maxiter=kmiter, metric=metric, verbose=2 )
    else:
        randomcentres = randomsample( X, ncluster )
        centres, xtoc, dist = kmeans( X, randomcentres,
            delta=kmdelta, maxiter=kmiter, metric=metric, verbose=2 )
    print "%.0f msec" % ((time() - t0) * 1000)

    # also ~/py/np/kmeans/test-kmeans.py

Some notes added 26mar 2012:

1) for cosine distance, first normalize all the data vectors to |X| = 1; then

cosinedistance( X, Y ) = 1 - X . Y = Euclidean distance |X - Y|^2 / 2

is fast. For bit vectors, keep the norms separately from the vectors instead of expanding out to floats (although some programs may expand for you). For sparse vectors, say 1 % of N, X . Y should take time O( 2 % N ), space O(N); but I don't know which programs do that.

2) Scikit-learn clustering gives an excellent overview of k-means, mini-batch-k-means ... with code that works on scipy.sparse matrices.

3) Always check cluster sizes after k-means. If you're expecting roughly equal-sized clusters, but they come out [44 37 9 5 5] % ... (sound of head-scratching).

Unfortunately no: scikit-learn current implementation of k-means only uses Euclidean distances.

It is not trivial to extend k-means to other distances and denis' answer above is not the correct way to implement k-means for other metrics.

Just use nltk instead where you can do this, e.g.

from nltk.cluster.kmeans import KMeansClusterer
NUM_CLUSTERS = <choose a value>
data = <sparse matrix that you would normally give to scikit>.toarray()

kclusterer = KMeansClusterer(NUM_CLUSTERS, distance=nltk.cluster.util.cosine_distance, repeats=25)
assigned_clusters = kclusterer.cluster(data, assign_clusters=True)

Yes you can use a difference metric function; however, by definition, the k-means clustering algorithm relies on the eucldiean distance from the mean of each cluster.

You could use a different metric, so even though you are still calculating the mean you could use something like the mahalnobis distance.

There is pyclustering which is python/C++ (so its fast!) and lets you specify a custom metric function

from pyclustering.cluster.kmeans import kmeans
from pyclustering.utils.metric import type_metric, distance_metric

user_function = lambda point1, point2: point1[0] + point2[0] + 2
metric = distance_metric(type_metric.USER_DEFINED, func=user_function)

# create K-Means algorithm with specific distance metric
start_centers = [[4.7, 5.9], [5.7, 6.5]];
kmeans_instance = kmeans(sample, start_centers, metric=metric)

# run cluster analysis and obtain results
kmeans_instance.process()
clusters = kmeans_instance.get_clusters()

Actually, i haven't tested this code but cobbled it together from a ticket and example code.

k-means of Spectral Python allows the use of L1 (Manhattan) distance.

Sklearn Kmeans uses the Euclidean distance. It has no metric parameter. This said, if you're clustering time series, you can use the tslearn python package, when you can specify a metric (dtw, softdtw, euclidean).

The Affinity propagation algorithm from the sklearn library allows you to pass the similarity matrix instead of the samples. So, you can use your metric to compute the similarity matrix (not the dissimilarity matrix) and pass it to the function by setting the "affinity" term to "precomputed".https://scikit-learn.org/stable/modules/generated/sklearn.cluster.AffinityPropagation.html#sklearn.cluster.AffinityPropagation.fit In terms of the K-Mean, I think it is also possible but I have not tried it. However, as the other answers stated, finding the mean using a different metric will be the issue. Instead, you can use PAM (K-Medoids) algorthim as it calculates the change in Total Deviation (TD), thus it does not rely on the distance metric. https://python-kmedoids.readthedocs.io/en/latest/#fasterpam

Yes, in the current stable version of sklearn (scikit-learn 1.1.3), you can easily use your own distance metric. All you have to do is create a class that inherits from sklearn.cluster.KMeans and overwrites its _transform method.

The below example is for the IOU distance from the Yolov2 paper.

import sklearn.cluster
import numpy as np

def anchor_iou(box_dims, centroid_box_dims):
    box_w, box_h = box_dims[..., 0], box_dims[..., 1]
    centroid_w, centroid_h = centroid_box_dims[..., 0], centroid_box_dims[..., 1]
    inter_w = np.minimum(box_w[..., np.newaxis], centroid_w[np.newaxis, ...])
    inter_h = np.minimum(box_h[..., np.newaxis], centroid_h[np.newaxis, ...])
    inter_area = inter_w * inter_h
    centroid_area = centroid_w * centroid_h
    box_area = box_w * box_h
    return inter_area / (
        centroid_area[np.newaxis, ...] + box_area[..., np.newaxis] - inter_area
    )

class IOUKMeans(sklearn.cluster.KMeans):
    def __init__(
        self,
        n_clusters=8,
        *,
        init="k-means++",
        n_init=10,
        max_iter=300,
        tol=1e-4,
        verbose=0,
        random_state=None,
        copy_x=True,
        algorithm="lloyd",
    ):
        super().__init__(
            n_clusters=n_clusters,
            init=init,
            n_init=n_init,
            max_iter=max_iter,
            tol=tol,
            verbose=verbose,
            random_state=random_state,
            copy_x=copy_x,
            algorithm=algorithm
        )

    def _transform(self, X):
        return anchor_iou(X, self.cluster_centers_)

rng = np.random.default_rng(12345)
num_boxes = 10
bboxes = rng.integers(low=0, high=100, size=(num_boxes, 2))

kmeans = IOUKMeans(num_clusters).fit(bboxes)

def distance_metrics(dist_metrics):
    kmeans_instance = kmeans(trs_data, initial_centers, metric=dist_metrics)

    label = np.zeros(210, dtype=int)
    for i in range(0, len(clusters)):
        for index, j in enumerate(clusters[i]):
            label[j] = i