是否可以使用scikit-learn K-Means Clustering指定您自己的距离函数？

Question

是否可以使用scikit-learn K-Means Clustering指定自己的距离函数？

Answer 1

这是一个使用20多个距离中的任何一个的小kmeans scipy.spatial.distance或用户功能 欢迎评论（到目前为止只有一个用户，还不够）; 尤其是你的N，dim，k，metric是什么？

#!/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

一些笔记添加了26mar 2012：

1）对于余弦距离，首先将所有数据向量归一化为| X | = 1;然后

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

很快。对于位向量，请将规范与向量分开 而不是扩展到浮动 （虽然有些程序可能会为您扩展）。 对于稀疏向量，比如N，X的1％。 Y应该花费时间O（2％N）， 空间O（N）;但我不知道哪个程序会这样做。

2） Scikit-learn clustering 给出了k-means，mini-batch-k-means的优秀概述...... 使用适用于scipy.sparse矩阵的代码。

3）在k-means之后总是检查簇大小。 如果您期望大致相同大小的群集，但它们会出现 [44 37 9 5 5] % ......（头疼的声音）。

Answer 2

不幸的是没有：scikit-learn k-means的当前实现仅使用欧几里德距离。

将k-means扩展到其他距离并不是一件轻而易举的事情，而且上面的回答并不是为其他指标实现k均值的正确方法。

Answer 3

只需使用nltk代替您可以执行此操作的地方，例如

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)

Answer 4

是的，你可以使用差异度量函数;然而，根据定义，k均值聚类算法依赖于距离每个聚类的均值的eucldiean距离。

你可以使用不同的指标，所以即使你仍在计算均值，你也可以使用像mahalnobis距离这样的东西。

Answer 5

k-means of Spectral Python允许使用L1（曼哈顿）距离。

Answer 6

有pyclustering是python / C ++（非常快！），可让您指定自定义指标功能

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()

实际上，我尚未测试此代码，而是将它们从a ticket和example code拼凑而成。

Answer 7

Sklearn Kmeans 使用欧几里德距离。它没有指标参数。这就是说，如果要对时间序列进行聚类，则可以在指定指标（{{1}，tslearn，{ {1}}。

Answer 8

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