是否可以使用scikit-learn K-Means聚类来指定自己的距离函数?
当前回答
python/ c++中有pyclustering(所以它很快!),可以让你指定一个自定义度量函数
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,我还没有测试这段代码,但它拼凑在一起从一个票和示例代码。
其他回答
这是一个用20多个距离中的任意一个的小kmeans scipy.spatial。距离,或者用户函数。 欢迎评论(到目前为止只有一个用户,还不够); 特别是,你的N,暗度,k度规是什么?
#!/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
2012年3月26日增加了一些说明:
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
是快。对于位向量,将范数与向量分开 而不是展开到浮点数 (尽管有些程序可能会为你扩展)。 对于稀疏向量,比如N, X的1%。Y需要时间O(2% N), 空间O (N);但我不知道哪个程序能做到。
2) Scikit-learn集群 给出了k-means, mini-batch-k-means… 使用在scipy上工作的代码。稀疏矩阵。
3)总是在k-means之后检查聚类大小。 如果你期待大致相同大小的集群,但是它们出来了 [44 37 9 5 5] %……(挠头的声音)。
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
不幸的是没有:scikit-learn目前实现的k-means只使用欧几里得距离。
将k-means扩展到其他距离并不是一件简单的事情,denis上面的回答并不是对其他度量实现k-means的正确方法。
是的,在当前稳定版本的sklearn (scikit-learn 1.1.3)中,您可以轻松地使用自己的距离度量。你所要做的就是创建一个继承自sklearn.cluster.KMeans的类,并覆盖它的_transform方法。
下面的例子是IOU与Yolov2论文的距离。
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)
python/ c++中有pyclustering(所以它很快!),可以让你指定一个自定义度量函数
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,我还没有测试这段代码,但它拼凑在一起从一个票和示例代码。
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