Question

将NxM特征向量作为numpy矩阵。是否有任何例程可以使用L1距离（曼哈顿距离）通过Kmeans算法对其进行聚类？

Answer 1

这是一个使用L1距离（曼哈顿距离）的Kmeans算法。一般来说，特征向量表示为列表，很容易转换为numpy矩阵。

    import random
    #Manhattan Distance
    def L1(v1,v2):
      if(len(v1)!=len(v2):
        print “error”
        return -1
      return sum([abs(v1[i]-v2[i]) for i in range(len(v1))])

    # kmeans with L1 distance. 
    # rows refers to the NxM feature vectors
    def kcluster(rows,distance=L1,k=4):# Cited from Programming Collective Intelligence 
        # Determine the minimum and maximum values for each point
        ranges=[(min([row[i] for row in rows]),max([row[i] for row in rows])) for i in range(len(rows[0]))]

        # Create k randomly placed centroids
        clusters=[[random.random( )*(ranges[i][1]-ranges[i][0])+ranges[i][0] for i in range(len(rows[0]))] for j in range(k)]

        lastmatches=None
        for t in range(100):
            print 'Iteration %d' % t
            bestmatches=[[] for i in range(k)]
            # Find which centroid is the closest for each row
            for j in range(len(rows)):
                row=rows[j]
                bestmatch=0
                for i in range(k):
                    d=distance(clusters[i],row)
                    if d<distance(clusters[bestmatch],row): 
                        bestmatch=i
                bestmatches[bestmatch].append(j)
            ## If the results are the same as last time, this is complete
            if bestmatches==lastmatches:
                break
            lastmatches=bestmatches

            # Move the centroids to the average of their members
            for i in range(k):
                avgs=[0.0]*len(rows[0])
                if len(bestmatches[i])>0:
                    for rowid in bestmatches[i]:
                        for m in range(len(rows[rowid])):
                            avgs[m]+=rows[rowid][m]
                    for j in range(len(avgs)):
                        avgs[j]/=len(bestmatches[i])
                    clusters[i]=avgs
        return bestmatches

Answer 2

我不认为这是在scipy中明确提供的，但你应该看看以下内容：

http://projects.scipy.org/scipy/ticket/612

Answer 3

下面有代码 is-it-possible-to-specify-your-own-distance-function-using-scikits-learn-k-means，它使用scipy.spatial.distance中的任意20个指标。也可以看看 L1-or-L.5-metrics-for-clustering;你能用L1和L2评论你的结果吗？

Answer 4

看看pyclustering。在这里，您可以找到可配置为使用 L1 距离的 k-means 的实现。但是你必须将 numpy 数组转换成一个列表。

如何安装pyclustering

pip3 install pyclustering

从pyclustering复制的代码片段

pip3 install pyclustering

from pyclustering.cluster.kmeans import kmeans, kmeans_visualizer
from pyclustering.cluster.center_initializer import kmeans_plusplus_initializer
from pyclustering.samples.definitions import FCPS_SAMPLES
from pyclustering.utils import read_sample

sample = read_sample(FCPS_SAMPLES.SAMPLE_TWO_DIAMONDS)

manhattan_metric = distance_metric(type_metric.MANHATTAN)
kmeans_instance = kmeans(sample, initial_centers, metric=manhattan_metric)
kmeans_instance.process()

k表示在python中具有L1距离

4 个答案: