scipy.cluster.hierarchy 教程 [英] Tutorial for scipy.cluster.hierarchy

问题描述

我试图了解如何操作层次结构集群，但文档太......技术性?......我无法理解它是如何工作的.

I'm trying to understand how to manipulate a hierarchy cluster but the documentation is too ... technical?... and I can't understand how it works.

是否有任何教程可以帮助我开始，逐步解释一些简单的任务?

Is there any tutorial that can help me to start with, explaining step by step some simple tasks?

假设我有以下数据集:

a = np.array([[0,   0  ],
              [1,   0  ],
              [0,   1  ],
              [1,   1  ], 
              [0.5, 0  ],
              [0,   0.5],
              [0.5, 0.5],
              [2,   2  ],
              [2,   3  ],
              [3,   2  ],
              [3,   3  ]])

我可以轻松地进行层次聚类并绘制树状图:

I can easily do the hierarchy cluster and plot the dendrogram:

z = linkage(a)
d = dendrogram(z)

• 现在，我该如何恢复特定的集群?假设在树状图中具有 [0,1,2,4,5,6] 元素的那个?
• 如何取回这些元素的值?

推荐答案

层次凝聚聚类 (HAC) 分为三个步骤:

There are three steps in hierarchical agglomerative clustering (HAC):

1. 量化数据(metric 参数)
2. 集群数据(method 参数)
3. 选择簇数

做

z = linkage(a)

将完成前两个步骤.由于您没有指定任何参数，它使用标准值

will accomplish the first two steps. Since you did not specify any parameters it uses the standard values

1. metric = '欧几里得'
2. method = 'single'

所以 z = links(a) 会给你一个 a 的单一链接层次凝聚聚类.这种聚类是一种解决方案的层次结构.从这个层次结构中，您可以获得有关数据结构的一些信息.你现在可以做的是:

So z = linkage(a) will give you a single linked hierachical agglomerative clustering of a. This clustering is kind of a hierarchy of solutions. From this hierarchy you get some information about the structure of your data. What you might do now is:

• 检查哪个 metric 是合适的，例如.G.cityblock 或 chebychev 将以不同的方式量化您的数据(cityblock、euclidean 和 chebychev 对应到 L1、L2 和 L_inf 范数)
• 检查methdos 的不同属性/行为(例如single、complete 和average)
• 检查如何确定集群的数量，例如.G.通过阅读有关它的维基
• 计算找到的解决方案(聚类)的索引，例如 剪影系数(使用该系数，您可以获得关于点/观测值与聚类分配的聚类的匹配程度的反馈).不同的索引使用不同的标准来限定聚类.
• Check which metric is appropriate, e. g. cityblock or chebychev will quantify your data differently (cityblock, euclidean and chebychev correspond to L1, L2, and L_inf norm)
• Check the different properties / behaviours of the methdos (e. g. single, complete and average)
• Check how to determine the number of clusters, e. g. by reading the wiki about it
• Compute indices on the found solutions (clusterings) such as the silhouette coefficient (with this coefficient you get a feedback on the quality of how good a point/observation fits to the cluster it is assigned to by the clustering). Different indices use different criteria to qualify a clustering.

从这里开始

import numpy as np
import scipy.cluster.hierarchy as hac
import matplotlib.pyplot as plt


a = np.array([[0.1,   2.5],
              [1.5,   .4 ],
              [0.3,   1  ],
              [1  ,   .8 ],
              [0.5,   0  ],
              [0  ,   0.5],
              [0.5,   0.5],
              [2.7,   2  ],
              [2.2,   3.1],
              [3  ,   2  ],
              [3.2,   1.3]])

fig, axes23 = plt.subplots(2, 3)

for method, axes in zip(['single', 'complete'], axes23):
    z = hac.linkage(a, method=method)

    # Plotting
    axes[0].plot(range(1, len(z)+1), z[::-1, 2])
    knee = np.diff(z[::-1, 2], 2)
    axes[0].plot(range(2, len(z)), knee)

    num_clust1 = knee.argmax() + 2
    knee[knee.argmax()] = 0
    num_clust2 = knee.argmax() + 2

    axes[0].text(num_clust1, z[::-1, 2][num_clust1-1], 'possible
<- knee point')

    part1 = hac.fcluster(z, num_clust1, 'maxclust')
    part2 = hac.fcluster(z, num_clust2, 'maxclust')

    clr = ['#2200CC' ,'#D9007E' ,'#FF6600' ,'#FFCC00' ,'#ACE600' ,'#0099CC' ,
    '#8900CC' ,'#FF0000' ,'#FF9900' ,'#FFFF00' ,'#00CC01' ,'#0055CC']

    for part, ax in zip([part1, part2], axes[1:]):
        for cluster in set(part):
            ax.scatter(a[part == cluster, 0], a[part == cluster, 1], 
                       color=clr[cluster])

    m = '
(method: {})'.format(method)
    plt.setp(axes[0], title='Screeplot{}'.format(m), xlabel='partition',
             ylabel='{}
cluster distance'.format(m))
    plt.setp(axes[1], title='{} Clusters'.format(num_clust1))
    plt.setp(axes[2], title='{} Clusters'.format(num_clust2))

plt.tight_layout()
plt.show()

给

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