如何在scipy创建的树状图中获得与颜色簇对应的平面聚类 [英] How to get flat clustering corresponding to color clusters in the dendrogram created by scipy

问题描述

使用发布的代码 ).但是，我看不到在哪里可以为 fcluster 提供与我在创建树状图时指定的相同的 cutoff.fcluster、t 中的阈值变量似乎必须是各种模糊的测量值(inconsistent、distance>、maxclust、monocrit).有任何想法吗?

解决方案

我认为您走对了路.让我们试试这个:

导入scipy将 scipy.cluster.hierarchy 导入为 schX = scipy.randn(100, 2) # 100 个二维观察d = sch.distance.pdist(X) # (100 选择 2) 对距离的向量L = sch.linkage(d, method='complete')ind = sch.fcluster(L, 0.5*d.max(), '距离')

ind 将为您提供 100 个输入观察中的每一个的聚类索引.ind 取决于你在linkage 中使用的method.尝试method=single、complete 和average.然后注意 ind 的不同之处.

示例:

在[59]中:L = sch.linkage(d, method='complete')在 [60]: sch.fcluster(L, 0.5*d.max(), 'distance')出[60]:数组([5, 4, 2, 2, 5, 5, 1, 5, 5, 2, 5, 2, 5, 5, 1, 1, 5, 5, 4, 2, 5, 2, 5,2, 5, 3, 5, 3, 5, 5, 5, 5, 5, 5, 5, 2, 2, 5, 5, 4, 1, 4, 5, 2, 1, 4,2, 4, 2, 2, 5, 5, 5, 2, 5, 5, 3, 5, 5, 4, 5, 4, 5, 3, 5, 3, 5, 5, 5,2, 3, 5, 5, 4, 5, 5, 2, 2, 5, 2, 2, 4, 1, 2, 1, 5, 2, 5, 5, 5, 1, 5,4, 2, 4, 5, 2, 4, 4, 2])在 [61] 中:L = sch.linkage(d, method='single')在 [62]: sch.fcluster(L, 0.5*d.max(), 'distance')出[62]:数组([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1, 1, 1, 1, 1, 1, 1, 1])

scipy.cluster.hierarchy 确实令人困惑.在您的链接中，我什至不认识自己的代码！

Using the code posted here, I created a nice hierarchical clustering:

Let's say the the dendrogram on the left was created by doing something like

Y = sch.linkage(D, method='average') # D is a distance matrix
cutoff = 0.5*max(Y[:,2])
Z = sch.dendrogram(Y, orientation='right', color_threshold=cutoff)

Now how do I get the indices of the members of each of the colored clusters? To simplify this situation, ignore the clustering on the top, and focus only on the dendrogram on the left of the matrix.

This information should be stored in the dendrogram Z stored variable. There is a function that should do just what I want called fcluster (see documentation here). However I cannot see where I can give fcluster the same cutoff as I specified in the creation of the dendrogram. It seems that the threshold variable in fcluster, t has to be in terms of various obscure measurements (inconsistent, distance, maxclust, monocrit). Any ideas?

解决方案

I think you're on the right track. Let's try this:

import scipy
import scipy.cluster.hierarchy as sch
X = scipy.randn(100, 2)     # 100 2-dimensional observations
d = sch.distance.pdist(X)   # vector of (100 choose 2) pairwise distances
L = sch.linkage(d, method='complete')
ind = sch.fcluster(L, 0.5*d.max(), 'distance')

ind will give you cluster indices for each of the 100 input observations. ind depends on what method you used in linkage. Try method=single, complete, and average. Then note how ind differs.

Example:

In [59]: L = sch.linkage(d, method='complete')

In [60]: sch.fcluster(L, 0.5*d.max(), 'distance')
Out[60]: 
array([5, 4, 2, 2, 5, 5, 1, 5, 5, 2, 5, 2, 5, 5, 1, 1, 5, 5, 4, 2, 5, 2, 5,
       2, 5, 3, 5, 3, 5, 5, 5, 5, 5, 5, 5, 2, 2, 5, 5, 4, 1, 4, 5, 2, 1, 4,
       2, 4, 2, 2, 5, 5, 5, 2, 5, 5, 3, 5, 5, 4, 5, 4, 5, 3, 5, 3, 5, 5, 5,
       2, 3, 5, 5, 4, 5, 5, 2, 2, 5, 2, 2, 4, 1, 2, 1, 5, 2, 5, 5, 5, 1, 5,
       4, 2, 4, 5, 2, 4, 4, 2])

In [61]: L = sch.linkage(d, method='single')

In [62]: sch.fcluster(L, 0.5*d.max(), 'distance')
Out[62]: 
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1])

scipy.cluster.hierarchy sure is confusing. In your link, I don't even recognize my own code!

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