我正在尝试使用AgglomerativeClustering 提供的children_ 属性构建树状图，但到目前为止我运气不佳。我不能使用scipy.cluster，因为scipy 中提供的凝聚集群缺少一些对我很重要的选项（例如指定集群数量的选项）。我将非常感谢那里的任何建议。
import sklearn.cluster
clstr = cluster.AgglomerativeClustering(n_clusters=2)
clusterer.children_
【问题讨论】：
标签： python plot cluster-analysis dendrogram
import numpy as np
from matplotlib import pyplot as plt
from scipy.cluster.hierarchy import dendrogram
from sklearn.datasets import load_iris
from sklearn.cluster import AgglomerativeClustering
def plot_dendrogram(model, **kwargs):
# Create linkage matrix and then plot the dendrogram
# create the counts of samples under each node
counts = np.zeros(model.children_.shape[0])
n_samples = len(model.labels_)
for i, merge in enumerate(model.children_):
current_count = 0
for child_idx in merge:
if child_idx < n_samples:
current_count += 1 # leaf node
else:
current_count += counts[child_idx - n_samples]
counts[i] = current_count
linkage_matrix = np.column_stack([model.children_, model.distances_,
counts]).astype(float)
# Plot the corresponding dendrogram
dendrogram(linkage_matrix, **kwargs)
iris = load_iris()
X = iris.data
# setting distance_threshold=0 ensures we compute the full tree.
model = AgglomerativeClustering(distance_threshold=0, n_clusters=None)
model = model.fit(X)
plt.title('Hierarchical Clustering Dendrogram')
# plot the top three levels of the dendrogram
plot_dendrogram(model, truncate_mode='level', p=3)
plt.xlabel("Number of points in node (or index of point if no parenthesis).")
plt.show()
请注意，目前（从 scikit-learn v0.23 开始）仅在使用 distance_threshold 参数调用 AgglomerativeClustering 时才有效，但从 v0.24 开始，您将能够通过设置 @987654325 来强制计算距离@ 为真 (see nightly build docs)。
【讨论】：
【讨论】：
plot_dendrogram() 函数在这里所做的——除了最后一行：scikit-learn.org/stable/auto_examples/cluster/… 最后一行中调用的dendrogram 函数是从scipy.cluster.hierarchy 导入的跨度>
plot_dendrogram sn-p 来回答这个问题。我很高兴看到文档有所改进。我现在在这里删除了我的赞成票。
这是一个simple function，用于从 sklearn 获取层次聚类模型并使用 scipy dendrogram 函数对其进行绘图。似乎sklearn中通常不直接支持图形功能。您可以找到与此 plot_dendrogram 代码 sn-p here 的拉取请求相关的有趣讨论。
我要澄清的是，您描述的用例（定义集群数量）在 scipy 中可用：在您使用 scipy 的linkage 执行层次聚类之后，您可以将层次结构切割成您想要使用的任何数量的集群fcluster 与 t 参数和 criterion='maxclust' 参数中指定的簇数。
【讨论】：
对于那些愿意放弃 Python 并使用强大的 D3 库的人来说，使用 d3.cluster()（或者，我猜是 d3.tree()）API 来实现一个不错的、可定制的结果并不是很困难。
查看jsfiddle 进行演示。
children_ 数组幸运地很容易用作 JS 数组，唯一的中间步骤是使用 d3.stratify() 将其转换为分层表示。具体来说，我们需要每个节点都有一个id 和一个parentId：
var N = 272; // Your n_samples/corpus size.
var root = d3.stratify()
.id((d,i) => i + N)
.parentId((d, i) => {
var parIndex = data.findIndex(e => e.includes(i + N));
if (parIndex < 0) {
return; // The root should have an undefined parentId.
}
return parIndex + N;
})(data); // Your children_
由于findIndex 行，您最终在这里至少会出现 O(n^2) 行为，但在您的 n_samples 变得巨大之前，这可能并不重要，在这种情况下，您可以预先计算更有效的索引。
除此之外，d3.cluster() 几乎就是即插即用。请参阅 mbostock 的 canonical block 或我的 JSFiddle。
注意对于我的用例，仅显示非叶节点就足够了；将样本/叶子可视化有点棘手，因为它们可能并不都明确地在 children_ 数组中。
【讨论】：
前段时间我遇到了完全相同的问题。我设法绘制该死的树状图的方法是使用软件包ete3。该软件包能够灵活地绘制具有各种选项的树。唯一的困难是将sklearn 的children_ 输出转换为Newick Tree format 可以被ete3 阅读和理解。此外，我需要手动计算树突的跨度，因为 children_ 没有提供该信息。这是我使用的代码的 sn-p。它计算 Newick 树，然后显示 ete3 Tree 数据结构。更多关于如何绘图的详细信息，请查看here
import numpy as np
from sklearn.cluster import AgglomerativeClustering
import ete3
def build_Newick_tree(children,n_leaves,X,leaf_labels,spanner):
"""
build_Newick_tree(children,n_leaves,X,leaf_labels,spanner)
Get a string representation (Newick tree) from the sklearn
AgglomerativeClustering.fit output.
Input:
children: AgglomerativeClustering.children_
n_leaves: AgglomerativeClustering.n_leaves_
X: parameters supplied to AgglomerativeClustering.fit
leaf_labels: The label of each parameter array in X
spanner: Callable that computes the dendrite's span
Output:
ntree: A str with the Newick tree representation
"""
return go_down_tree(children,n_leaves,X,leaf_labels,len(children)+n_leaves-1,spanner)[0]+';'
def go_down_tree(children,n_leaves,X,leaf_labels,nodename,spanner):
"""
go_down_tree(children,n_leaves,X,leaf_labels,nodename,spanner)
Iterative function that traverses the subtree that descends from
nodename and returns the Newick representation of the subtree.
Input:
children: AgglomerativeClustering.children_
n_leaves: AgglomerativeClustering.n_leaves_
X: parameters supplied to AgglomerativeClustering.fit
leaf_labels: The label of each parameter array in X
nodename: An int that is the intermediate node name whos
children are located in children[nodename-n_leaves].
spanner: Callable that computes the dendrite's span
Output:
ntree: A str with the Newick tree representation
"""
nodeindex = nodename-n_leaves
if nodename<n_leaves:
return leaf_labels[nodeindex],np.array([X[nodeindex]])
else:
node_children = children[nodeindex]
branch0,branch0samples = go_down_tree(children,n_leaves,X,leaf_labels,node_children[0])
branch1,branch1samples = go_down_tree(children,n_leaves,X,leaf_labels,node_children[1])
node = np.vstack((branch0samples,branch1samples))
branch0span = spanner(branch0samples)
branch1span = spanner(branch1samples)
nodespan = spanner(node)
branch0distance = nodespan-branch0span
branch1distance = nodespan-branch1span
nodename = '({branch0}:{branch0distance},{branch1}:{branch1distance})'.format(branch0=branch0,branch0distance=branch0distance,branch1=branch1,branch1distance=branch1distance)
return nodename,node
def get_cluster_spanner(aggClusterer):
"""
spanner = get_cluster_spanner(aggClusterer)
Input:
aggClusterer: sklearn.cluster.AgglomerativeClustering instance
Get a callable that computes a given cluster's span. To compute
a cluster's span, call spanner(cluster)
The cluster must be a 2D numpy array, where the axis=0 holds
separate cluster members and the axis=1 holds the different
variables.
"""
if aggClusterer.linkage=='ward':
if aggClusterer.affinity=='euclidean':
spanner = lambda x:np.sum((x-aggClusterer.pooling_func(x,axis=0))**2)
elif aggClusterer.linkage=='complete':
if aggClusterer.affinity=='euclidean':
spanner = lambda x:np.max(np.sum((x[:,None,:]-x[None,:,:])**2,axis=2))
elif aggClusterer.affinity=='l1' or aggClusterer.affinity=='manhattan':
spanner = lambda x:np.max(np.sum(np.abs(x[:,None,:]-x[None,:,:]),axis=2))
elif aggClusterer.affinity=='l2':
spanner = lambda x:np.max(np.sqrt(np.sum((x[:,None,:]-x[None,:,:])**2,axis=2)))
elif aggClusterer.affinity=='cosine':
spanner = lambda x:np.max(np.sum((x[:,None,:]*x[None,:,:]))/(np.sqrt(np.sum(x[:,None,:]*x[:,None,:],axis=2,keepdims=True))*np.sqrt(np.sum(x[None,:,:]*x[None,:,:],axis=2,keepdims=True))))
else:
raise AttributeError('Unknown affinity attribute value {0}.'.format(aggClusterer.affinity))
elif aggClusterer.linkage=='average':
if aggClusterer.affinity=='euclidean':
spanner = lambda x:np.mean(np.sum((x[:,None,:]-x[None,:,:])**2,axis=2))
elif aggClusterer.affinity=='l1' or aggClusterer.affinity=='manhattan':
spanner = lambda x:np.mean(np.sum(np.abs(x[:,None,:]-x[None,:,:]),axis=2))
elif aggClusterer.affinity=='l2':
spanner = lambda x:np.mean(np.sqrt(np.sum((x[:,None,:]-x[None,:,:])**2,axis=2)))
elif aggClusterer.affinity=='cosine':
spanner = lambda x:np.mean(np.sum((x[:,None,:]*x[None,:,:]))/(np.sqrt(np.sum(x[:,None,:]*x[:,None,:],axis=2,keepdims=True))*np.sqrt(np.sum(x[None,:,:]*x[None,:,:],axis=2,keepdims=True))))
else:
raise AttributeError('Unknown affinity attribute value {0}.'.format(aggClusterer.affinity))
else:
raise AttributeError('Unknown linkage attribute value {0}.'.format(aggClusterer.linkage))
return spanner
clusterer = AgglomerativeClustering(n_clusters=2,compute_full_tree=True) # You can set compute_full_tree to 'auto', but I left it this way to get the entire tree plotted
clusterer.fit(X) # X for whatever you want to fit
spanner = get_cluster_spanner(clusterer)
newick_tree = build_Newick_tree(clusterer.children_,clusterer.n_leaves_,X,leaf_labels,spanner) # leaf_labels is a list of labels for each entry in X
tree = ete3.Tree(newick_tree)
tree.show()
【讨论】：