在Python中打印每个层次聚类步骤中的所有聚类和样本

Print all clusters and samples at each step of hierarchical clustering in python(在Python中打印每个层次聚类步骤中的所有聚类和样本)

问题描述

我使用Scipy库执行层次聚类并创建树形图。以下是简单的代码和生成的树状图：

import numpy as np
from scipy.cluster.hierarchy import dendrogram, linkage
from matplotlib import pyplot as plt

X = np.array([[5, 3],
              [10, 15],
              [15, 12],
              [24, 10],
              [30, 30],
              [85, 70],
              [71, 80],
              [60, 78],
              [70, 55],
              [80, 91]])
linkage_matrix = linkage(X, "single")
_ = dendrogram(linkage_matrix,)

我需要在集群过程的每个步骤打印属于每个集群的所有集群和样本。以下是上述数据和树状图的预期输出：

[{0}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}]
[{0}, {1, 2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}]
[{0}, {1, 2, 3}, {4}, {5}, {6}, {7}, {8}, {9}]
[{0}, {1, 2, 3}, {4}, {5}, {6, 7}, {8}, {9}]
[{0, 1, 2, 3}, {4}, {5}, {6, 7}, {8}, {9}]
[{0, 1, 2, 3}, {4}, {5}, {6, 7, 9}, {8}]
[{0, 1, 2, 3}, {4}, {5, 6, 7, 9}, {8}]
[{0, 1, 2, 3, 4}, {5, 6, 7, 9}, {8}]
[{0, 1, 2, 3, 4}, {5, 6, 7, 9, 8}]
[{0, 1, 2, 3, 4, 5, 6, 7, 9, 8}]

请注意，如果有使用Scikit-Learn agglomerative clustering的解决方案也可以。

推荐答案

使用同一模块中的cut_tree函数，指定簇数为Cut条件。遗憾的是，在每个元素都是它自己的集群的情况下，它不会被切断，但这种情况很容易添加。

此外，从cut_tree返回的矩阵的形状是这样的，即每个列表示某个切点处的组。所以我转置了矩阵，但您也可以相应地调整for循环。

import numpy as np
from scipy.cluster.hierarchy import dendrogram, linkage, to_tree, cut_tree
from matplotlib import pyplot as plt

X = np.array([[5, 3],
              [10, 15],
              [15, 12],
              [24, 10],
              [30, 30],
              [85, 70],
              [71, 80],
              [60, 78],
              [70, 55],
              [80, 91]])
linkage_matrix = linkage(X, "single")
clusters = cut_tree(linkage_matrix, n_clusters=range(1, X.shape[0]))
print(clusters)
# insert column for the case, where every element is its own cluster
clusters = np.insert(clusters, clusters.shape[1], range(clusters.shape[0]), axis=1)
# transpose matrix
clusters = clusters.T
print(clusters)
for row in clusters[::-1]:
    # create empty dictionary
    groups = {}
    for i, g in enumerate(row):
        if g not in groups:
            # add new key to dict and assign empty set
            groups[g] = set([])
        # add to set of certain group
        groups[g].add(i)
    print(list(groups.values()))

更好的解决方案

不是两个for循环和cut_tree，而是使用集合上的操作和LINKINK_MATRATE中的信息。For循环以线性时间复杂性运行，但最耗时的将是print语句

对于大约30_000个样本，打印到文件将创建大约30 GB的大文件。

linkage_matrix = linkage(X, "single")

dct = dict([(i, {i}) for i in range(X.shape[0])])
print(list(dct.values()))
for i, row in enumerate(linkage_matrix, X.shape[0]):
    dct[i] = dct[row[0]].union(dct[row[1]])
    del dct[row[0]]
    del dct[row[1]]
    print(list(dct.values()))

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