如何在python中提取树状图中的点之间的距离? [英] How can I extract the distances between points within a dendogram in python?

问题描述

我在 python 中执行分层聚类，并获得树状图.我想知道是否有一种方法可以提取最近点之间的距离，例如这里:7 和 8 之间的距离(最近的一个)，然后是 0 和 1 之间的距离等等，为了生成我使用的图:

I am performing and hierarchical clustering in python and I obtain the dendogram plot. I was wondering if there is a way to extract the distances between closest point for example here: distances between 7 and 8 (the closest one), then distances between 0 and 1 and so on, To produce the plot I've used the function:

linkage_matrix= linkage(dfP, method="single") 

cluster_dict = dendrogram (linkage_matrix)

推荐答案

什么时候做

Z = hierarchy.linkage(X, method='single')

在 Z 矩阵中，您拥有所需的一切:cluster1、cluster2、距离、集群中的元素数.

in Z matrix you have everything you need: cluster1, cluster2, distance, number of elements in the cluster.

例如

import numpy as np
import pandas as pd
from scipy.cluster import hierarchy
import matplotlib.pyplot as plt
import seaborn as sns

X = np.array([662., 877., 255., 412., 996., 295., 468., 268.,
                   400., 754., 564., 138., 219., 869., 669.])

Z = hierarchy.linkage(X, method='single')
plt.figure()
dn = hierarchy.dendrogram(Z)

我们有 Z

array([[  2.,   5., 138.,   2.],
       [  3.,   4., 219.,   2.],
       [  0.,   7., 255.,   3.],
       [  1.,   8., 268.,   4.],
       [  6.,   9., 295.,   6.]])

因为我们只有 6 个元素，所以 0 到 5 是单个元素，从 6 开始它们是元素簇

since we have only 6 elements, 0 to 5 are single elements, from 6 on they are clusters of elements

• 6 是 2 个元素的第一个簇 (2,5)
• 7 是 2 个元素的第二个簇 (3,4)
• 8 是第三个簇 (0,7)，即 (0,(3,4)) 的 3 个元素
• 9 是第四个簇 (1,8)，即 (1,(0,(3,4))) 的 4 个元素

然后我们有 (6,9) 即 ((2,5),(1,(0,(3,4)))) 的 6 个元素

then we have (6,9) i.e. ((2,5),(1,(0,(3,4)))) of 6 elements

clusters = {
    0: '0',
    1: '1',
    2: '2',
    3: '3',
    4: '4',
    5: '5',
    6: '2,5',
    7: '3,4',
    8: '0,3,4',
    9: '1,0,3,4',
}

现在我们可以构建一个df来显示热图

now we can build a df to display the heatmap

# init the DataFrame
df = pd.DataFrame(
    columns=Z[:,0].astype(int), 
    index=Z[:,1].astype(int)
)

df.columns = df.columns.map(clusters)
df.index = df.index.map(clusters)

# populate the diagonal
for i, d in enumerate(Z[:,2]):
    df.iloc[i, i] = d

# fill NaN
df.fillna(0, inplace=True)
# mask everything but diagonal
mask = np.ones(df.shape, dtype=bool)
np.fill_diagonal(mask, 0)

# plot the heatmap
sns.heatmap(df, 
            annot=True, fmt='.0f', cmap="YlGnBu", 
            mask=mask)
plt.show()

我将 X 定义为距离数组.这些是元素间距离的幂零下三角矩阵的值，按列.

I defined X as an array of distances. These are the values of the nilpotent lower triangular matrix of distances between elements, by column.

我们可以验证

# number of elements
n = (np.sqrt(8 * X.size + 1) + 1) / 2
n
6.0

我们有 n=6 个元素，这是距离的幂零下三角矩阵

we have n=6 elements and here's the nilpotent lower triangular matrix of distances

# init the DataFrame
df = pd.DataFrame(columns=range(int(n)), index=range(int(n)))
# populate the DataFrame
idx = 0
for c in range(int(n)-1):
    for r in range(c+1, int(n)):
        df.iloc[r, c] = X[idx]
        idx += 1
# fill NaNs and mask
df.fillna(0, inplace=True)
mask = np.zeros_like(df)
mask[np.triu_indices_from(mask)] = True
# plot the matrix
sns.heatmap(df, annot=True, fmt='.0f', cmap="YlGnBu", mask=mask)
plt.show()

如何为聚类距离对角矩阵自动填充地图字典.

How to automatically populate the map dictionary for the clusters distances diagonal matrix.

首先我们必须计算元素的数量(仅当 X 是距离数组时才需要)，正如我们之前看到的

First we have to calculate the number of elements (needed only if X is an array of distances) as we saw earlier

# number of elements
n = (np.sqrt(8 * X.size + 1) + 1) / 2

然后，我们可以遍历Z矩阵来填充字典

then, we can loop through Z matrix to populate the dictionary

# clusters of single elements
clusters = {i: str(i) for i in range(int(n))}
# loop through Z matrix
for i, z in enumerate(Z.astype(int)):
    # cluster number
    cluster_num = int(n+i)
    # elements in clusters
    cluster_names = [clusters[z[0]], clusters[z[1]]]
    cluster_elements = [str(i) for i in cluster_names]
    # update the dictionary
    clusters.update({cluster_num: ','.join(cluster_elements)})

我们有

clusters

{0: '0',
 1: '1',
 2: '2',
 3: '3',
 4: '4',
 5: '5',
 6: '2,5',
 7: '3,4',
 8: '0,3,4',
 9: '1,0,3,4',
 10: '2,5,1,0,3,4'}

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