如何使用 nltk (python) 获得 K 均值簇的单个质心

【问题标题】：How do I obtain individual centroids of K mean cluster using nltk (python)如何使用 nltk (python) 获得 K 均值簇的单个质心
【发布时间】：2020-04-19 15:56:23
【问题描述】：

我使用 nltk 来执行 k 均值聚类，因为我想将距离度量更改为余弦距离。但是，如何获得所有簇的质心？

kclusterer = KMeansClusterer(8, distance = nltk.cluster.util.cosine_distance, repeats = 1)
predict = kclusterer.cluster(features, assign_clusters = True)
centroids = kclusterer._centroid
df_clustering['cluster'] = predict
#df_clustering['centroid'] = centroids[df_clustering['cluster'] - 1].tolist()
df_clustering['centroid'] = centroids

我正在尝试在 pandas 数据帧上执行 k 均值聚类，并希望每个数据点的聚类质心坐标位于数据帧列“质心”中。

提前谢谢你！

【问题讨论】：

标签： python nltk k-means

【解决方案1】：

import pandas as pd
import numpy as np

# created dummy dataframe with 3 feature
df = pd.DataFrame([[1,2,3],[50, 51,52],[2.0,6.0,8.5],[50.11,53.78,52]], columns = ['feature1', 'feature2','feature3'])
print(df)

obj = KMeansClusterer(2, distance = nltk.cluster.util.cosine_distance) #giving number of cluster 2
vectors = [np.array(f) for f in df.values]

df['predicted_cluster'] = obj.cluster(vectors,assign_clusters = True))

print(obj.means())
#OP
[array([50.055, 52.39 , 52.   ]), array([1.5 , 4.  , 5.75])] #which is going to be mean of three feature for 2 cluster, since number of cluster that we passed is 2

 #now if u want the cluster center in pandas dataframe 
 df['centroid'] = df['predicted_cluster'].apply(lambda x: obj.means()[x])

【讨论】：