如何自动化集群的数量？

Question

Edit: I accept that my question has been closed for being similar but I think the answers have provided valuable knowledge for others so this should be open.

我一直在玩下面的脚本：

from sklearn.feature_extraction.text import TfidfVectorizer
from sklearn.cluster import KMeans
from sklearn.metrics import adjusted_rand_score
import textract
import os

folder_to_scan = '/media/sf_Documents/clustering'
dict_of_docs = {}

# Gets all the files to scan with textract
for root, sub, files in os.walk(folder_to_scan):
    for file in files:
        full_path = os.path.join(root, file)
        print(f'Processing {file}')
        try:
            text = textract.process(full_path)
            dict_of_docs[file] = text
        except Exception as e:
            print(e)


vectorizer = TfidfVectorizer(stop_words='english')
X = vectorizer.fit_transform(dict_of_docs.values())

true_k = 3
model = KMeans(n_clusters=true_k, init='k-means++', max_iter=100, n_init=1)
model.fit(X)

print("Top terms per cluster:")
order_centroids = model.cluster_centers_.argsort()[:, ::-1]
terms = vectorizer.get_feature_names()
for i in range(true_k):
    print("Cluster %d:" % i,)
    for ind in order_centroids[i, :10]:
        print(' %s' % terms[ind],)

它扫描一个包含扫描文档图像的文件夹，提取文本然后对文本进行聚类。我知道事实上有 3 种不同类型的文档，所以我将 true_k 设置为 3。但是如果我有一个未知文档文件夹，其中可能有 1 到 100 种不同文档类型。

Answer 1

这是一个不稳定的领域，因为很难衡量 "good" 您的聚类算法在没有任何基本事实标签的情况下如何工作。为了进行自动选择，您需要有一个指标来比较 KMeans 对 n_clusters.

的不同值的执行情况。

一个流行的选择是剪影得分。您可以找到有关它的更多详细信息 here。这是 scikit-learn 文档：

The Silhouette Coefficient is calculated using the mean intra-cluster distance (a) and the mean nearest-cluster distance (b) for each sample. The Silhouette Coefficient for a sample is (b - a) / max(a, b). To clarify, b is the distance between a sample and the nearest cluster that the sample is not a part of. Note that Silhouette Coefficient is only defined if number of labels is 2 <= n_labels <= n_samples - 1.

因此，您只能计算 n_clusters >= 2 的剪影分数（不幸的是，鉴于您的问题描述，这可能是您的限制）。

这就是你在虚拟数据集上使用它的方式（你可以根据你的代码调整它，只是为了有一个可重现的例子）：

from sklearn.datasets import load_iris
from sklearn.cluster import KMeans
from sklearn.metrics import silhouette_score

iris = load_iris()
X = iris.data

sil_score_max = -1 #this is the minimum possible score

for n_clusters in range(2,10):
  model = KMeans(n_clusters = n_clusters, init='k-means++', max_iter=100, n_init=1)
  labels = model.fit_predict(X)
  sil_score = silhouette_score(X, labels)
  print("The average silhouette score for %i clusters is %0.2f" %(n_clusters,sil_score))
  if sil_score > sil_score_max:
    sil_score_max = sil_score
    best_n_clusters = n_clusters

这将 return:

The average silhouette score for 2 clusters is 0.68
The average silhouette score for 3 clusters is 0.55
The average silhouette score for 4 clusters is 0.50
The average silhouette score for 5 clusters is 0.49
The average silhouette score for 6 clusters is 0.36
The average silhouette score for 7 clusters is 0.46
The average silhouette score for 8 clusters is 0.34
The average silhouette score for 9 clusters is 0.31

因此你将有 best_n_clusters = 2（注意：实际上，Iris 有三个类...）

如何自动化集群的数量？

How do I automate the number of clusters?

python

cluster-analysis

machine-learning

scikit-learn