大小限制的聚類問題

1 主要方法

# 導入庫 from size_constrained_clustering import fcm, equal, minmax, shrinkage,da # by default it is euclidean distance, but can select others from sklearn.metrics.pairwise import haversine_distances from sklearn.datasets import make_blobs import numpy as np import matplotlib.pyplot as plt

1.1 Fuzzy C-means Algorithm

和KMeans類似，不過利用了歸屬概率（membership probability）進行計算，而不是直接的0或者1

n_samples = 2000 n_clusters = 4 centers = [(-5, -5), (0, 0), (5, 5), (7, 10)] X, _ = make_blobs(n_samples=n_samples, n_features=2, cluster_std=1.0,centers=centers, shuffle=False, random_state=42) #生成數據集，每個主句兩個特征值，一共2000個樣本，四個分類，同時設定了聚類中心點的位置model = fcm.FCM(n_clusters)# use other distance function: e.g. haversine distance # model = fcm.FCM(n_clusters, distance_func=haversine_distances)model.fit(X)centers = model.cluster_centers_ ''' array([[ 0.06913083, 0.07352352],[-5.01038079, -4.98275774],[ 6.99974221, 10.01169349],[ 4.98686053, 5.0026792 ]]) 模型擬合之后，樣本聚類的中心點 '''labels = model.labels_ #模型擬合后，每個樣本的類別 plt.figure(figsize=(10,10))colors=['red','green','blue','yellow']for i,color in enumerate(colors):color_tmp=np.where(labels==i)[0]plt.scatter(X[color_tmp,0],X[color_tmp,1],c=color,label=i)plt.legend() plt.scatter(centers[:,0],centers[:,1],s=1000,c='black')

1. 2 Same Size Contrained KMeans Heuristics

利用啟發式的方法獲取等大聚類結果

n_samples = 2000 n_clusters = 4 X = np.random.rand(n_samples, 2) # use minimum cost flow framework to solve model = equal.SameSizeKMeansHeuristics(n_clusters)model.fit(X) centers = model.cluster_centers_ labels = model.labels_ import matplotlib.pyplot as plt plt.figure(figsize=(10,10)) colors=['red','green','blue','yellow'] for i,color in enumerate(colors):color_tmp=np.where(labels==i)[0]plt.scatter(X[color_tmp,0],X[color_tmp,1],c=color,label=i) plt.legend() plt.scatter(centers[:,0],centers[:,1],s=1000,c='black')

1.3 Same Size Contrained KMeans Inspired by Minimum Cost Flow Problem：

將聚類轉換為分配問題，并用最小費用流的思路進行求解

n_samples = 2000 n_clusters = 4 X = np.random.rand(n_samples, 2) # use minimum cost flow framework to solve model = equal.SameSizeKMeansMinCostFlow(n_clusters)model.fit(X) centers = model.cluster_centers_ labels = model.labels_ plt.figure(figsize=(10,10)) colors=['red','green','blue','yellow']for i,color in enumerate(colors):color_tmp=np.where(labels==i)[0]plt.scatter(X[color_tmp,0],X[color_tmp,1],c=color,label=i)plt.legend() plt.scatter(centers[:,0],centers[:,1],s=1000,c='black')

1.4 Minimum and Maximum Size Constrained KMeans Inspired by Minimum Cost Flow Problem

將聚類轉換為分配問題，并用最小費用流的思路進行求解，加入最小和最大聚類規模限制

n_samples = 2000 n_clusters = 4 X = np.random.rand(n_samples, 2) # use minimum cost flow framework to solvemodel = minmax.MinMaxKMeansMinCostFlow(n_clusters, size_min=200, size_max=800)model.fit(X) centers = model.cluster_centers_ labels = model.labels_plt.figure(figsize=(10,10)) colors=['red','green','blue','yellow'] for i,color in enumerate(colors):color_tmp=np.where(labels==i)[0]plt.scatter(X[color_tmp,0],X[color_tmp,1],c=color,label=i) plt.legend() plt.scatter(centers[:,0],centers[:,1],s=1000,c='black')

1.5 Deterministic Annealling Algorithm:

輸入目標每類規模比例，獲得相應聚類規模的結果。

n_samples = 2000 n_clusters = 4 X = np.random.rand(n_samples, 2) # use minimum cost flow framework to solve model = da.DeterministicAnnealing(n_clusters, distribution=[0.1, 0.2,0.4, 0.3])model.fit(X) centers = model.cluster_centers_ labels = model.labels_plt.figure(figsize=(10,10)) colors=['red','green','blue','yellow'] for i,color in enumerate(colors):color_tmp=np.where(labels==i)[0]plt.scatter(X[color_tmp,0],X[color_tmp,1],c=color,label=i) plt.legend() plt.scatter(centers[:,0],centers[:,1],s=1000,c='black')

總結

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