二分kmeans python實現

標籤：

今天要對一個1000個個記錄，每個記錄有n個屬性的文本進行聚類，採用的是二分k均值方法。

演算法思想：

我參考了Pang-Ning Tan資料採礦導論裡P317

相對於kmeans的優點是不受其初始質心的影響。

#coding utf-8
#python 3.4
#2015-4-3
#Fitz Yin  
#yinruyi.hm@gmail.com
from sklearn.cluster import KMeansimport numpy as npdef makedict(f):    #建立行號和每行資料間的字典關係    a = [line.split() for line in f]    data_dict = {}    for i in range(len(a)):        data_dict[i] = a[i]    return data_dictdef kmeans(data):    #kmeans演算法    data = np.array(data)    computer=KMeans(n_clusters=2)    computer.fit(data)    labels = computer.labels_    one_class = []    zero_class = []    for i in range(len(labels)):        if labels[i] == 1:            one_class.append(i)#0類的行號        else:            zero_class.append(i)#1類的行號    centers = computer.cluster_centers_#找到中心    cohesion_0,cohesion_1 = -1,-1#初始化，自己和自己的cos是1    for i in zero_class:        cohesion_0 += judge_cos(data[i],centers[0])#0類cos評價    for i in one_class:        cohesion_1 += judge_cos(data[i],centers[1])#1類cos評價    return zero_class,one_class,cohesion_0,cohesion_1def judge_cos(x,y):    #cos評價函數    af,bf,ab = 0,0,0    for i in range(len(x)):        af = float(x[i])*float(x[i])        bf = float(y[i])*float(y[i])        ab = float(x[i])*float(y[i])    if af == 0 or bf == 0:        print(‘error‘)        return 0        #本例中不出現全是0情況    else:        cos_value = ab/(np.sqrt(af)*np.sqrt(bf))        return cos_valuedef gettransdict(split_set,split_number):    #建立kmeans計算的矩陣和原來矩陣 兩個行號之間的字典關係    a = split_set[split_number][0]    transdict = {}    for i in range(len(a)):        transdict[i] = a[i]    return transdictdef getsplitset(split_set,split_number):    #簇中去掉要分的簇    new_split_set = []    for i in range(len(split_set)):        if i == split_number:            pass        else:            new_split_set.append(split_set[i])    return new_split_setdef getsplitnumber(split_set):    #找尋待分簇的編號    split_number = 0    temp = []    for i in range(len(split_set)):        temp.append(split_set[i][1])    for i in range(len(temp)):        if temp[split_number] > temp[i]:            split_number = i    return split_numberdef main():    f = open(‘train.txt‘,‘r‘,encoding=‘utf-8‘).readlines()    data_dict = makedict(f)    k = 3#分類個數    #sse = 0.001    split_set = [[[i for i in range(1000)],0]]#此處1000是行號    split_number = 0#需要分類的簇標號    while len(split_set) != k:        transdict = gettransdict(split_set,split_number)#轉換字典        array2kmeans = [data_dict[i] for i in split_set[split_number][0]]#擷取二分kmeans計算矩陣        zero_class,one_class,cohesion_0,cohesion_1 = kmeans(array2kmeans)        real_zero_class = [transdict[i] for i in zero_class]#分裂後的簇0        real_one_class = [transdict[i] for i in one_class]#分裂後的簇1        split_set = getsplitset(split_set,split_number)#將總的簇中去掉分的大的簇        split_set.append([real_zero_class,cohesion_0])        split_set.append([real_one_class,cohesion_1])#總的簇中加入分完的小簇        split_number = getsplitnumber(split_set)#擷取下一個迴圈待分的簇編號    print(split_set)    #[[[行號1類],sse1],[[行號2類],sse2],[[行號三類],sse3]]if __name__ == ‘__main__‘:    main()

 

二分kmeans python實現