求一个商汤今晚算法题的题解~~~

我觉得找离群点有点类似KMEAN算法，考完后做出来，但不确定对不对 from numpy import * def distEclud(vecA, vecB): x = sqrt(pow(vecA[0]-vecB[0],2)+pow(vecA[1]-vecB[1],2)) return x def Kmeans_b(dataSet,N): clusterAssment = [[0,0] for i in range(N)] # 创建一个簇分配结果矩阵，第0列记录输入哪一个簇，第1列记录存储误差 # 使用第一个作为质心 centroids = dataSet[0] clusterChanged = True K = int(N * 0.3) past_centroids = centroids while clusterChanged: clusterChanged = False for i in range(N): clusterAssment[i][1] = distEclud(dataSet[i],centroids) arg_clusterAssment = argsort(clusterAssment,axis=0) # 修改属于哪一类 for x in range(N): if x<K: clusterAssment[arg_clusterAssment[x][1]][0] = 1 else: clusterAssment[arg_clusterAssment[x][1]][0] = 0 # 更新质心 temp = [] for x in clusterAssment: if x[0]==1: temp.append(x) centroids = sum(temp,axis=0)/K if distEclud(centroids,past_centroids) <= 0.001: clusterChanged = True past_centroids = centroids for i in clusterAssment: print(i[0]) N = 20 data =[[1, 1], [0, 0], [0, 1], [-200, -100], [-200, 0], [-200, 100], [-100, -100], [-100, 0], [-100, 100], [0, -100], [0, 100], [100, -100], [100, 0], [100, 100], [200, -100], [200, 0], [200, 100], [2, 1], [2, 0], [1, 0]] # N = int(input()) # data = [] # for i in range(N): # data.append(list(map(int,input().split()))) Kmeans_b(data,N)