Posted in Python onMarch 16, 2014
简单实现平面的点K均值分析,使用欧几里得距离,并用pylab展示。
import pylab as pl #calc Euclid squire def calc_e_squire(a, b): return (a[0]- b[0]) ** 2 + (a[1] - b[1]) **2 #init the 20 point a = [2,4,3,6,7,8,2,3,5,6,12,10,15,16,11,10,19,17,16,13] b = [5,6,1,4,2,4,3,1,7,9,16,11,19,12,15,14,11,14,11,19] #define two k_value k1 = [6,3] k2 = [6,1] #defint tow cluster sse_k1 = [] sse_k2 = [] while True: sse_k1 = [] sse_k2 = [] for i in range(20): e_squire1 = calc_e_squire(k1, [a[i], b[i]]) e_squire2 = calc_e_squire(k2, [a[i], b[i]]) if (e_squire1 <= e_squire2): sse_k1.append(i) else: sse_k2.append(i) #change k_value k1_x = sum([a[i] for i in sse_k1]) / len(sse_k1) k1_y = sum([b[i] for i in sse_k1]) / len(sse_k1) k2_x = sum([a[i] for i in sse_k2]) / len(sse_k2) k2_y = sum([b[i] for i in sse_k2]) / len(sse_k2) if k1 != [k1_x, k1_y] or k2 != [k2_x, k2_y]: k1 = [k1_x, k1_y] k2 = [k2_x, k2_y] else: break kv1_x = [a[i] for i in sse_k1] kv1_y = [b[i] for i in sse_k1] kv2_x = [a[i] for i in sse_k2] kv2_y = [b[i] for i in sse_k2] pl.plot(kv1_x, kv1_y, 'o') pl.plot(kv2_x, kv2_y, 'or') pl.xlim(1, 20) pl.ylim(1, 20) pl.show()
python实现k均值算法示例(k均值聚类算法)
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