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。

复制代码 代码如下:

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()