Exercise 11.9 - K-means clustering by hand

import math 
def distance(a,b): 
   return math.sqrt((a[0]-b[0])**2+(a[1]-b[1])**2) 
data=[[1,0],[3,0],[5,0],[7,0],[9,0],[11,0],[13,0],[15,0],[17,0], 
     [1,3],[3,3],[5,3],[7,3],[9,3],[11,3],[13,3],[15,3],[17,3]] 
c1=[9,3] 
c2=[11,3] 
membership=[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1] 
for i in range(10): 
   c1i=0 
   c1x=0 
   c1y=0 
   c2i=0 
   c2x=0 
   c2y=0 
   for j in range(len(data)): 
       if (distance(data[j],c1)<=distance(data[j],c2)): 
           membership[i]=1 
           c1i=c1i+1 
           c1x=c1x+data[j][0] 
           c1y=c1y+data[j][1] 
       else: 
           membership[i]=2 
           c2i=c2i+1 
           c2x=c2x+data[j][0] 
           c2x=c2x+data[j][1] 
   c1[0]=c1x/c1i 
   c1[1]=c1y/c1i 
   c2[0]=c2x/c2i 
   c2[1]=c2y/c2i 
print(c1) 
print(c2)

The result is:

[5.0, 1.5] 
[15.5, 0.0]