KMeans的C++及Python实现
1. C++实现
#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>
#include <ctime>
using namespace std;
const int k = 3;
const int dims = 4;
const int dataNum = 150;
typedef vector<double> Tuple;
void doKmeans(vector<Tuple>& tuples);
void assignTuples(vector<Tuple> clusters[],vector<Tuple> tuples,Tuple means[]);
double getDist(const Tuple& t1, const Tuple& t2);
double getVal(vector<Tuple> clusters[], Tuple means[]);
Tuple updateMeans(const vector<Tuple>& cluster_i);
void print(vector<Tuple> clustes[]);
int main()
{
char filename[] = "bezdekIris.data";
fstream file(filename); //打开存放样本数据的文件
if (!file)
{
cout << "Cannot open the file" << endl;
return 0;
}
vector<Tuple> tuples;
int pos = 0;
while (!file.eof())
{
string str;
getline(file,str);
stringstream ss(str);
Tuple tuple(dims+1,0);
tuple[0] = pos + 1;
for (int i = 1; i <= dims; i++)
ss >> tuple[i];
tuples.push_back(tuple);
pos++;
}
//
doKmeans(tuples);
system("pause");
return 0;
}
void doKmeans(vector<Tuple>& tuples)
{
cout << "初始化......" << endl;
//初始化k个聚类中心
vector<Tuple> clusters[k];
Tuple means[k];
srand((unsigned)time(NULL));
for (int i = 0; i < k; i++)
{
int temp = rand()%tuples.size();
means[i] = tuples[temp];
}
//将样本分配到与其最近的聚类中心
assignTuples(clusters,tuples,means);
//计算初始整体误差平方和
double newVal = getVal(clusters,means);
double oldVal = -1;
int t = 0;
while ((abs(newVal - oldVal)) > 1)
{
cout << "开始第" << t + 1 << "次迭代......" << endl;
cout << "更新聚类中心..." << endl;
for (int i = 0; i < k; i++)
means[i] = updateMeans(clusters[i]);
cout << "计算新的整体误差平方和..." << endl;
oldVal = newVal;
newVal = getVal(clusters,means);
}
print(clusters);
}
void assignTuples(vector<Tuple> clusters[], vector<Tuple> tuples,Tuple means[])
{
for (int i = 0; i < tuples.size(); i++)
{
int label = 0;
double dist = getDist(tuples[i],means[0]);
for (int j = 1; j < k; j++)
{
double temp = getDist(tuples[i],means[j]);
if (temp < dist)
{
dist = temp;
label = j;
}
}
clusters[label].push_back(tuples[i]);
}
}
double getDist(const Tuple& t1, const Tuple& t2)
{
double sum = 0;
for (int i = 1; i <= dims; i++)
sum += (t1[i] - t2[i]) * ((t1[i] - t2[i]));
return sum;
}
double getVal(vector<Tuple> clusters[], Tuple means[])
{
double val = 0;
for (int i = 0; i < k; i++)
{
vector<Tuple> t = clusters[i];
for (int j = 0; j < t.size(); j++)
val += getDist(t[j],means[i]);
}
return val;
}
Tuple updateMeans(const vector<Tuple>& cluster_i)
{
Tuple t(dims+1,0);
for (int i = 0; i < cluster_i.size(); i++)
for (int j = 1; j <= dims; j++)
t[j] += cluster_i[i][j];
for (int i = 1; i <= dims; i++)
t[i] /= cluster_i.size();
return t;
}
void print(vector<Tuple> clustes[])
{
for (int i = 0; i < k; i++)
{
cout << "第" << i+1 << "簇：" << endl;
vector<Tuple> t = clustes[i];
for (int j = 0; j < t.size(); j++)
{
for (int d = 0; d <= dims; d++)
cout << t[j][d] << " ";
cout << endl;
}
}
}
Python实现：
（1）kmeans.py
from numpy import *
import pdb
import matplotlib.pyplot as plt
def createCenter(dataSet,k):
n = shape(dataSet)[0]
d = shape(dataSet)[1]
centroids = zeros((k,d))
for i in range(k):
c = int(random.uniform(0,n-1)) #浮点数
centroids[i,:] = dataSet[c,:]
return centroids
def getDist(vecA,vecB):
return sqrt(sum(power(vecA - vecB,2)))
def kmeans(dataSet, k):
n = shape(dataSet)[0]
clusterAssment = mat(zeros((n,2)))
centroids = createCenter(dataSet,k)
clusterChnaged = True
while clusterChnaged:
clusterChnaged = False
for i in range(n):
minDist = inf
minIndex = -1
for j in range(k):
distJI = getDist(dataSet[i,:],centroids[j,:])
if distJI < minDist:
minDist = distJI
minIndex = j
if clusterAssment[i,0] != minIndex: #收敛条件：分配结果不再变化
clusterChnaged = True
clusterAssment[i,:] = minIndex,minDist**2
#更新质心的位置
for i in range(k):
ptsdataSet = dataSet[nonzero(clusterAssment[:,0].A == i)[0]]
centroids[i,:] = mean(ptsdataSet,axis = 0) #沿矩阵的列方向进行矩阵计算
return centroids,clusterAssment
def print_result(dataSet,k,centroids,clusterAssment):
n,d = dataSet.shape
if d !=2:
print "Cannot draw!"
return 1
mark = ['or', 'ob', 'og', 'ok', '^r', '+r', 'sr', 'dr', '<r', 'pr']
if k > len(mark):
print "Sorry your k is too large"
return 1
for i in range(n):
markIndex = int(clusterAssment[i,0])
plt.plot(dataSet[i, 0],dataSet[i, 1],mark[markIndex])
mark = ['Dr', 'Db', 'Dg', 'Dk', '^b', '+b', 'sb', 'db', '<b', 'pb']
# draw the centroids
for i in range(k):
plt.plot(centroids[i, 0], centroids[i, 1], mark[i], markersize = 12)
plt.show()
(2)主函数实现：
from numpy import *
import matplotlib.pyplot as plt
import kmeans
dataSet = []
file = open('E:\\ZForWorks\\MLPython\\kmeansP\\testSet.txt')
for line in file.readlines():
strline = line.strip().split('\t')
sline = map(float,strline)
dataSet.append(sline)
dataSet = mat(dataSet)
k = 4
kcentroids, clusterAssment = kmeans.kmeans(dataSet,k)
kmeans.print_result(dataSet,k,kcentroids,clusterAssment)
结果：
print_result只针对二维的样本数据
测试数据：
1.658985 4.285136
-3.453687 3.424321
4.838138 -1.151539
-5.379713 -3.362104
0.972564 2.924086
-3.567919 1.531611
0.450614 -3.302219
-3.487105 -1.724432
2.668759 1.594842
-3.156485 3.191137
3.165506 -3.999838
-2.786837 -3.099354
4.208187 2.984927
-2.123337 2.943366
0.704199 -0.479481
-0.392370 -3.963704
2.831667 1.574018
-0.790153 3.343144
2.943496 -3.357075
-3.195883 -2.283926
2.336445 2.875106
-1.786345 2.554248
2.190101 -1.906020
-3.403367 -2.778288
1.778124 3.880832
-1.688346 2.230267
2.592976 -2.054368
-4.007257 -3.207066
2.257734 3.387564
-2.679011 0.785119
0.939512 -4.023563
-3.674424 -2.261084
2.046259 2.735279
-3.189470 1.780269
4.372646 -0.822248
-2.579316 -3.497576
1.889034 5.190400
-0.798747 2.185588
2.836520 -2.658556
-3.837877 -3.253815
2.096701 3.886007
-2.709034 2.923887
3.367037 -3.184789
-2.121479 -4.232586
2.329546 3.179764
-3.284816 3.273099
3.091414 -3.815232
-3.762093 -2.432191
3.542056 2.778832
-1.736822 4.241041
2.127073 -2.983680
-4.323818 -3.938116
3.792121 5.135768
-4.786473 3.358547
2.624081 -3.260715
-4.009299 -2.978115
2.493525 1.963710
-2.513661 2.642162
1.864375 -3.176309
-3.171184 -3.572452
2.894220 2.489128
-2.562539 2.884438
3.491078 -3.947487
-2.565729 -2.012114
3.332948 3.983102
-1.616805 3.573188
2.280615 -2.559444
-2.651229 -3.103198
2.321395 3.154987
-1.685703 2.939697
3.031012 -3.620252
-4.599622 -2.185829
4.196223 1.126677
-2.133863 3.093686
4.668892 -2.562705
-2.793241 -2.149706
2.884105 3.043438
-2.967647 2.848696
4.479332 -1.764772
-4.905566 -2.911070