好的，所以我必须说 OpenCV 提供的示例 K-means 算法程序相当混乱。即使花了整个下午都没有得到完整的画面。这些是我想问的几个问题：
1）我如何将给定的图像转换为单列矩阵，因为 K-means 函数只接受这样的矩阵作为输入？我知道我必须使用 CvMat 函数，但无法弄清楚到底如何。
2）是否可以根据颜色强度进行聚类，使用一些预先确定的强度作为种子值？
最后但并非最不重要的一点，它会很高；如果有人可以提供任何详细解释 K-means 的链接，您将不胜感激。我已经经历了willowgarage和aishack的解释，仍然存在疑问。提前致谢 ！！
这正是我想要做的：假设这是提供的图像
我的代码的输出应该有点像这样：
如您所见，在第二张图像中，由于阴影的影响被移除，我们得到了具有明确颜色层的图像。
现在为此我应用以下方法首先我根据图像的相应 LAB 值选择种子颜色。然后在获得种子值后，我尝试使用 K-means 聚类将相似的颜色聚类成一个确定的颜色层。（如上图所示）。
//Aim:To implement Kmeans clustering algorithm.
//Program
import java.util.*;
class k_means
{
static int count1,count2,count3;
static int d[];
static int k[][];
static int tempk[][];
static double m[];
static double diff[];
static int n,p;
static int cal_diff(int a) // This method will determine the cluster in which an element go at a particular step.
{
int temp1=0;
for(int i=0;i<p;++i)
{
if(a>m[i])
diff[i]=a-m[i];
else
diff[i]=m[i]-a;
}
int val=0;
double temp=diff[0];
for(int i=0;i<p;++i)
{
if(diff[i]<temp)
{
temp=diff[i];
val=i;
}
}//end of for loop
return val;
}
static void cal_mean() // This method will determine intermediate mean values
{
for(int i=0;i<p;++i)
m[i]=0; // initializing means to 0
int cnt=0;
for(int i=0;i<p;++i)
{
cnt=0;
for(int j=0;j<n-1;++j)
{
if(k[i][j]!=-1)
{
m[i]+=k[i][j];
++cnt;
}}
m[i]=m[i]/cnt;
}
}
static int check1() // This checks if previous k ie. tempk and current k are same.Used as terminating case.
{
for(int i=0;i<p;++i)
for(int j=0;j<n;++j)
if(tempk[i][j]!=k[i][j])
{
return 0;
}
return 1;
}
public static void main(String args[])
{
Scanner scr=new Scanner(System.in);
/* Accepting number of elements */
System.out.println("Enter the number of elements ");
n=scr.nextInt();
d=new int[n];
/* Accepting elements */
System.out.println("Enter "+n+" elements: ");
for(int i=0;i<n;++i)
d[i]=scr.nextInt();
/* Accepting num of clusters */
System.out.println("Enter the number of clusters: ");
p=scr.nextInt();
/* Initialising arrays */
k=new int[p][n];
tempk=new int[p][n];
m=new double[p];
diff=new double[p];
/* Initializing m */
for(int i=0;i<p;++i)
m[i]=d[i];
int temp=0;
int flag=0;
do
{
for(int i=0;i<p;++i)
for(int j=0;j<n;++j)
{
k[i][j]=-1;
}
for(int i=0;i<n;++i) // for loop will cal cal_diff(int) for every element.
{
temp=cal_diff(d[i]);
if(temp==0)
k[temp][count1++]=d[i];
else
if(temp==1)
k[temp][count2++]=d[i];
else
if(temp==2)
k[temp][count3++]=d[i];
}
cal_mean(); // call to method which will calculate mean at this step.
flag=check1(); // check if terminating condition is satisfied.
if(flag!=1)
/*Take backup of k in tempk so that you can check for equivalence in next step*/
for(int i=0;i<p;++i)
for(int j=0;j<n;++j)
tempk[i][j]=k[i][j];
System.out.println("\n\nAt this step");
System.out.println("\nValue of clusters");
for(int i=0;i<p;++i)
{
System.out.print("K"+(i+1)+"{ ");
for(int j=0;k[i][j]!=-1 && j<n-1;++j)
System.out.print(k[i][j]+" ");
System.out.println("}");
}//end of for loop
System.out.println("\nValue of m ");
for(int i=0;i<p;++i)
System.out.print("m"+(i+1)+"="+m[i]+" ");
count1=0;count2=0;count3=0;
}
while(flag==0);
System.out.println("\n\n\nThe Final Clusters By Kmeans are as follows: ");
for(int i=0;i<p;++i)
{
System.out.print("K"+(i+1)+"{ ");
for(int j=0;k[i][j]!=-1 && j<n-1;++j)
System.out.print(k[i][j]+" ");
System.out.println("}");
}
}
}
/*
Enter the number of elements
8
Enter 8 elements:
2 3 6 8 12 15 18 22
Enter the number of clusters:
3
At this step
Value of clusters
K1{ 2 }
K2{ 3 }
K3{ 6 8 12 15 18 22 }
Value of m
m1=2.0 m2=3.0 m3=13.5
At this step
Value of clusters
K1{ 2 }
K2{ 3 6 8 }
K3{ 12 15 18 22 }
Value of m
m1=2.0 m2=5.666666666666667 m3=16.75
At this step
Value of clusters
K1{ 2 3 }
K2{ 6 8 }
K3{ 12 15 18 22 }
Value of m
m1=2.5 m2=7.0 m3=16.75
At this step
Value of clusters
K1{ 2 3 }
K2{ 6 8 }
K3{ 12 15 18 22 }
Value of m
m1=2.5 m2=7.0 m3=16.75
The Final Clusters By Kmeans are as follows:
K1{ 2 3 }
K2{ 6 8 }
K3{ 12 15 18 22 } */