我是新来的,我很抱歉这个“问题”对你们中的一些人来说有点太愚蠢了。我必须在大学做一个关于 k-mean 聚类的 C++ 项目,我需要一些帮助。这是代码。这是工作。现在,我必须单独构建 G-Matrix。在代码中,我得到以下信息:
4.30 0.50 * 1 *
3.54 0.50 * 1 *
0.71 3.20 * 0 *
0.71 4.61 * 0 *
质心坐标为:
4.50
3.50
质心坐标为:
1.50
1.00
这很好,但我需要1,1,0,0
一个额外的 G 矩阵,如下所示:
A B C D
1 1 0 0 ->c1
0 0 1 1 ->c2
A,B,C,D
点在哪里?c1
是c2
质心。知道如何显示这个吗?
这是我的代码:
float dmin, dpoint;
float sum[2][2];
int cluster[4], count[4], group;
float flips;
const int rows = 4;
const int columns = 2;
const int crows = 2;
const int ccolumns = 2;
// initialize the points
int point[rows][columns]={{1,1},{2,1},{4,3},{5,4}};
// initialize the centroids
double centroid [crows][ccolumns] = {{1,1},{2,1}};
// ...
for (i = 0; i<4; i++) cluster[i] = 0;
// until there is no change of clusters belonging to each pattern, continue
flips = 4;
while (flips>0) {
flips = 0;
for (j = 0; j < 2; j++)
{
count[j] = 0;
for (i = 0; i < 2; i++)
sum[j][i] = 0;
}
// now, we need to calculate the distance
for (i = 0; i < 4; i++) {
dmin = 2; group = cluster[i];
for (j = 0; j < 2; j++)
{
dpoint = 0.0;
dpoint += sqrt(pow((point[i][0] - centroid[j][0]),2)+pow((point[i][1] - centroid[j][1]),2));
fprintf(stdout, "%5.2f ", dpoint); // Show the value of the distance
if (dpoint < dmin) {
group = j;
dmin = dpoint;
}
}
// now, we need to calculate the G matrix (1 or 0)
fprintf(stdout, " * %d *\n", group); // displays 0 or 1 (to which cluster it belongs)
if (cluster[i] != group)
{
flips++;
cluster[i] = group; // repeat this process until G(n)=G(n+1)
}
count[cluster[i]]++;
for (j = 0; j < 2; j++)
sum[cluster[i]][j] += point[i][j];
}
// now, display the coordinates of the centroid
for (i = 0; i < 2; i++) {
fprintf(stderr," The coordinates of the centroid are: \n");
for (j = 0; j < 2; j++) {
centroid[i][j] = sum[i][j]/count[i];
fprintf(stderr, "%5.2f \n", centroid[i][j]);
}
}
}
}
谢谢你的帮助!