Question

使用

X = [1.8421 4.6076;
     5.6586 4.8;
     6.3526 3.2909;
     2.904  4.6122;
     3.232  4.9399;
     1.2479 4.9327]

还有

centroids = [3  3;
             6  2;
             8  5]

我正在尝试找到x中每个点最近的质心。我正在用八度编程，这是我的代码

K = size(centroids, 1);
idx = zeros(size(X,1), 1);   %idx is the vector storing the index of the closest centroid
for e = 1: size(X,1)
    difference(1, :) = X(e,:) - centroids(1,:);
    min_distance = sum(difference(1,:).^2);
    for j = 2:K
        difference(j, :) = X(e,:) - centroids(j,:);
        distance = sum(difference.^2);
        if distance<min_distance
            min_distance = distance;
            idx(e) = centroids(j);
        endif
    endfor
endfor

代码有效，但我只能得到

idx = 0 0 0

x的前三个条目

Answer 1

计算时，您需要保持操作数大小的一致性距离并正确设置idx：

进行此更改后，输出为：

K = size(centroids, 1);
difference=zeros(size(centroids));
idx = zeros(size(X,1), 1);   %idx is the vector storing the index of the closest centroid
for e = 1: size(X,1)
    difference(1, :) = X(e,:) - centroids(1,:);
    min_distance = sum(difference(1,:).^2);
    idx(e)=1;
    for j = 2:K
        difference(j, :) = X(e,:) - centroids(j,:);
        distance = sum(difference(j,:).^2);
        if (distance<min_distance)
            min_distance = distance;
            idx(e) = j;
        endif
    endfor
endfor
idx

Answer 2

使用 Octave 时，您可能会发现 min 函数很有用；)

[minimum_values, minimum_index] = min(k);

您可以为 idx 使用 minimum_index 值，例如：

idx(i) = minimum_index;

因此，您可以将其进一步减少到：

[minimum_values, idx(i)] = min(k);

在k中找到最接近的质心意味着聚类

2 个答案: