我目前正在做一个机器学习课程，我的代码有一个小问题。我正在创建一个函数，它将返回最接近训练示例X的聚类质心的索引，其中initial_centroids = [3 3; X1 -3 = [1.8421，4.6076; 5.6586，4.8; 6.3526，3.2909]
我的想法是做一个for循环，计算X1和质心1/2/3之间的距离，然后选择最小的一个，返回最小的索引。
调用函数时，我的答案（已提供）应为[1 3 2]。
我收到的是[1 0 0]。
我相信我的for循环有一个错误，因为idx之前被设置为0的向量，而现在只有第一个0被改为1。
我似乎找不到错误，我是编程新手，我不想看这些练习的现成解决方案，所以我希望有人至少能指导我正确的方向，不需要直接的解决方案。

function idx = findClosestCentroids(X, centroids)
%FINDCLOSESTCENTROIDS computes the centroid memberships for every example
%   idx = FINDCLOSESTCENTROIDS (X, centroids) returns the closest centroids
%   in idx for a dataset X where each row is a single example. idx = m x 1 
%   vector of centroid assignments (i.e. each entry in range [1..K])
%

% Set K
K = size(centroids, 1);
tempidx = zeros(1,2);

% You need to return the following variables correctly.
idx = zeros(size(X,1), 1);
for i = 1:X;
    C = [(X(i,:)-centroids(1,:)).^2;(X(i,:)-centroids(2,:)).^2;(X(i,:)-centroids(3,:)).^2];
    Ctot = sum(C(i),2);
    [res,dx] = min(Ctot);
    idx(i,:) = dx;
end;
% ====================== YOUR CODE HERE ======================
% Instructions: Go over every example, find its closest centroid, and store
%               the index inside idx at the appropriate location.
%               Concretely, idx(i) should contain the index of the centroid
%               closest to example i. Hence, it should be a value in the 
%               range 1..K
%
% Note: You can use a for-loop over the examples to compute this.
%


The lines that call function and return [1 0 0] (should be [1 3 2] are:
idx = findClosestCentroids(X, initial_centroids);
fprintf('Closest centroids for the first 3 examples: %d %d %d', idx(1:3))

我试图改变的是：

for i = 1:X;
    C = [(X(i,:)-centroids(1,:)).^2;(X(i,:)-centroids(2,:)).^2;(X(i,:)-centroids(3,:)).^2];
    Ctot(i) = sum(C(i),2);
    [res,dx] = min(Ctot(i));
    idx(i,:) = dx(i);
end;

还是不工作，返回[100]....
我想也许我指的不是迭代。