octave - 下标索引必须是实数正整数或逻辑数

Question

function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
%GRADIENTDESCENT Performs gradient descent to learn theta
%   theta = GRADIENTDESENT(X, y, theta, alpha, num_iters) updates theta by 
%   taking num_iters gradient steps with learning rate alpha

% Initialize some useful values  
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);

for iter = 1:num_iters

% ====================== YOUR CODE HERE ======================
% Instructions: Perform a single gradient step on the parameter vector
%               theta. 
%
% Hint: While debugging, it can be useful to print out the values
%       of the cost function (computeCost) and gradient here.
%

    hypothesis = x*theta;
    theta_0 = theta(1) - alpha(1/m)*sum((hypothesis-y)*x);
    theta_1 = theta(2) - alpha(1/m)*sum((hypothesis-y)*x);
    theta(1) = theta_0;
    theta(2) = theta_1;








% ============================================================

% Save the cost J in every iteration    
    J_history(iter) = computeCost(X, y, theta);

end

end

我不断收到此错误

error: gradientDescent: subscript indices must be either positive integers less than 2^31 or logicals

在这条线上，就在第一个 theta 和 = 之间

theta_0 = theta(1) - alpha(1/m)*sum((hypothesis-y)*x);

我对 Octave 非常陌生，所以请放轻松，并且
先感谢您。
这是来自第 2 周机器学习类(class)的类(class)

Answer 1

99% 确定你的错误在 topsig 指出的那一行，你有 alpha(1/m)
如果您为您的函数提供输入值示例以及您希望作为输出看到的内容，这将有所帮助，但我从您的评论中假设
% taking num_iters gradient steps with learning rate alpha
那个alpha是常数，不是函数。因此，您的线路是 alpha(1/m)中间没有任何运算符。当您编制索引时， Octave 会看到这一点 alpha值为 1/m .

即，如果你有一个数组

x = [3 4 5]
x*(2) = [6 8 10]  %% two times each element in the array
x(2) = [4]  %% second element in the array

你所做的似乎没有意义，因为 'm = length(y)' 将输出一个标量，所以

x = [3 4 5]; m = 3;
x*(1/m) = x*(1/3) = [1 1.3333 1.6666]  %% element / 3
x(1/m) = ___error___  %% the 1/3 element in the array makes no sense

请注意，对于某些错误，它始终表示错误的位置在赋值运算符处(行首的等号)。如果它指向那里，您通常必须在行中的其他地方查找实际错误。在这里，它因为您试图应用非整数下标而对您大喊大叫 (1/m)