我試圖通過計(jì)算函數(shù)中的值來獲得梯度下降。并在我的代碼中出錯(cuò)def gradient_descent(X, y, theta, alpha, num_iters):m = len(y)cost_history = np.zeros(num_iters)theta_history = np.zeros((num_iters,2))for i in range(num_iters):    prediction = np.reshape(np.dot(np.transpose(theta), X),97)    theta = theta -(1/m)*alpha*( X.T.dot((prediction - y)))    theta_history[i,:] =theta.T    J_history[i]  = cal_cost(theta,X,y)return theta, J_history"""Args----X (numpy mxn array) - The example inputs, first column is expected   to be all 1's.y (numpy m array) - A vector of the correct outputs of length mtheta (numpy nx1 array) - An array of the set of theta parameters   to evaluatealpha (float) - The learning rate to use for the iterative gradient   descentnum_iters (int) - The number of gradient descent iterations to performReturns-------theta (numpy nx1 array) - The final theta parameters discovered after    out gradient descent.J_history (numpy num_itersx1 array) - A history of the calculated    cost for each iteration of our descent."""以下是我傳遞給函數(shù)和變量的參數(shù)theta = np.zeros( (2, 1) )iterations = 1500;alpha = 0.01theta, J = gradient_descent(X, y, theta, alpha, iterations)錯(cuò)誤信息是：ValueError：形狀（97,2）和（97，）未對(duì)齊：2（dim 1）！= 97（dim 0）