Python机器学习：梯度下降法008如何确定梯度计算的准确性 调试梯度下降法

#调试梯度import numpy as npimport matplotlib.pyplot as pltnp.random.seed(666)X = np.random.random(size = (1000,10))true_theta = np.arange(1,12,dtype = float)X_b = np.hstack([np.ones((len(X),1)),X])y = X_b.dot(true_theta) + np.random.normal(size = 1000)

print(true_theta)

[ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.]

def J(theta,X_b,y):try:return np.sum((y - X_b.dot(theta)) ** 2) / len(X_b)except:return float('inf')def dJ_math(theta,X_b,y):return X_b.T.dot(X_b.dot(theta) - y) * 2. / len(y)def dJ_debug(theta,X_b,y,epsilon = 0.01):res = np.empty(len(theta))for i in range(len(theta)):theta_1 = theta.copy()theta_1[i] += epsilontheta_2 = theta.copy()theta_2[i] -= epsilonres[i] = (J(theta_1,X_b,y) - J(theta_2,X_b,y)) / (2 * epsilon)return resdef gradient_descent(dJ,X_b,initial_theta,eta,n_iters = 1e4,epsilon = 1e-4):theta = initial_thetacur_iter = 0while cur_iter < n_iters:gradient = dJ(theta,X_b,y)last_theta = thetatheta = theta - eta * gradientif (np.abs(J(theta,X_b,y) - J(last_theta,X_b,y)) < epsilon):breakcur_iter += 1return theta

X_b = np.hstack([np.ones((len(X),1)),X])initial_theta = np.zeros(X_b.shape[1])eta = 0.1theta = gradient_descent(dJ_debug,X_b,initial_theta,eta)print(theta)

[ 1.60733388 1.98991873 2.80925629 4.00526045 4.9565647 5.82284661 6.86831506 7.92353791 8.75087035 9.900878 10.80269625]