调试梯度下降法

如图，关于梯度的调试，通过不求导的方式来确定梯度计算的准确性。

在某点的两边各取一点，通过式

来得到某点的梯度。

#%% 如何调试梯度

import numpy as np
import matplotlib.pyplot as plt

np.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)

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] += epsilon
        theta_2 = theta.copy()
        theta_2[i] -= epsilon
        res[i] = (J(theta_1,X_b,y) - J(theta_2,X_b,y)) / (2 * epsilon)
    return res

def gradient_descent(dJ,X_b,y,initial_theta,eta,n_iters = 1e4,epsilon=1e-8):
    theta = initial_theta
    cur_iter = 0    # 初值设为0

    while cur_iter < n_iters:
        gradient = dJ(theta,X_b,y)
        last_theta = theta
        theta = theta - eta * gradient
        if(abs(J(theta,X_b,y) - J(last_theta,X_b,y)) < epsilon):
            break

        cur_iter += 1

    return theta

initial_theta = np.zeros(X_b.shape[1])
eta = 0.01

%time theta = gradient_descent(dJ_debug,X_b,y,initial_theta,eta)
%time theta = gradient_descent(dJ_math,X_b,y,initial_theta,eta)

通过代码最后两行验证，结果基本为第9行设置的 1-11 的值
结果为

array([ 1.1251597 ,  2.05312521,  2.91522497,  4.11895968,  5.05002117,
        5.90494046,  6.97383745,  8.00088367,  8.86213468,  9.98608331,
       10.90529198])