python梯度下降算法

x = [(2, 0, 3), (1, 0, 3), (1, 1, 3), (1, 4, 2), (1, 2, 4)]
y = [5, 6, 8, 10, 11]
epsilon = 0.002
alpha = 0.02
diff = [0, 0]
max_itor = 1000
error0 = 0
error1 = 0
cnt = 0
m = len(x)
theta0 = 0
theta1 = 0
theta2 = 0
while True:
    cnt += 1
    for i in range(m):
        diff[0] = (theta0 * x[i][0] + theta1 * x[i][1] + theta2 * x[i][2]) - y[i]
        theta0 -= alpha * diff[0] * x[i][0]
        theta1 -= alpha * diff[0] * x[i][1]
        theta2 -= alpha * diff[0] * x[i][2]
    error1 = 0
    for lp in range(len(x)):
        error1 += (y[lp] - (theta0 + theta1 * x[lp][1] + theta2 * x[lp][2])) ** 2 / 2
    if abs(error1 - error0) < epsilon:
        break
    else:
        error0 = error1
print('theta0 : %f, theta1 : %f, theta2 : %f, error1 : %f' % (theta0, theta1, theta2, error1))
print('Done: theta0 : %f, theta1 : %f, theta2 : %f' % (theta0, theta1, theta2))
print('迭代次數: %d' % cnt)

作者：ChenBD