画出线性回归图每个点到直线的距离

Question

怎么画出每个点到直线的距离

import numpy as np
import matplotlib.pyplot as plt
from scipy import stats

# 构造训练数据
x = np.arange(0., 10., 0.2)
m = len(x)
x0 = np.full(m, 1.0)
input_data = np.vstack([x0, x]).T
target_data = 2 * x + 5 + np.random.randn(m)

# 终止条件
loop_max = 10000  # 最大迭代次数
epsilon = 1e-3    # 收敛条件最小值

# 初始化权值
np.random.seed(0)
theta = np.random.randn(2)
alpha = 0.001  # 步长
diff = 0.
error = np.zeros(2)
count = 0  # 循环次数
finish = 0  # 终止标志

# 迭代
while count < loop_max:
    count += 1
    # 在标准梯度下降中，权值更新的每一步对多个样例求和，需要更多的计算
    sum_m = np.zeros(2)
    for i in range(m):
        dif = (np.dot(theta, input_data[i]) - target_data[i]) * input_data[i]
        # 当alpha取值过大时,sum_m会在迭代过程中会溢出
        sum_m = sum_m + dif

    # 注意步长alpha的取值,过大会导致振荡
    theta = theta - alpha * sum_m

    # 判断是否已收敛
    if np.linalg.norm(theta - error) < epsilon:
        finish = 1
        break
    else:
        error = theta

    print('迭代次数 = %d' % count, '\t w:', theta)
print('迭代次数 = %d' % count, '\t w:', theta)

# 用scipy线性回归进行检查
slope, intercept, r_value, p_value, slope_std_error = stats.linregress(x, 
                                                                       target_data)
print('截距 = %s 斜率 = %s' % (intercept, slope))

# 用plot进行展示
plt.plot(x, target_data, 'b*')
plt.plot(x, theta[1] * x + theta[0], 'r')
plt.xlabel("x")
plt.ylabel("y")
plt.show()

Answer 1

这是代码块

import matplotlib.pyplot as plt

import numpy as np

# 生成示例数据
np.random.seed(0)
x = np.random.rand(20)
y = 2 * x + 1 + np.random.randn(20) * 0.1

# 计算回归线
coef = np.polyfit(x, y, 1)
line_x = np.linspace(0, 1, 2)
line_y = np.polyval(coef, line_x)

# 绘制回归图和数据点
plt.figure()
plt.scatter(x, y)

# 绘制回归线
plt.plot(line_x, line_y, c='g')

# 绘制每个数据点到回归线的距离
for i in range(len(x)):
    x_val = x[i]
    y_val = y[i]
    y_pred = np.polyval(coef, x_val)
    plt.plot([x_val, x_val], [y_val, y_pred], c='r')

plt.show()