【线性回归】| Linear Regression实现示例

# 首先观察一下数据的分布
import pandas as pd

# 读取数据集
df = pd.read_csv('./data.csv')
df.head(5)

import matplotlib.pyplot as plt
import numpy as np
%matplotlib inline

# 利用这些点在平面坐标上进行绘图,将数据样本转化成数组
points = np.array(df)

# 提取列表中的第一列为x的值
X = points[:,0]
# 提取列表中的第二列为y的值
Y = points[:,1]

# 绘制散点图
plt.scatter(X,Y)

<matplotlib.collections.PathCollection at 0x1cf2d5e2448>

# 思路：列出损失函数 loss = ∑i（wxi + b - yi）**2
# 对于每一轮的迭代都会更新参数w与b，更新方程为：w' = w - lr*（dloss/dw）
# 一次梯读下降对w，b进行更新
b_start = 0
w_start = 0
iteration = 900
learning_rate = 0.0001
N = len(points)
# 一步的梯度下降，对b，w进行更新并返回
def Step_GradientDescent(b,w):
    lr = learning_rate
    # 求b，w的偏导数值
    for i in range(0,len(points)):
        dw = 2*X[i]*(w*X[i]+b-Y[i])/N
        db = 2*(w*X[i]+b-Y[i])/N
    # 进行更新
    w_updata = w - lr*dw
    b_updata = b - lr*db
    return b_updata,w_updata
# 实现梯度下降算法
def Total_GradientDescent(b,w):
    iteration_times = iteration
    for i in range(0,iteration_times):
        b,w = Step_GradientDescent(b,w)
    return b,w
# 计算方差
def Deviation_value(b,w):
    loss = 0
    for i in range(0,len(points)):
        loss += (w*X[i]+b-Y[i])**2
    return loss
def main():
    b = b_start
    w = w_start
    print("Before__b:{0},w:{1},Deviation value:{2}".format(b,w,Deviation_value(b,w)))
    b,w = Total_GradientDescent(b,w)
    print("After__b:{0},w:{1},Deviation value:{2}".format(b,w,Deviation_value(b,w)))
    return b,w
b,w = main()

Before__b:0,w:0,Deviation value:556510.7834490552
After__b:0.0574654741275659,w:1.4440202845195163,Deviation value:11557.105373357175

# 根据以上结果，绘制出一条直线，与前述的散点图进行拟合
# axline((0, 0), (1, 1), linewidth=4, color='r')
plt.subplot(111)
plt.scatter(X,Y)
plt.axline((0,0.0574654741275659),slope=1.4440202845195163,linewidth=2, color='blue')

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