机器学习之简单线性回归(最小二乘法)
本帖最后由 zyt温柔发 于 2020-3-13 11:34 编辑0.引入依赖
import numpy as np
import matplotlib.pyplot as plt
1.导入数据(data.csv)(data.csv成绩与学习时间的100×2矩阵)
points = np.genfromtxt("data.csv",delimiter=',')#提取points中的两列数据，分别作为x，yx = points[:,0]y = points[:,1]# 用plt画出散点图plt.scatter(x,y)plt.show()
2.定义损失函数
# 损失函数是系数的函数，另外还要传入数据的x，y
def computer_cost(w,b,points):
total_cost = 0
M = len(points)
# 逐点计算平方损失误差，计算平均数
for i in range(M):
x = points
y = points
total_cost+= (y-w*x-b)**2
return total_cost/M ### 1//3取整 1/3浮点型
3.定义算法拟合函数
# 先定义一个求平均数的函数
def average(data):
sum = 0
num = len(data)
for i in range(num):
sum+= data
return sum/num
# 定义核心拟合函数
def fit(points):
M = len(points)
x_bar = average(points[:,0])
sum_yx = 0
sum_x2 = 0
sum_delta = 0
for i in range(M):
x = points
y = points
sum_yx+=y*(x-x_bar)
sum_x2+=x**2
# 根据公式计算w
w = sum_yx/(sum_x2-M*(x_bar**2))
for i in range(M):
x = points
y = points
sum_delta+=(y-w*x)
b = sum_delta/M
return w,b
4.测试
w,b = fit(points)
print('w is:',w)
print('b is:',b)
cost = computer_cost(w,b,points)
print("cost is:",cost)
w is: 1.3224310227553846
b is: 7.991020982269173
cost is: 110.25738346621313
5.画出拟合曲线
plt.scatter(x,y)
# 针对每一个x，计算出预测的y值
pred_y = w*x+b
plt.plot(x,pred_y,c='r')
plt.show()
数据： 没有bug吧？ 牛啊。。。。
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