轉載自線性拟合_星-耀的部落格-CSDN部落格_線性拟合
令 ∑xi=c，∑(xi)^2=d，∑yi=e，∑(xi*yi)=h，
則解得 b=(m*h-c*e)/(m*d-c^2)，a=(e-c*b)/m，
即：
#線性拟合：y=a+bx
import numpy as np
x_array = np.array([13,15,16,21,22,23,25,29,30,31,36,40,42,55,60,62,64,70,72,100,130])#x_array,y_array是我們要拟合的資料
y_array = np.array([11,10,11,12,12,13,13,12,14,16,17,13,14,22,14,21,21,24,17,23,34])
m = len(x_array) #方程個數
sum_x = np.sum(x_array)
sum_y = np.sum(y_array)
sum_xy = np.sum(x_array * y_array)
sum_xx = np.sum(x_array **2 )
a=(sum_y*sum_xx-sum_x*sum_xy)/(m*sum_xx-(sum_x)**2)
b=(m*sum_xy-sum_x*sum_y)/(m*sum_xx-(sum_x)**2)
print("p = {:.4f} + {:.4f}x".format(a,b))
輸出結果：p = 8.2084 + 0.1795x
資料可視化展示：