python线性回归_python向量线性回归

这是我尝试仅使用numpy和线性代数执行线性回归的尝试&＃xff1a;

def linear_function(w , x , b):

return np.dot(w , x) &＃43; b

x &＃61; np.array([[1, 1,1],[0, 0,0]])

y &＃61; np.array([0,1])

w &＃61; np.random.uniform(-1,1,(1 , 3))

print(w)

learning_rate &＃61; .0001

xT &＃61; x.T

yT &＃61; y.T

for i in range(30000):

h_of_x &＃61; linear_function(w , xT , 1)

loss &＃61; h_of_x - yT

if i % 10000 &＃61;&＃61; 0:

print(loss , w)

w &＃61; w &＃43; np.multiply(-learning_rate , loss)

linear_function(w , x , 1)

这会导致错误&＃xff1a;

ValueError Traceback (most recent call last)

in ()

24 if i % 10000 &＃61;&＃61; 0:

25 print(loss , w)

---> 26 w &＃61; w &＃43; np.multiply(-learning_rate , loss)

27

28 linear_function(w , x , 1)

ValueError: operands could not be broadcast together with shapes (1,3) (1,2)

这似乎可以减少训练集的维数&＃xff1a;

import numpy as np

def linear_function(w , x , b):

return np.dot(w , x) &＃43; b

x &＃61; np.array([[1, 1],[0, 0]])

y &＃61; np.array([0,1])

w &＃61; np.random.uniform(-1,1,(1 , 2))

print(w)

learning_rate &＃61; .0001

xT &＃61; x.T

yT &＃61; y.T

for i in range(30000):

h_of_x &＃61; linear_function(w , xT , 1)

loss &＃61; h_of_x - yT

if i % 10000 &＃61;&＃61; 0:

print(loss , w)

w &＃61; w &＃43; np.multiply(-learning_rate , loss)

linear_function(w , x , 1)

print(linear_function(w , x[0] , 1))

print(linear_function(w , x[1] , 1))

哪个返回&＃xff1a;

[[ 0.68255806 -0.49717912]]

[[ 1.18537894 0. ]] [[ 0.68255806 -0.49717912]]

[[ 0.43605474 0. ]] [[-0.06676614 -0.49717912]]

[[ 0.16040755 0. ]] [[-0.34241333 -0.49717912]]

[ 0.05900769]

[ 1.]

[0.05900769]& [1.]接近培训示例,因此看来此实现是正确的.引发错误的实现有什么问题&＃xff1f;我还没有实现2->的维度扩展. 3正确吗&＃xff1f;