python线性拟合数据(深度学习笔记一)

给定样本x数据:

[1, 3, 5, 2], [2, 5, 4, 9], [1, 3, 7, 9]

对应y数据:

[27, 58, 67, 102]

下面用python求线性曲线方程:

import numpy as np

# y=2x1+ 7x2+3x3+8 
x = np.array([[1, 3, 5, 2], [2, 5, 4, 9], [1, 3, 7, 9]])
y = np.array([[27, 58, 67, 102]])
w = np.array([0, 0, 0]).reshape(1, 3)
b = 0
'''
for i in range(4):
    print(x[0][i]*2+x[1][i]*7+x[2][i]*3+8)
'''
'''定义传播函数 由于不涉及到二分类等问题 所以不使用sigmoid函数  使用线性回归 取平方差的和  '''

def process(w, b, X, Y):
    m = X.shape[1]
    A = np.dot(w, X) + b  # (1,4)
    assert (A.shape == (1, 4))
    cost = 1 / m * np.sum((A - Y) ** 2)  # 越小则越接近实际值
    dw = 2 / m * np.dot((A - Y), X.T)
    db = 2 / m * np.sum(A - Y)
    assert (dw.shape == w.shape)
    assert (cost.shape == ())
    dao = {'dw': dw, 'db': db}
    return dao, cost


alpha = 0.01  # 学习率
itemra_Num =10001  # 迭代次数
for i in range(itemra_Num):
    dao, cost = process(w, b, x, y)
    dw = dao['dw']
    db = dao['db']
    w = w - alpha * dw
    b = b - alpha * db
    if i % 1000 == 0:
        print(w)
        print(b)
        print(cost)

如下结果:

[[ 3.7   7.65  7.94]]
1.27
4746.5
[[ 2.30019751  7.26012923  2.81900947]]
6.6797835089
0.14539597224
[[ 2.06694472  7.0580094   2.95963871]]
7.70558877116
0.00723054126807
[[ 2.01492882  7.01293623  2.99099934]]
7.93434563782
0.000359574795807
[[ 2.00332916  7.00288481  2.99799283]]
7.98535893046
1.78816535286e-05
[[ 2.00074241  7.00064332  2.9995524 ]]
7.99673500876
8.89254576922e-07
[[ 2.00016556  7.00014346  2.99990018]]
7.99927189966
4.42226274712e-08
[[ 2.00003692  7.00003199  2.99997774]]
7.999837632
2.1991911329e-09
[[ 2.00000823  7.00000713  2.99999504]]
7.99996379157
1.09365768502e-10
[[ 2.00000184  7.00000159  2.99999889]]
7.99999192544
5.43875935097e-12
[[ 2.00000041  7.00000035  2.99999975]]
7.99999819935
2.70469485079e-13

从数据来看 成本函数一直在递减 说明 方向是正确的 整体上系数也越来越接近(2,7,3)+8