标签：差值   oss   pre   结果   第一步   and   组合   nump   技术分享   

BP 算法推导过程

一.FP过程(前向-计算预测值)

定义sigmoid激活函数

import numpy as np
def sigmoid(z):
    return 1.0 / (1 + np.exp(-z))

输入层值和 标签结果

l = [5.0, 10.0]
y = [0.01,0.99]
alpha=0.5

初始化 w,b 的值

w = [0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4, 0.45, 0.5, 0.55, 0.6, 0.65]
b = [0.35, 0.65]

计算隐层的结果

\[ h1 = Sigmod( Net_{h1}) =Sigmod(w1*l1+ w2*l2+b1*1) \]

h1 = sigmoid(w[0] * l[0] + w[1] * l[1] + b[0])
h2 = sigmoid(w[2] * l[0] + w[3] * l[1] + b[0])
h3 = sigmoid(w[4] * l[0] + w[5] * l[1] + b[0])
[h1,h2,h3]
#[0.9129342275597286, 0.9791636554813196, 0.9952742873976046]

[0.9129342275597286, 0.9791636554813196, 0.9952742873976046]

计算输出层的结果

\[ o1 = Sigmod( Net_{o1}) =Sigmod(w7*h1+ w9*h2+w11*h3+b2*1) \]

o1 = sigmoid(w[6] * h1 + w[8] * h2+ w[10] * h3 + b[1])
o2 = sigmoid(w[7] * h1 + w[9] * h2+ w[11] * h3 + b[1])
[o1,o2]
#[0.8910896614765176, 0.9043299248500164]

[0.8910896614765176, 0.9043299248500164]

计算整体loss误差值 (平方损失函数)

\[ loss=E_{total}=E_{o1}+E_{o2}=\frac{1}{2}*(y_1-o1)^2+\frac{1}{2}*(y_2-o2)^2 \]

E_total=1/2*(y[0]-o1)**2+1/2*(y[1]-o2)**2
E_total
#0.3918291766685041

0.3918291766685041

二.BP过程 (反向-更新参数值)

更新w7的值:

w7的值只和o1的损失函数有关系,所以整体Loss对w7求偏导,E_o2 会约掉
target 代表目标值

\[ Loss=E_{total}=E_{o1}+E_{o2} \]

\[ E_{o1}=\frac{1}{2}*(target_{o1}-out_{o1})^2 =\frac{1}{2}*(target_{o1}-Sigmod(net_{o1}))^2 \]

\[ E_{o1} =\frac{1}{2}*(target_{o1}-Sigmod(w7*h1+ w9*h2+w11*h3+b2*1))^2 \]

同过复合函数链式求导发展的如下公式

\[ \text{整体损失 对w7 求偏导 : } \frac{\partial E_{total}}{\partial w7} =\frac{\partial E_{total}}{\partial out_{o1}} *\frac{\partial out_{o1}}{\partial net_{o1}} *\frac{\partial net_{o1}}{\partial w7} \]

第一步: E_total 对 Out_o1 求导

\[ \text{原函数 : } E_{total} =\frac{1}{2}(target_{o1}-out_{o1})^2+E_{o2} \]

\[ \frac{\partial E_{total}}{\partial out_{o1}}=2*\frac{1}{2}*(target_{o1}-out_{o1})^{2-1}*-1+0 \]

(y[0]-o1)*(-1) #0.8910896614765176

0.8810896614765176

第二步: Out_o1 对 net_o1 求导
详见sigmod 函数求导公式

\[ \text{原函数 } out_{o1}=\frac{1}{1+e^{-net_{o1}}} \]

\[ \frac{\partial out_{o1}}{\partial net_{o1}}=out_{o1}(1-out_o) \]

o1*(1-o1) #0.09704887668618288

0.09704887668618288

第三步: Net_o1 对 w7 求导

\[ \text{原函数 : } net_{o1} =w7*out_{h1}+ w9*out_{h2}+w11*out_{h3}+b2*1 \]

\[ \frac{\partial net_{o1}}{\partial w7}=1*out_{h1}*w7^{(1-1)}+0+0+0 \]

1*h1*1 #0.9129342275597286

0.9129342275597286

组合整个求导过程

\[ \text{整体损失 对w7 求偏导 : } \frac{\partial E_{total}}{\partial w7} =-1*(target_{o1}-out_{o1})*out_{o1}*(1-out_{o1})*out_{h1} \]

(o1-y[0])*o1*(1-o1)*h1 #0.07806387550033887

0.07806387550033887

第四部分 更新前向分布算法 更新w7的值
η指 学习率

\[ w7^+=w7+Δw7=w7-η*\frac{\partial E_{total}}{\partial w7} \]

# w[6] = w[6]-alpha*(o1-y[0])*o1*(1-o1)*h1 # w7
# w[7] = w[7]-alpha*(o2-y[1])*o2*(1-o2)*h1 #w8
# w[8] = w[8]-alpha*(o1-y[0])*o1*(1-o1)*h2 #w9
# w[9] = w[9]-alpha*(o2-y[1])*o2*(1-o2)*h2 #w10
# w[10]=w[10]-alpha*(o1-y[0])*o1*(1-o1)*h3 #w11
# w[11]=w[11]-alpha*(o2-y[1])*o2*(1-o2)*h3 #w12

得到w7-w12的更新公式(后面一起更新)

# #提取公共部分
#t1=(o1-y[0])*o1*(1-o1)
#t2=(o2-y[1])*o2*(1-o2)

# w[6] = w[6]-alpha*t1*h1 #w7
# w[7] = w[7]-alpha*t2*h1 #w8
# w[8] = w[8]-alpha*t1*h2 #w9
# w[9] = w[9]-alpha*t2*h2 #w10
# w[10]=w[10]-alpha*t1*h3 #w11
# w[11]=w[11]-alpha*t2*h3 #w12
# print("新的w7的值:",w[6])
# w[6:]

更新w1的值:

\[ \text{整体损失 对w1 求偏导 : } \frac{\partial E_{total}}{\partial w1} =\frac{\partial E_{total}}{\partial out_{h1}} *\frac{\partial out_{h1}}{\partial net_{h1}} *\frac{\partial net_{h1}}{\partial w1} \]

\[ \text{展开 : } \frac{\partial E_{total}}{\partial w1}= (\frac{\partial E_{o1}}{\partial out_{h1}}+\frac{\partial E_{o2}}{\partial out_{h1}}) *\frac{\partial out_{h1}}{\partial net_{h1}} *\frac{\partial net_{h1}}{\partial w1} \]

第一步: E_o1对 out_h1 求偏导 and E_o2对 out_h1

\[ \frac{\partial E_{o1}}{\partial out_{h1}}= \frac{\partial E_{o1}}{\partial out_{o1}} *\frac{\partial out_{o1}}{\partial net_{o1}} *\frac{\partial net_{o1}}{\partial out_{h1}} \]

\[ \frac{\partial E_{o1}}{\partial out_{h1}}= -(target_{o1}-out_{o1}) *out_{o1}(1-out_{o1}) *w7 \]

-(y[0]-o1)*o1*(1-o1)*w[6] # E_o1 对out_h1  0.03420350476244207

0.03420350476244207

-(y[1]-o2)*o2*(1-o2)*w[7] # E_o1 对out_h1 -0.003335375074384934

-0.003335375074384934

第二步 out_h1 对net_h1 求偏导(前面计算过)

h1*(1-h1)

0.07948532370965024

第三步 net_h1 对w1 求篇导

\[ \text{原函数 : } net_{h1} =w1*l1+ w2*l2+b1*1 \]

l[0]

5.0

最终整合 w1 的更新公式得

\[ \frac{\partial E_{total}}{\partial w1}=[ -(target_{o1}-out_{o1})*out_{o1}(1-out_o1)*w7 -(target_{o2}-out_{o2})*out_{o2}(1-out_o2)*w8] *out_{o1}*(1-out_{o1}) *l1 \]

#w[0] = w[0]-alpha* w(-1*(y[0]-o1)*o1*(1-o1)*w[6]+-1*(y[1]-o2)*o2*(1-o2)*w[7])*h1*(1-h1)*l[0] # w1的更新值
# 提取公共部分



t1=(o1-y[0])*o1*(1-o1)
t2=(o2-y[1])*o2*(1-o2)

w[0] = w[0] - alpha * (t1 * w[6] + t2 * w[7]) * h1 * (1 - h1) * l[0]
w[1] = w[1] - alpha * (t1 * w[6] + t2 * w[7]) * h1 * (1 - h1) * l[1]
w[2] = w[2] - alpha * (t1 * w[8] + t2 * w[9]) * h2 * (1 - h2) * l[0]
w[3] = w[3] - alpha * (t1 * w[8] + t2 * w[9]) * h2 * (1 - h2) * l[1]
w[4] = w[4] - alpha * (t1 * w[10]+ t2 *w[11]) * h3 * (1 - h3) * l[0]
w[5] = w[5] - alpha * (t1 * w[10]+ t2 *w[11]) * h3 * (1 - h3) * l[1]

w[6] = w[6]-alpha*t1*h1 
w[7] = w[7]-alpha*t2*h1 
w[8] = w[8]-alpha*t1*h2 
w[9] = w[9]-alpha*t2*h2 
w[10]=w[10]-alpha*t1*h3 
w[11]=w[11]-alpha*t2*h3
print("进行一次跌倒更新后的结果")
print(w)

进行一次跌倒更新后的结果
[0.0938660917985833, 0.13773218359716657, 0.19802721973428622, 0.24605443946857242, 0.2994533791079845, 0.3489067582159689, 0.3609680622498306, 0.4533833089635062, 0.4581364640581681, 0.5536287533891512, 0.5574476639638248, 0.653688458944847]

输入值为 0.01,0.99 的第一次迭代对照结果

0.0938660917985833, 0.13773218359716657, 0.19802721973428622, 0.24605443946857242, 0.2994533791079845, 0.3489067582159689, 0.3609680622498306, 0.4533833089635062, 0.4581364640581681, 0.5536287533891512, 0.5574476639638248, 0.653688458944847

输入值为 0.00,1.00 的第一次迭代对照结果

0.09386631682087375, 0.13773263364174748, 0.1980267403252208, 0.24605348065044158, 0.2994531447534454, 0.34890628950689084, 0.3605250660434654, 0.4537782320399227, 0.4576613303938861, 0.5540523264259203, 0.556964712705892, 0.6541190012244457

BP 算法推导过程

标签：差值   oss   pre   结果   第一步   and   组合   nump   技术分享   

原文地址：https://www.cnblogs.com/zhuimengzhe/p/10259259.html