python 实现结论不确定性合成

最新推荐文章于 2024-11-24 02:47:18 发布

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最新推荐文章于 2024-11-24 02:47:18 发布

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分类专栏：人工智能原理文章标签： python 机器学习人工智能不确定性推理

本文链接：https://blog.youkuaiyun.com/qq_36949278/article/details/122047995

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4 篇文章

订阅专栏

本文介绍了一种基于几率函数和概率函数的不确定性合成计算方法，并通过具体例题展示了计算流程。涉及几率函数、概率函数的转换及条件概率计算。

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参数说明

计算中所涉及到的说明见下图：

在这里插入图片描述

图1 参数说明

计算公式

计算中用到的相关公式罗列如下：

$\frac{{P(x)}}{{P(\neg x)}} = \frac{{P(x)}}{{1 - P(x)}},O(x) \in [0,\infty )，几率函数$

$\frac{{O(x)}}{{1 + O(x)}}, P(x) \in [0,1]，概率函数$

$P(Hi∣Ej)=LSj×P(Hi)(LSj−1)×P(Hi)+1P({H_i}|{E_j}) = \frac{{L{S_j} \times P({H_i})}}{{(L{S_j} - 1) \times P({H_i}) + 1}}$

$\left\{ {\begin{array}{l} {P(H|\neg E) + \frac{{P(H) - P(H|\neg E)}}{{P(E)}} \times P(E|S),0 \le P(E|S) < P(E)}\\ \\ {P(H) + \frac{{P(H|E) - P(H)}}{{1 - P(E)}} \times [P(E|S) - P(E)],P(E) \le P(E|S) \le 1} \end{array}} \right.$

$O(H∣S1,S2,...,Sn)=O(H∣S1)O(H)O(H∣S2)O(H)...O(H∣Sn)O(H)O(H)O(H|{S_1},{S_2},...,{S_n}) = \frac{{O(H|{S_1})}}{{O(H)}}\frac{{O(H|{S_2})}}{{O(H)}}...\frac{{O(H|{S_n})}}{{O(H)}}O(H)$

例题

通过下图例题实现结论不确定性合成：

图2 例题

计算流程

（1）计算 $O(H_1|S_1)$

$P(E1∣S1)>P(E1)，P(H1∣E1)=LS1×P(H1)(LS1−1)×P(H1)+1P(H1∣S1)=P(H1)+P(H1∣E1)−P(H1)1−P(E1)×(P(E1∣S1)−P(E1))O(H1|S1)=P(H1∣S1)1−P(H1∣S1){P(E}_1\left|S_1\right)>P(E_1)，\\ P(H_1|E_1)=\frac{{LS}_1\times P(H_1)}{\left({LS}_1-1\right)\times P\left(H_1\right)+1}\\ P(H_1|S_1)=P\left(H_1\right)+\frac{P(H_1|E_1)-P\left(H_1\right)}{1-P(E_1)}\times(P(E_1\left|S_1\right)-P\left(E_1\right))\\ O\left(H_1\middle| S_1\right)=\frac{P(H_1|S_1)}{1-P(H_1|S_1)}$

（2）计算 $O(H_1|S_2)$

$P(E2∣S2)>P(E2)，P(H1∣E2)=LS2×P(H1)(LS2−1)×P(H1)+1P(H1∣S2)=P(H1)+P(H1∣E2)−P(H1)1−P(E2)×(P(E2∣S2)−P(E2))O(H1|S2)=P(H1∣S2)1−P(H1∣S2){P(E}_2\left|S_2\right)>P(E_2)，\\ P(H_1|E_2)=\frac{{LS}_2\times P(H_1)}{\left({LS}_2-1\right)\times P\left(H_1\right)+1}\\ P(H_1|S_2)=P\left(H_1\right)+\frac{P(H_1|E_2)-P\left(H_1\right)}{1-P(E_2)}\times(P(E_2\left|S_2\right)-P\left(E_2\right))\\ O\left(H_1\middle| S_2\right)=\frac{P(H_1|S_2)}{1-P(H_1|S_2)}$

（3）计算 $P(H_1|S_1,S_2)$

$O(H1)=P(H1)1−P(H1)O(H1|S1,S2)=O(H1∣S1)O(H1)×O(H1∣S2)O(H1)×O(H1)P(H1|S1,S2)=O(H1∣S1,S2)1+O(H1∣S1,S2)O\left(H_1\right)=\frac{P(H_1)}{1-P(H_1)}\\ O\left(H_1\middle| S_1,S_2\right)=\frac{O{(H}_1|S_1)}{O\left(H_1\right)}\times\frac{O{(H}_1|S_2)}{O\left(H_1\right)}\times O\left(H_1\right)\\ P\left(H_1\middle| S_1,S_2\right)=\frac{O{(H}_1|S_1,S_2)}{1+O{(H}_1|S_1,S_2)}$

（4）计算 $O(H2|S1,S2)O\left(H_2\middle| S_1,S_2\right)$

$P(H1|S1,S2)>P(H2)，P(H2|H1)=LS4×P(H2)(LS4−1)×P(H2)+1P(H2|S1,S2)=P(H2)+P(H2∣H1)−P(H2)1−P(H1)×(P(H1|S1,S2)−P(H1))O(H2|S1,S2)=P(H1∣S1,S2)1−P(H1∣S1,S2)P\left(H_1\middle| S_1,S_2\right)>P(H_2)，\\ P\left(H_2\middle| H_1\right)=\frac{{LS}_4\times P(H_2)}{\left({LS}_4-1\right)\times P\left(H_2\right)+1}\\ P\left(H_2\middle| S_1,S_2\right)=P\left(H_2\right)+\frac{P(H_2|H_1)-P\left(H_2\right)}{1-P(H_1)}\times(P\left(H_1\middle| S_1,S_2\right)-P\left(H_1\right))\\ O\left(H_2\middle| S_1,S_2\right)=\frac{P{(H}_1|S_1,S_2)}{1-P{(H}_1|S_1,S_2)}$

（5）计算 $O(H_2|S_3)$

$P(E3∣S3)<P(E3)，P(H2∣¬E3)=LN3×P(H2)(LN3−1)×P(H2)+1P(H2∣S3)=P(H2∣¬E3)+P(H2)−P(H2∣¬E3)P(E3)×P(E3∣S3)O(H2|S3)=P(H2∣S3)1−P(H2∣S3){P(E}_3\left|S_3\right)<P(E_3)，\\ P(H_2|{\lnot E}_3)=\frac{{LN}_3\times P(H_2)}{\left({LN}_3-1\right)\times P\left(H_2\right)+1}\\ P(H_2|S_3)=P(H_2|{\lnot E}_3)+\frac{P\left(H_2\right)-P(H_2|{\lnot E}_3)}{P(E_3)}\times P(E_3|S_3)\\ O\left(H_2\middle| S_3\right)=\frac{P(H_2|S_3)}{1-P(H_2|S_3)}$

（6）计算 $P(H_2|S_1,S_2,S_3)$

$O(H1∣S1,S2,S3)1+O(H1∣S1,S2,S3)O\left (H_2\right)=\frac{P(H_2)}{1-P(H_2)}\\ O\left(H_2\middle| S_1,S_2,S_3\right)=\frac{{O(H}_2|S_1,S_2)}{O\left(H_2\right)}\times\frac{{O(H}_2|S_3)}{O\left(H_2\right)}\times O\left(H_2\right)\\ P(H_2|S_1,S_2,S_3)=\ \frac{{O(H}_1|S_1,S_2,S_3)}{1+{O(H}_1|S_1,S_2,S_3)}$

代码实现

PE1 = PE2 = PE3 = 0.6
PHs = [0.091, 0.01]
PExSs = [0.84, 0.68, 0.36]
LSs = [2, 100, 200, 50]
LNs = [0.00001, 0.0001, 0.001, 0.1]

def PLS(LS, PH):
    PHxLS = (LS*PH)/((LS-1)*PH+1)
    return PHxLS

def PHS(PH, PE, PHxE, PExS):
    if 0 < PExS < PE:
        PHxS = PHxE + (PH-PHxE)/PE*(PExS)
    elif PE <= PExS <= 1:
        PHxS = PH + (PHxE-PH)/(1-PE)*(PExS-PE)
    return PHxS

def OHS(PHxS):
    OHxS = PHxS/(1-PHxS)
    return OHxS

print('（1）计算O(H1|S1)')
PH1xE1 = PLS(LSs[0], PHs[0])
PH1xS1 = PHS(PHs[0], PE1, PH1xE1, PExSs[0])
OH1xS1 = OHS(PH1xS1)
print('P(H1|E1):', PH1xE1, '\nP(H1|S1):', PH1xS1, '\nO(H1|S1):', OH1xS1)
print('（2）计算O(H1|S2)')
PH1xE2 = PLS(LSs[1], PHs[0])
PH1xS2 = PHS(PHs[0], PE2, PH1xE2, PExSs[1])
OH1xS2 = OHS(PH1xS2)
print('P(H1|E2):', PH1xE2, '\nP(H1|S2):', PH1xS2, '\nO(H1|S2):', OH1xS2)
print('（3）计算P(H1|S1,S2)')
OH1 = OHS(PHs[0])
OH1xS1S2 = (OH1xS1/OH1)*(OH1xS2/OH1)*OH1
PH1xS1S2 = OH1xS1S2/(1+OH1xS1S2)
print('O(H1):', OH1, '\nO(H1|S1,S2):', OH1xS1S2, '\nP(H1|S1,S2):', PH1xS1S2)
print('（4）计算O(H2|S1,S2)')
PH2xH1 = PLS(LSs[3], PHs[1])
PH2xS1S2 = PHS(PHs[1], PHs[0], PH2xH1, PH1xS1S2)
OH2xS1S2 = OHS(PH2xS1S2)
print('P(H2|H1):', PH2xH1, '\nP(H2|S1,S2):', PH2xS1S2, '\nO(H1|S1,S2):', OH2xS1S2)
print('（5）计算O(H2|S3)')
PH2x0E3 = PLS(LNs[2], PHs[1])
PH2xS3 = PHS(PHs[1], PE3, PH2x0E3, PExSs[2])
OH2xS3 = OHS(PH2xS3)
print('P(H2|非E3):', PH2x0E3, '\nP(H2|S3):', PH2xS3, '\nO(H2|S3):', OH2xS3)
print('（6）计算P(H1|S1,S2,S3)')
OH2 = OHS(PHs[1])
OH2xS1S2S3 = (OH2xS1S2/OH2)*(OH2xS3/OH2)*OH2
PH2xS1S2S3 = OH2xS1S2S3/(1+OH2xS1S2S3)
print('O(H2):', OH2, '\nO(H1|S1,S2,S3):', OH2xS1S2S3, '\nP(H1|S1,S2,S3):', PH2xS1S2S3)