Basic Probability Assignment Generation Method and Application Based on Cloud Model
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摘要: 针对证据理论应用中基本概率赋值(BPA)生成模型难以确定问题,该文提出一种基于云模型的BPA生成方法。首先基于样本属性的正态云模型构建单子集命题的BPA模型函数,并将复合子集的模型函数表示为高斯函数乘积融合。其次提出一种根据测试样本动态度量属性权重的方法来兼顾信息源的可靠性。最后,用属性权重修正模型函数输出的结果得到BPA。鸢尾花等数据集分类识别实验表明,该方法识别准确性高,且适用于样本较少的情况。Abstract: Basic Probability Assignment (BPA) has no fixed generative model in the application of evidence theory. To solve this problem, a BPA generation method based on cloud model is proposed. Firstly, based on the normal cloud model of the sample attributes, the BPA model function of the single subset proposition is constructed, and the model function of the composite subset is expressed as Gaussian function product fusion. Secondly, a method of dynamically measuring attribute weights based on test samples is proposed to take into account the reliability of information sources. Finally, the BPA is obtained by modifying the output result of the model function with attribute weights. The classification and recognition experiments of iris and other data sets show that this method has high recognition accuracy and is suitable for situations with fewer samples.
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表 1 SL属性下的模型参数
参数 $ \mu _1^1 $ $ \mu _2^1 $ $ \mu _3^1 $ $ \mu _{12}^1 $ $ \mu _{13}^1 $ $ \mu _{23}^1 $ $ \mu _{123}^1 $ $ {\text{Ex}} $ 5.006 5.974 6.588 5.295 5.362 6.223 5.514 $ {\text{En}} $ 0.339 0.520 0.630 0.284 0.299 0.401 0.259 $ {\text{Sg}} $ 0.297 0.087 0.754 0.174 表 2 SL属性下计算的BPA
BPA $ {\theta _1} $ $ {\theta _2} $ $ {\theta _3} $ $ {\theta _1}{\theta _2} $ $ {\theta _1}{\theta _3} $ $ {\theta _2}{\theta _3} $ $\varTheta$ $ {P_1}(\theta ) $ 0.849 0.331 0.088 0.281 0.075 0.029 0.025 $P'_1 (\theta )$ 0.506 0.197 0.053 0.167 0.045 0.017 0.015 $ {m_1}(\theta ) $ 0.451 0.176 0.047 0.149 0.040 0.016 0.122 表 3 属性权重
属性 SL SW PL PW $ {p^j} $ 1 2 2 2 $ {q^j} $ 2 3 3 3 $ {w^j} $ 0.891 0.710 0.919 0.954 表 4 BPA及其最大信度值对应焦元
BPA $ {\theta _1} $ $ {\theta _2} $ $ {\theta _3} $ $ {\theta _1}{\theta _2} $ $ {\theta _1}{\theta _3} $ $ {\theta _2}{\theta _3} $ $\varTheta$ 最大信度对应焦元 $ {m_1}(\theta ) $ 0.451 0.176 0.047 0.149 0.040 0.016 0.122 $ {\theta _1} $ $ {m_2}(\theta ) $ 0.028 0.196 0.132 0.027 0.019 0.130 0.469 $\varTheta$ $ {m_3}(\theta ) $ 0 0.889 0.018 0 0 0.013 0.081 $ {\theta _2} $ $ {m_4}(\theta ) $ 0 0.803 0.078 0 0 0.073 0.046 $ {\theta _2} $ $ m(\theta ) $ 0.003 0.986 0.008 0 0 0.002 0 $ {\theta _2} $ -
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