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  • Kernel Principle Component Analysis

    • 算法特征:
      ①. 数据中心化; ②. 极大化投影方差; ③. 固定投影矢量长度
    • 算法推导:
      Part Ⅰ: PCA
      PCA特征提取方程如下:
      egin{equation}
      h(x) = W^mathrm{T}(x - ar{x})
      label{eq_1}
      end{equation}
      其中, $W$为单位投影矢量, $ar{x} = sumlimits_{i}x^{(i)} / n$为样本均值.
      PCA原始优化问题如下:
      egin{equation}
      egin{split}
      maxquad &left| frac{A^{mathrm{T}}W}{|W|_2} ight|_2^2   \
      mathrm{s.t.}quad &| W |_2^2 = 1
      end{split}
      label{eq_2}
      end{equation}
      其中, $A = [x^{(1)} - ar{x}, x^{(2)} - ar{x}, cdots, x^{(n)} - ar{x}]$.
      不失一般性, 将优化问题( ef{eq_2})等效转换如下:
      egin{equation}
      egin{split}
      minquad &-| A^{mathrm{T}}W |_2^2   \
      mathrm{s.t.}quad &| W |_2^2 = 1
      end{split}
      label{eq_3}
      end{equation}
      根据优化问题( ef{eq_3})KKT条件中关于原变量$W$的稳定性条件有:
      egin{equation}
      AA^mathrm{T}W = eta W
      label{eq_4}
      end{equation}
      其中, $eta$、$W$分别为协方差矩阵$AA^mathrm{T}$的本征值与本征矢. 由于$AA^mathrm{T} succeq 0$, 有$etageq 0$. 进一步可将优化问题( ef{eq_3})转换如下:
      egin{equation}
      egin{split}
      maxquad &eta   \
      mathrm{s.t.}quad &| W |_2^2 = 1
      end{split}
      label{eq_5}
      end{equation}
      因此, 所求单位投影矢量$W$即为协方差矩阵$AA^mathrm{T}$最大本征值$eta$所对应的本征矢.
      Part Ⅱ: KPCA
      现进行如下变数变换,
      egin{equation}
      W = Aalpha
      label{eq_6}
      end{equation}
      代入式( ef{eq_4})并左乘$A^mathrm{T}$有:
      egin{equation}
      A^mathrm{T}AA^mathrm{T}Aalpha = A^mathrm{T}Aetaalpha quadRightarrowquad A^mathrm{T}Aalpha = etaalpha
      label{eq_7}
      end{equation}
      为增强模型的特征提取能力, 引入kernel trick, 则式( ef{eq_7})转换如下:
      egin{equation}
      K(A^mathrm{T}, A)alpha = etaalpha
      label{eq_8}
      end{equation}
      其中, $K$代表kernel矩阵, 则$alpha$为此kernel矩阵最大本征值$eta$所对应的本征矢. 内部元素$K_{ij} = k(x^{(i)}, x^{(j)})$由kernel函数决定. 常见的kernel函数如下:
      ①. 多项式核: $k(x^{(i)}, x^{(j)}) = (x^{(i)}cdot x^{(j)} + 1)^n$
      ②. 高斯核: $k(x^{(i)}, x^{(j)}) = mathrm{e}^{-mu| x^{(i)} - x^{(j)} |_2^2}$
      结合式( ef{eq_1})、式( ef{eq_6})及kernel trick可得最终特征提取方程:
      egin{equation}
      h(x) = alpha^mathrm{T}K(A^mathrm{T}, x - ar{x})
      label{eq_9}
      end{equation}
    • 代码实现:
        1 # Kernal Princlple Component Analysis之实现
        2 
        3 import numpy
        4 from matplotlib import pyplot as plt
        5 
        6 numpy.random.seed(0)
        7 
        8 
        9 def getOriData(type1Size=100, type2Size=100):
       10     '''
       11     生成特征提取数据集
       12     '''
       13     theta1 = numpy.random.uniform(0, 2 * numpy.pi, type1Size)
       14     theta2 = numpy.random.uniform(0, 2 * numpy.pi, type2Size)
       15     
       16     R1 = numpy.random.uniform(0, 0.3, type1Size).reshape((-1, 1))
       17     R2 = numpy.random.uniform(0.7, 1, type2Size).reshape((-1, 1))
       18     X1 = R1 * numpy.hstack((numpy.cos(theta1).reshape((-1, 1)), numpy.sin(theta1).reshape((-1, 1))))
       19     X2 = R2 * numpy.hstack((numpy.cos(theta1).reshape((-1, 1)), numpy.sin(theta1).reshape((-1, 1))))
       20     return X1, X2
       21 
       22 
       23 class KPCA(object):
       24     
       25     def __init__(self, X, mu=1):
       26         '''
       27         X: 特征提取数据集, 1行代表1个样本
       28         '''
       29         self.__X = X                          # 特征提取数据集
       30         self.__mu = mu                        # gaussian kernel之超参数
       31         
       32         self.__mean = numpy.mean(X, axis=0)
       33         self.__A = (X - self.__mean).T
       34         
       35         
       36     def get_XTra(self, featureCnt=1):
       37         '''
       38         获取所有样本特征提取结果
       39         featureCnt: 待提取特征数
       40         '''
       41         XTra = list()
       42         for xOri in self.__X:
       43             xTra = self.get_xTra(xOri, featureCnt)
       44             XTra.append(xTra)
       45         XTra = numpy.array(XTra)
       46         return XTra
       47         
       48         
       49     def get_xTra(self, xOri, featureCnt=1):
       50         '''
       51         获取指定样本特征提取结果
       52         '''
       53         A = self.__A
       54         mu = self.__mu
       55         mean = self.__mean.reshape((-1, 1))
       56         if not hasattr(self, "eigVals"):
       57             KAA = self.__get_KAA(A, mu)
       58             self.__calc_eigs(KAA)
       59         eigVecs = numpy.real(self.eigVecs)
       60         
       61         x = numpy.array(xOri).reshape((-1, 1))
       62         xTra = list()
       63         for i in range(featureCnt):
       64             alpha = eigVecs[:, i:i+1]
       65             hVal = self.__get_hVal(x, mean, alpha, A, mu)
       66             xTra.append(hVal)
       67         xTra = numpy.array(xTra)
       68         return xTra
       69             
       70     
       71     def __get_hVal(self, x, mean, alpha, A, mu):
       72         KAx = self.__get_KAx(A, x - mean, mu)
       73         hVal = numpy.matmul(alpha.T, KAx)[0, 0]
       74         return hVal
       75         
       76     
       77     def __calc_eigs(self, KAA):
       78         oriEigVals, oriEigVecs = numpy.linalg.eig(KAA)
       79         seqArr = numpy.argsort(-oriEigVals)
       80         self.eigVals = oriEigVals[seqArr]
       81         self.eigVecs = oriEigVecs[:, seqArr]
       82     
       83     
       84     def __get_KAx(self, A, x, mu):
       85         KAx = numpy.zeros((A.shape[1], 1))
       86         for rowIdx in range(KAx.shape[0]):
       87             x1 = A[:, rowIdx:rowIdx+1]
       88             val = self.__calc_gaussian(x1, x, mu)
       89             KAx[rowIdx, 0] = val
       90         return KAx
       91     
       92         
       93     def __get_KAA(self, A, mu):
       94         KAA = numpy.zeros((A.shape[1], A.shape[1]))
       95         for rowIdx in range(KAA.shape[0]):
       96             for colIdx in range(rowIdx + 1):
       97                 x1 = A[:, rowIdx:rowIdx+1]
       98                 x2 = A[:, colIdx:colIdx+1]
       99                 val = self.__calc_gaussian(x1, x2, mu)
      100                 KAA[rowIdx, colIdx] = KAA[colIdx, rowIdx] = val
      101         return KAA
      102         
      103         
      104     def __calc_gaussian(self, x1, x2, mu):
      105         val = numpy.math.exp(-mu * numpy.linalg.norm(x1 - x2) ** 2)
      106         # val = (numpy.sum(x1 * x2) + 1)
      107         return val
      108         
      109 
      110 class KPCAPlot(object):
      111     
      112     @classmethod
      113     def show_XTra(cls, X1Ori, X2Ori, mu=1):
      114         XOri = numpy.vstack((X1Ori, X2Ori))
      115         kpcaObj = KPCA(XOri, mu)
      116         XTra = kpcaObj.get_XTra(2)
      117         
      118         X1Tra = XTra[:X1Ori.shape[0], :]
      119         X2Tra = XTra[X1Ori.shape[0]:, :]
      120         
      121         fig = plt.figure(figsize=(10, 3))
      122         ax1 = plt.subplot(1, 2, 1)
      123         ax2 = plt.subplot(1, 2, 2)
      124         
      125         ax1.scatter(X1Ori[:, 0], X1Ori[:, 1], c="red", s=5, label="cluster 0")
      126         ax1.scatter(X2Ori[:, 0], X2Ori[:, 1], c="green", s=5, label="cluster 1")
      127         ax1.set(title="Ori Data", xlabel="$x_1$", ylabel="$x_2$")
      128         
      129         ax2.scatter(X1Tra[:, 0], X1Tra[:, 1], c="red", s=5, label="cluster 0")
      130         ax2.scatter(X2Tra[:, 0], X2Tra[:, 1], c="green", s=5, label="cluster 1")
      131         ax2.set(title="Tra Data", xlabel="$c_1$", ylabel="$c_2$")
      132         
      133         ax1.legend()
      134         ax2.legend()
      135         fig.tight_layout()
      136         fig.savefig("./show_XTra.png", dpi=100)
      137         
      138         
      139 
      140 if __name__ == "__main__":
      141     X1, X2 = getOriData(100, 100)
      142     KPCAPlot.show_XTra(X1, X2, 1)
      View Code
    • 结果展示:
      左侧为特征提取数据集分布情况, 右侧为该数据集经过KPCA映射后取前2个主成分的分布情况. 很显然, 在合适的超参数选取下, 经过KPCA映射后, 原始非线性可分的数据集转变为线性可分.
    • 使用建议:
      ①. 实对称矩阵不同本征值的本征矢彼此正交.
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  • 原文地址:https://www.cnblogs.com/xxhbdk/p/13463888.html
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