机器学习笔记主成分分析（PCA）

àì夳堔傛蜴生んèń 2022-03-22 07:54 233阅读 0赞

今天看了关于主成分分析算法，然后加上自己的理解写成笔记，作为以后复习的资料。

主成分分析是通过减少一定的精度来减少数据的维度，而这里主要的目标是怎么在给定的需要减少的维度的条件下，使得精度下降的最少。主成分分析法是通过寻找数据的主要特征，减少维度。目前主要的方法有最大化方差法和最小化平方误差，这两种方法从不同的两种角度去分析这个问题，首先看最大化方差法：

最大化方差法的只要思想是最大化数据在某一个方向上的方差，而这个方向就是主成分对应的方向。为了便于计算，我们假设这个方向为单位方向向量w，假设我们有样本数据![\\small \\\{ \{s\_1\},\{s\_2\}, \\cdots \{s\_n\}\\\}][small _ _s_1_s_2_ _cdots _s_n]，都是列向量， 为了便于计算将样本进行中心化得到，![\\small \\mu = \\frac\{1\}\{n\}\\sum\\limits\_\{i = 1\}^n \{\{s\_i\}\}][small _mu _ _frac_1_n_sum_limits_i _ 1_n _s_i]![\\small \\\{ \{x\_1\},\{x\_2\}, \\cdots \{x\_n\}\\\} = \\\{ \{s\_1\} - \\mu ,\{s\_2\} - \\mu , \\cdots ,\{s\_3\} - \\mu \\\}][small _ _x_1_x_2_ _cdots _x_n_ _ _ _s_1_ - _mu _s_2_ - _mu _ _cdots _s_3_ - _mu] 。我们先算样本![\\small x\_\{i\}][small x_i]在w上的投影长度，得到![\\small \{l\_i\} = x\_i^Tw][small _l_i_ _ x_i_Tw]，这样我们就能得到所有样本在w上投影的方差：

![\\small \{\\sigma ^2\}(x) = \\frac\{1\}\{n\}\\sum\\limits\_\{i = 1\}^n \{(x\_i^T\} w\{)^2\} = \\frac\{1\}\{n\}\\sum\\limits\_\{i = 1\}^n \{(x\_i^T\} w\{)^T\}(x\_i^Tw) = \\frac\{1\}\{n\}\\sum\\limits\_\{i = 1\}^n \{\{w^T\}\} \{x\_i\}x\_i^Tw = \{w^T\}(\\frac\{1\}\{n\}\\sum\\limits\_\{i = 1\}^n \{\{x\_i\}x\_i^T)w\}][small _sigma _2_x_ _ _frac_1_n_sum_limits_i _ 1_n _x_i_T_ w_2_ _ _frac_1_n_sum_limits_i _ 1_n _x_i_T_ w_T_x_i_Tw_ _ _frac_1_n_sum_limits_i _ 1_n _w_T_ _x_i_x_i_Tw _ _w_T_frac_1_n_sum_limits_i _ 1_n _x_i_x_i_T_w]

这样问题就转化为带约束的优化问题：

![\\small \\left\\\{ \{\\begin\{array\}\{\*\{20\}\{c\}\} \{\\max \\\{ \{w^T\}Cw\\\} \}\\\\ \{s.t.\{w^T\}w = 1\} \\end\{array\}\} \\right.][small _left_ _begin_array_20_c_ _max _ _w_T_Cw_ _ _s.t._w_T_w _ 1_ _end_array_ _right.]

引入拉格朗日乘子得到：

![\\small L(w,C,\\lambda ) = - \{w^T\}Cw + \\lambda \{w^T\}w][small L_w_C_lambda _ _ - _w_T_Cw _ _lambda _w_T_w]

对w求导得到![\\small Cw = \\lambda w][small Cw _ _lambda w]，所以![\\small \{\\sigma ^2\}(x) = \{w^T\}Cw = \\lambda \{w^T\}w = \\lambda][small _sigma _2_x_ _ _w_T_Cw _ _lambda _w_T_w _ _lambda]。通过分析，我们可以看出，我们寻找的最大主成分的方向向量为样本的协方差矩阵最大特征值对应的特征向量。如果我们需要将样本降到d维的空间，则可以求出样本协方差矩阵，然后将特征值进行排序，然后将前d个特征值对应的特征向量求出得到转化矩阵，这样就能得到降维后的样本。

--------------------

下面是另一种方法，这种方法从另一个角度来分析这个问题，与上面寻找一个方向不同，然后根据所有方向确定绛维后的超平面，这种方法是直接寻找超平面，然后通过最小化样本到超平面的距离来得到最优的超平面。于是我们可以得到所有样本到超平面的距离为：

![\\small L = \\sum\\limits\_\{i = 1\}^n \{\\left\\| \{\{x\_i\} - \{\{\\bar x\}\_i\}\} \\right\\|\}][small L _ _sum_limits_i _ 1_n _left_ _x_i_ - _bar x_i_ _right]

这里的![\\small \{\\bar x\_i\}][small _bar x_i]表示![\\small x\_\{i\}][small x_i]在超平面上的投影向量，假设超平面有d个标准正交基组成![\\small W = \\\{ \{w\_1\},w\{\}\_2, \\cdots \{w\_d\}\\\}][small W _ _ _w_1_w_2_ _cdots _w_d]，d为绛维后的维度。现在，我们需要求解出样本在超平面的投影向量![\\small \{\\bar x\_i\}][small _bar x_i]，可以通过求出![\\small x\_\{i\}][small x_i]在每一个基上的投影，然后将这些投影相加就能得到![\\small x\_\{i\}][small x_i]在由这些基组成的超平面的投影向量

![\\small \{\\bar x\_i\} = \\sum\\limits\_\{j = 1\}^d \{(w\_j^\{\\rm\{T\}\}\{x\_i\})\{w\_j\}\}][small _bar x_i_ _ _sum_limits_j _ 1_d _w_j_rm_T_x_i_w_j]

跟上面方法一样，我们将问题转化为带约束的优化问题：

![\\small \\left\\\{ \{\\begin\{array\}\{\*\{20\}\{c\}\} \{\\mathop \{\{\\mathop\{\\rm argmin\}\\nolimits\} \}\\limits\_W \\sum\\limits\_\{i = 1\}^n \{\{\{\\left\\| \{\{x\_i\} - \{\{\\bar x\}\_i\}\} \\right\\|\}^2\}\} \}\\\\ \{s.t.\{W^T\}W = I\} \\end\{array\}\} \\right.][small _left_ _begin_array_20_c_ _mathop _mathop_rm argmin_nolimits_ _limits_W _sum_limits_i _ 1_n _left_ _x_i_ - _bar x_i_ _right_2_ _ _s.t._W_T_W _ I_ _end_array_ _right.]

这里我们是最小化距离的平方，便于后面的计算。先来求解优化目标

![\\small \{\\left\\| \{\{x\_i\} - \{\{\\bar x\}\_i\}\} \\right\\|^2\} = \{(\{x\_i\} - \{\\bar x\_i\})^T\}(\{x\_i\} - \{\\bar x\_i\}) = \{x\_i\}^T\{x\_i\} - 2\{x\_i\}^T\{\\bar x\_i\} + \{\\bar x\_i\}^T\{\\bar x\_i\}][small _left_ _x_i_ - _bar x_i_ _right_2_ _ _x_i_ - _bar x_i_T_x_i_ - _bar x_i_ _ _x_i_T_x_i_ - 2_x_i_T_bar x_i_ _ _bar x_i_T_bar x_i]

通过上面的式子可以得到：

![\\small x\_i^T\{\\bar x\_i\} = x\_i^T\\sum\\limits\_\{j= 1\}^d \{(w\_j^T\} \{x\_i\})\{w\_j\} = \\sum\\limits\_\{j = 1\}^d ( w\_j^T\{x\_i\})x\_i^T\{w\_j\} = \\sum\\limits\_\{j = 1\}^d \{w\_j^T\} \{x\_i\}x\_i^T\{w\_j\}= tr(\{W^T\}\{x\_i\}x\_i^TW)][small x_i_T_bar x_i_ _ x_i_T_sum_limits_j_ 1_d _w_j_T_ _x_i_w_j_ _ _sum_limits_j _ 1_d _ w_j_T_x_i_x_i_T_w_j_ _ _sum_limits_j _ 1_d _w_j_T_ _x_i_x_i_T_w_j_ tr_W_T_x_i_x_i_TW]

![\\small \\bar x\_i^T\{\\bar x\_i\} = (\\sum\\limits\_\{j = 1\}^d \{(w\_j^T\} \{x\_i\})\{w\_j\}\{)^T\}(\\sum\\limits\_\{j = 1\}^d \{(w\_j^T\} \{x\_i\})\{w\_j\}) = \\sum\\limits\_\{j = 1\}^d \{\\sum\\limits\_\{k = 1\}^d \{((w\_j^T\} \} \{x\_i\})\{w\_j\}\{)^T\}((w\_k^T\{x\_j\})\{w\_k\})][small _bar x_i_T_bar x_i_ _ _sum_limits_j _ 1_d _w_j_T_ _x_i_w_j_T_sum_limits_j _ 1_d _w_j_T_ _x_i_w_j_ _ _sum_limits_j _ 1_d _sum_limits_k _ 1_d _w_j_T_ _ _x_i_w_j_T_w_k_T_x_j_w_k]

因为当![\\small i\\neq j][small i_neq j]时，![\\small \{w\_i\}\{w\_j\} = 1][small _w_i_w_j_ _ 1]，当![\\small i=j][small i_j]时，![\\small \{w\_i\}\{w\_j\} = 0][small _w_i_w_j_ _ 0]，所以上面的式子可以写成：

![\\small \\bar x\_i^T\{\\bar x\_i\} = \\sum\\limits\_\{j = 1\}^d \{w\_j^T\} \{x\_i\}x\_i^T\{w\_j\} = tr(\{W^T\}\{x\_i\}x\_i^TW)][small _bar x_i_T_bar x_i_ _ _sum_limits_j _ 1_d _w_j_T_ _x_i_x_i_T_w_j_ _ tr_W_T_x_i_x_i_TW]

结合上面的式子，可以得到：

![\\small \\sum\\limits\_\{i = 1\}^n \{\{\{\\left\\| \{\{x\_i\} - \{\{\\bar x\}\_i\}\} \\right\\|\}^2\}\} = \\sum\\limits\_\{i = 1\}^n \{( - tr(\{W^T\}\} \{x\_i\}x\_i^TW) + \{x\_i\}x\_i^T) = - tr(\{W^T\}X\{X^T\}W) + X\{X^T\}][small _sum_limits_i _ 1_n _left_ _x_i_ - _bar x_i_ _right_2_ _ _sum_limits_i _ 1_n _ - tr_W_T_ _x_i_x_i_TW_ _ _x_i_x_i_T_ _ - tr_W_T_X_X_T_W_ _ X_X_T]

因为![\\small XX^\{T\}][small XX_T]为一个常数，对优化结果没有影响，所以优化问题转化为：

![\\small \\left\\\{ \{\\begin\{array\}\{\*\{20\}\{c\}\} \{\\mathop \{\\arg \\max \}\\limits\_W tr(\{W^T\}X\{X^T\}W)\}\\\\ \{\{W^T\}W = I\} \\end\{array\}\} \\right.][small _left_ _begin_array_20_c_ _mathop _arg _max _limits_W tr_W_T_X_X_T_W_ _W_T_W _ I_ _end_array_ _right.]

最后就和前面第一种方法的求解达到一致，我们依次求解W中的每一个分量时，就是上面的最大化方差的方法，虽然两种方法考虑的角度不一样，但是最终的结果都化为求解带约束的优化问题。

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