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====== PlayGround ====== | ====== PlayGround ====== | ||
- | 以下是对课件的补充和扩展: | ||
- | 2.不适用于非高斯噪声污染的数据集。\\ | ||
- | 设数据$m\times n$矩阵$M=L+S$,$L$为潜在的低秩矩阵,$S$噪声矩阵。如果$S$非高斯噪声,例如稀疏且幅值不定的噪声,那么PCA将失效。此时,宜用改进的模型RPCA(Robust PCA)。 | ||
- | RPCA通过以下目标函数求解: | ||
- | <jsmath>\min_{L,S}\|L\|_{*}+\lambda \|S\|_{l_{1}} \;\; s.t. \; M=L+S</jsmath> | ||
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- | $\lambda$是权重参数,通常设为$\lambda=\frac{1}{\sqrt{\max\{m,n\}}}$。 $\|\cdot\|_{*}$为核范数,即矩阵奇异值之和。$\|\cdot\|_{l_{1}}$为一范数,即矩阵元素绝对值之和。\\ | ||
- | 此凸函数具有唯一最小值。使用ALM(Augmented Lagrange Multiplier)求解,最小化增强的拉格朗日函数: | ||
- | <jsmath>\min_{L,S,Y}l(L,S,Y)=\min_{L,S,Y}\|L\|_{*}+\lambda \|S\|_{l_{1}}+ tr\{Y^{T}(M-L-S)\}+\frac{\mu}{2}\|M-L-S\|_{F}^{2}</jsmath> | ||
- | $\mu$为另一权重,可取值$\mu=nm/(4\|M\|_{1})$。$\|\cdot\|_{F}$为Frobenius范数,即矩阵元素平方和开根号。\\ | ||
- | 通过迭代计算求最优值:\\ | ||
- | 1.初始化:$S_{0}=Y_{0}=0$,$k=0$\\ | ||
- | 2.while $\|M-L-S\|_{F}>10^{-7}\|M\|_{F}$ do\\ | ||
- | 3. $L_{k+1}=D_{1/\mu}(M-S_{k}-Y_{k}/\mu)$\\ | ||
- | 4. $S_{k+1}=S_{\lambda/\mu}(M-L_{k+1}+Y_{k}/\mu)$\\ | ||
- | 5. $Y_{k+1}=Y_{k}+\mu (M-L_{k+1}-S_{k+1})$\\ | ||
- | 6. $k=k+1$\\ | ||
- | 7.end while\\ | ||
- | 8.输出$L,S$\\ | ||
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- | $x^2 | ||
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- | ===== JSMath test ===== | ||
- | You can simple input inline latex syntax like: $x^2+\frac{1}{b}$. | ||
- | Have fun. <jsmath>\lim_{n\to\infty}\sum_{i=1}^{n} \frac{1}{i}=\infty</jsmath> | ||
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- | <note important>浙江大学2008-2009版权所有,如需转载或引用,请与[[zhx@cad.zju.edu.cn | 作者联系]]。</note> |