问题描述
我刚刚浏览了有关SVM的Wikipedia页面，这一行引起了我的注意:如果使用的核是高斯径向基函数，则相应的特征空间就是无限尺寸的希尔伯特空间." http://en.wikipedia.org/wiki/Support_vector_machine#Nonlinear_classification
I've just run through the Wikipedia page about SVMs, and this line caught my eyes:"If the kernel used is a Gaussian radial basis function, the corresponding feature space is a Hilbert space of infinite dimensions." http://en.wikipedia.org/wiki/Support_vector_machine#Nonlinear_classification
根据我的理解，如果我在SVM中应用高斯内核，则在选择界标作为训练示例时，结果特征空间将为m维(其中m是训练样本的数量)，并且您正在测量特定示例与所有带有高斯核的示例之间的相似性".因此，对于一个示例，您将拥有与训练示例一样多的相似性值.这些将成为新的特征向量，这些特征向量将成为m维向量，而不是无限维.
In my understanding, if I apply Gaussian kernel in SVM, the resulting feature space will be m-dimensional (where m is the number of training samples), as you choose your landmarks to be your training examples, and you're measuring the "similarity" between a specific example and all the examples with the Gaussian kernel. As a consequence, for a single example you'll have as many similarity values as training examples. These are going to be the new feature vectors which are going to m-dimensional vectors, and not infinite dimensionals.
有人可以向我解释我想念什么吗?
Could somebody explain to me what do I miss?
谢谢，丹尼尔
推荐答案
其他答案是正确的，但在这里并没有真正讲出正确的故事.重要的是，您是正确的.如果您有m个不同的训练点，则高斯径向基核使SVM在m维空间中运行.我们说径向基核映射到一个无穷大的空间，因为您可以使m随心所欲，并且其运行的空间不断扩大.
The other answers are correct but don't really tell the right story here. Importantly, you are correct. If you have m distinct training points then the gaussian radial basis kernel makes the SVM operate in an m dimensional space. We say that the radial basis kernel maps to a space of infinite dimension because you can make m as large as you want and the space it operates in keeps growing without bound.
但是，其他内核(如多项式内核)不具有此维度随训练样本数量缩放的属性.例如，如果您有1000个2D训练样本并且使用< x，y&^ 2的多项式内核，则SVM将在3维空间而不是1000维空间中运行.
However, other kernels, like the polynomial kernel do not have this property of the dimensionality scaling with the number of training samples. For example, if you have 1000 2D training samples and you use a polynomial kernel of <x,y>^2 then the SVM will operate in a 3 dimensional space, not a 1000 dimensional space.
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