支持多对象分类的向量机 [英] Support vector machines for mutliple object categorization

问题描述

我试图使用线性SVM进行多类对象类别识别。到目前为止，我已经理解的是，主要有两种方法使用 - 一个vs全（OVA）和一个vs一（OVO）。
但是我很难理解它的实现。我的意思是我认为使用的步骤是：

I am trying to use linear SVMs for multi-class object category recognition. So far what I have understood is that there are mainly two approaches used- one-vs-all(OVA) and one-vs-one(OVO). But I am having difficulty understanding its implementation. I mean the steps that I think is used are:

• 首先，特征描述符从我们说SIFT准备。


• 接下来为一个特定对象类别（比如汽车）准备一个SVM分类器模型，我将50张汽车图像作为正向训练集，总共50个从每个类别随机抽取的休息类别的图像（此部分是否正确？）。我为所有这样的类别（例如5个）准备这样的模型。


• 接下来当我有一个输入图像，我需要输入图像到所有5个模型，然后检查他们这些模型的值（+ 1 / -1）？

推荐答案

在一对多的方法中，必须检查所有5个型号。然后你可以采取最有信心值的决定。 LIBSVM 给出概率估计。

In one-vs-all approach, you have to check for all 5 models. Then you can take the decision with the most confidence value. LIBSVM gives probability estimates.

在一对一的方法中，你可以占大多数。例如，您测试1对2，1对3，1对4和1对5。您将其分为1在3个案件。你对其他4类也一样。假设其他四个类的值是 [0，1，1，2] 。因此，类1获得的次数最多，使该类作为最终类。在这种情况下，您还可以进行总概率估计。取最大值。这将工作，除非一对分类变得极端错误。例如，在1对4中，它将4（真实类是1）分类为置信0.7。然后只是因为这一个决定，你的总概率估计可能会出现并给出错误的结果。这个问题可以通过实验来检查。

In one-vs-one approach, you can take the majority. For example, you test 1 vs. 2, 1 vs. 3, 1 vs. 4 and 1 vs. 5. You classify it as 1 in 3 cases. You do the same for other 4 classes. Suppose for other four classes the values are [0, 1, 1, 2]. Therefore, class 1 was obtained most number of times, making that class as the final class. In this case, you could also do total of probability estimates. Take the maximum. That would work unless in one pair the classification goes extremely wrong. For example, in 1 vs. 4, it classifies 4 (true class is 1) with a confidence 0.7. Then just because of this one decision, your total of probability estimates may shoot up and give wrong results. This issue can be examined experimentally.

LIBSVM使用一个对一个。您可以在此处查看推理。您可以阅读这篇论文，他们可以保护一个或所有分类方法，并得出结论这是不一定比一个比一个。

LIBSVM uses one vs. one. You can check the reasoning here. You can read this paper too where they defend one vs. all classification approach and conclude that it is not necessarily worse than one vs. one.

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