I use the libsvm tool to do the classification. but the number of the support vector turns out to be zero, which means there is no support vector. Why? what does it mean and what are the main reasons?
It means, that parameters and data used for training led to the trivial model (always equal to one of the classes). So what are the most probable reasons?
Bad choice of SVM parameters (C, gamma, degree, depending on used kernel)
Wrong data (inconsistent, check what exactly are you feeding as X and Y matrices)
Related
I studying on SVM and Support Vector recently. for example if I select Hard Linear SVM in a two dimensional classification problem with n Data, then result consist k=2 Support Vector. if I add another labeled data in previous data and retrain SVM. what's the maximum Number of SV?
I think N+1. but I need some proof. anyone help?
There is no maximum bound on the number of support vectors, and the whole data set can be selected as support vectors. The proof is fairly simple (i.e.: left as an exercise for the reader).
Assuming that N is the size of your training set, and giving the fact that all of them can be selected as support vectors then N is the maximum amount of SVs for that particular scenario.
You might also want to take a look on this: What is the relation between the number of Support Vectors and training data and classifiers performance?
I'm using WEKA/LibSVM to train a classifier for a term extraction system. My data is not linearly separable, so I used an RBF kernel instead of a linear one.
I followed the guide from Hsu et al. and iterated over several values for both c and gamma. The parameters which worked best for classifying known terms (test and training material differ of course) are rather high, c=2^10 and gamma=2^3.
So far the high parameters seem to work ok, yet I wonder if they may cause any problems further on, especially regarding overfitting. I plan to do another evaluation by extracting new terms, yet those are costly as I need human judges.
Could anything still be wrong with my parameters, even if both evaluation turns out positive? Do I perhaps need another kernel type?
Thank you very much!
In general you have to perform cross validation to answer whether the parameters are all right or do they lead to the overfitting.
From the "intuition" perspective - it seems like highly overfitted model. High value of gamma means that your Gaussians are very narrow (condensed around each poinT) which combined with high C value will result in memorizing most of the training set. If you check out the number of support vectors I would not be surprised if it would be the 50% of your whole data. Other possible explanation is that you did not scale your data. Most ML methods, especially SVM, requires data to be properly preprocessed. This means in particular, that you should normalize (standarize) the input data so it is more or less contained in the unit sphere.
RBF seems like a reasonable choice so I would keep using it. A high value of gamma is not necessary a bad thing, it would depends on the scale where your data lives. While a high C value can lead to overfitting, it would also be affected by the scale so in some cases it might be just fine.
If you think that your dataset is a good representation of the whole data, then you could use crossvalidation to test your parameters and have some peace of mind.
When using SVMlight or LIBSVM in order to classify phrases as positive or negative (Sentiment Analysis), is there a way to determine which are the most influential words that affected the algorithms decision? For example, finding that the word "good" helped determine a phrase as positive, etc.
If you use the linear kernel then yes - simply compute the weights vector:
w = SUM_i y_i alpha_i sv_i
Where:
sv - support vector
alpha - coefficient found with SVMlight
y - corresponding class (+1 or -1)
(in some implementations alpha's are already multiplied by y_i and so they are positive/negative)
Once you have w, which is of dimensions 1 x d where d is your data dimension (number of words in the bag of words/tfidf representation) simply select the dimensions with high absolute value (no matter positive or negative) in order to find the most important features (words).
If you use some kernel (like RBF) then the answer is no, there is no direct method of taking out the most important features, as the classification process is performed in completely different way.
As #lejlot mentioned, with linear kernel in SVM, one of the feature ranking strategies is based on the absolute values of weights in the model. Another simple and effective strategy is based on F-score. It considers each feature separately and therefore cannot reveal mutual information between features. You can also determine how important a feature is by removing that feature and observe the classification performance.
You can see this article for more details on feature ranking.
With other kernels in SVM, the feature ranking is not that straighforward, yet still feasible. You can construct an orthogonal set of basis vectors in the kernel space, and calculate the weights by kernel relief. Then the implicit feature ranking can be done based on the absolute value of weights. Finally the data is projected into the learned subspace.
If I have 200 features, and if each feature can have a value ranging from 0 to infinity, should I scale the feature values to be in the range [0-1] before I go ahead and train a LibSVM on top of it?
Now, suppose I did scale the values, and after training the model if I get one vector with its values or the features as input, how do I scale these values of the input test vector before classifying it?
Thanks
Abhishek S
You should store the ranges of you feature-values used for training. Then when you extract a feature-value from an unknown instance, use the particular range for scaling.
Use the formula (here for the range [-1.0 , 1.0]):
double scaled_val = -1.0 + (1.0 - -1.0) * (extracted_val - vmin)/(vmax-vmin);
The Guide provided at libsvm website explains the scaling well:
"2.2 Scaling
Scaling before applying SVM is very important. Part 2 of Sarle's Neural Networks
FAQ Sarle (1997) explains the importance of this and most of considerations also apply
to SVM. The main advantage of scaling is to avoid attributes in greater numeric
ranges dominating those in smaller numeric ranges. Another advantage is to avoid
numerical diculties during the calculation. Because kernel values usually depend on
the inner products of feature vectors, e.g. the linear kernel and the polynomial kernel,
large attribute values might cause numerical problems. We recommend linearly
scaling each attribute to the range [-1; +1] or [0; 1].
Of course we have to use the same method to scale both training and testing
data."
If you've got infinite feature values, you're not going to be able to use LIBSVM anyway.
More practically, scaling is generally useful so the kernel doesn't have to deal with large numbers, so I would say go for it and scale. It's not a requirement, though.
And as Anony-Mousse implied in the comments, please try running experiments with and without scaling so you can see the difference.
Now, suppose I did scale the values, and after training the model if I get one vector with its values or the features as input, how do I scale these values of the input test vector before classifying it?
You don't need to scale again. You already did that in the pre-training step (i.e. data processing).
I have read through a lot of papers and understand the basic concept of a support vector machine at a very high level. You give it a training input vector which has a set of features and bases on how the "optimization function" evaluates this input vector lets call it x, (lets say we're talking about text classification), the text associated with the input vector x is classified into one of two pre-defined classes, this is only in the case of binary classification.
So my first question is through this procedure described above, all the papers say first that this training input vector x is mapped to a higher (maybe infinite) dimensional space. So what does this mapping achieve or why is this required? Lets say the input vector x has 5 features so who decides which "higher dimension" x is going to be mapped to?
Second question is about the following optimization equation:
min 1/2 wi(transpose)*wi + C Σi = 1..n ξi
so I understand that w has something to do with the margins of the hyperplane from the support vectors in the graph and I know that C is some sort of a penalty but I dont' know what it is a penalty for. And also what is ξi representing in this case.
A simple explanation of the second question would be much appreciated as I have not had much luck understanding it by reading technical papers.
When they talk about mapping to a higher-dimensional space, they mean that the kernel accomplishes the same thing as mapping the points to a higher-dimensional space and then taking dot products there. SVMs are fundamentally a linear classifier, but if you use kernels, they're linear in a space that's different from the original data space.
To be concrete, let's talk about the kernel
K(x, y) = (xy + 1)^2 = (xy)^2 + 2xy + 1,
where x and y are each real numbers (one-dimensional). Note that
(x^2, sqrt(2) x, 1) • (y^2, sqrt(2) y, 1) = x^2 y^2 + 2 x y + 1
has the same value. So K(x, y) = phi(x) • phi(y), where phi(a) = (a^2, sqrt(2), 1), and doing an SVM with this kernel (the inhomogeneous polynomial kernel of degree 2) is the same as if you first mapped your 1d points into this 3d space and then did a linear kernel.
The popular Gaussian RBF kernel function is equivalent to mapping your points into an infinite-dimensional Hilbert space.
You're the one who decides what feature space it's mapped into, when you pick a kernel. You don't necessarily need to think about the explicit mapping when you do that, though, and it's important to note that the data is never actually transformed into that high-dimensional space explicitly - then infinite-dimensional points would be hard to represent. :)
The ξ_i are the "slack variables". Without them, SVMs would never be able to account for training sets that aren't linearly separable -- which most real-world datasets aren't. The ξ in some sense are the amount you need to push data points on the wrong side of the margin over to the correct side. C is a parameter that determines how much it costs you to increase the ξ (that's why it's multiplied there).
1) The higher dimension space happens through the kernel mechanism. However, when evaluating the test sample, the higher dimension space need not be explicitly computed. (Clearly this must be the case because we cannot represent infinite dimensions on a computer.) For instance, radial basis function kernels imply infinite dimensional spaces, yet we don't need to map into this infinite dimension space explicitly. We only need to compute, K(x_sv,x_test), where x_sv is one of the support vectors and x_test is the test sample.
The specific higher dimensional space is chosen by the training procedure and parameters, which choose a set of support vectors and their corresponding weights.
2) C is the weight associated with the cost of not being able to classify the training set perfectly. The optimization equation says to trade-off between the two undesirable cases of non-perfect classification and low margin. The ξi variables represent by how much we're unable to classify instance i of the training set, i.e., the training error of instance i.
See Chris Burges' tutorial on SVM's for about the most intuitive explanation you're going to get of this stuff anywhere (IMO).