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?
Related
I have a classification problem and my current feature vector does not seem to hold enough information.
My training set has 10k entries and I am using a SVM as classifier (scikit-learn).
What is the maximum reasonable feature vector size (how many dimension)?
(Training and evaluation using Labtop CPU)
100? 1k? 10k? 100k? 1M?
The thing is not how many features should it be for a certain number of cases (i.e. entries) but rather the opposite:
It’s not who has the best algorithm that wins. It’s who has the most data. (Banko and Brill, 2001)
Banko and Brill in 2001 made a comparison among 4 different algorithms, they kept increasing the Training Set Size to millions and came up with the above-quoted conclusion.
Moreover, Prof. Andrew Ng clearly covered this topic, and I’m quoting here:
If a learning algorithm is suffering from high variance, getting more training data is likely to help.
If a learning algorithm is suffering from high bias, getting more training data will not (by itself) help much
So as a rule of thumb, your data cases must be greater than the number of features in your dataset taking into account that all features should be informative as much as possible (i.e. the features are not highly collinear (i.e. redundant)).
I read once in more than one place and somewhere in Scikit-Learn Documentation, that the number of inputs (i.e. samples) must be at least the square size of the number of features (i.e. n_samples > n_features ** 2 ).
Nevertheless, for SVM in particular, the number of features n v.s number of entries m is an important factor to specify the type of kernel to use initially, as a second rule of thumb for SVM in particular (also according to Prof. Andrew Ng):
If thr number of features is much greater than number of entries (i.e. n is up to 10K and m is up to 1K) --> use SVM without a kernel (i.e. "linear kernel") or use Logistic Regression.
If the number of features is small and if the number of entries is intermediate (i.e. n is up to 1K and m is up to 10K) --> use SVM with Gaussian kernel.
If the number of feature is small and if the number of entries is much larger (i.e. n is up to 1K and m > 50K) --> Create/add more features, then use SVM without a kernel or use Logistic Regression.
I want to find the opinion of a sentence either positive or negative. For example talk about only one sentence.
The play was awesome
If change it to vector form
[0,0,0,0]
After searching through the Bag of words
bad
naughty
awesome
The vector form becomes
[0,0,0,1]
Same for other sentences. Now I want to pass it to the machine learning algorithm for training it. How can I train the network using these multiple vectors? (for finding the opinion of unseen sentences) Obviously not! Because the input is fix in neural network. Is there any way? The above procedure is just my thinking. Kindly correct me if I am wrong. Thanks in advance.
Since your intuitive input format is "Sentence". Which is, indeed, a string of tokens with arbitrary length. Abstracting sentences as token series is not a good choice for many existing algorithms only works on determined format of inputs.
Hence, I suggest try using tokenizer on your entire training set. This will give you vectors of length of the dictionary, which is fixed for given training set.
Because when the length of sentences vary drastically, then size of the dictionary always keeps stable.
Then you can apply Neural Networks(or other algorithms) to the tokenized vectors.
However, vectors generated by tokenizer is extremely sparse because you only work on sentences rather than articles.
You can try LDA (supervised, not PCA), to reduce the dimension as well as amplify the difference.
That will keep the essential information of your training data as well as express your data at fixed size, while this "size" is not too large.
By the way, you may not have to label each word by its attitude since the opinion of a sentence also depends on other kind of words.
Simple arithmetics on number of opinion-expressing words many leave your model highly biased. Better label the sentences and leave the rest job to classifiers.
For the confusions
PCA and LDA are Dimensional Reduction techniques.
difference
Let's assume each tuple of sample is denoted as x (1-by-p vector).
p is too large, we don't like that.
Let's find a matrix A(p-by-k) in which k is pretty small.
So we get reduced_x = x*A, and most importantly, reduced_x must
be able to represent x's characters.
Given labeled data, LDA can provide proper A that can maximize
distance between reduced_x of different classes, and also minimize
the distance within identical classes.
In simple words: compress data, keep information.
When you've got
reduced_x, you can define training data: (reduced_x|y) where y is
0 or 1.
I work on classifying some reviews (paragraphs) consists of multiple sentences. I classified them with bag-of-word features in Weka via libSVM. However, I had another idea which I don't know how to implement :
I thought creating syntactical and shallow-semantics based features per sentence in the reviews is worth to try. However, I couldn't find any way to encode those features sequentially, since a paragraph's sentence size varies. The reason that I wanted to keep those features in an order is that the order of sentence features may give a better clue for classification. For example, if I have two instances P1 (with 3 sentences) and P2 (2 sentences), I would have a space like that (assume each sentence has one binary feature as a or b):
P1 -> a b b /classX
P2 -> b a /classY
So, my question is that whether I can implement that classification of different feature sizes in feature space or not? If yes, is there any kind of classifier that I can use in Weka, scikit-learn or Mallet? I would appreciate any responses.
Thanks
Regardless of the implementation, an SVM with the standard kernels (linear, poly, RBF) requires fixed-length feature vectors. You can encode any information in those feature vectors by encoding as booleans; e.g. collect all syntactical/semantic features that occur in your corpus, then introduce booleans that represent that "feature such and such occurred in this document". If it's important to capture the fact that these features occur in multiple sentences, count them and use put the frequency in the feature vector (but be sure to normalize your frequencies by document length, as SVMs are not scale-invariant).
In case you are classifying textual data, I would suggest looking at "Rational Kernels" which are made on weighted finite transducers for classifying natural language texts. Rational Kernels can be applied on varied length vectors and are already implemented as an open source project (OpenFST).
It is the library's problem, since SVM itself does not require fixed-length feature vectors, it only need a kernel function, if you can provide a kernel function with varied length vector, it should be OK for SVM
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.
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I am using LibSVM to classify some documents. The documents seem to be a bit difficult to classify as the final results show. However, I have noticed something while training my models. and that is: If my training set is for example 1000 around 800 of them are selected as support vectors.
I have looked everywhere to find if this is a good thing or bad. I mean is there a relation between the number of support vectors and the classifiers performance?
I have read this previous post but I am performing a parameter selection and also I am sure that the attributes in the feature vectors are all ordered.
I just need to know the relation.
Thanks.
p.s: I use a linear kernel.
Support Vector Machines are an optimization problem. They are attempting to find a hyperplane that divides the two classes with the largest margin. The support vectors are the points which fall within this margin. It's easiest to understand if you build it up from simple to more complex.
Hard Margin Linear SVM
In a training set where the data is linearly separable, and you are using a hard margin (no slack allowed), the support vectors are the points which lie along the supporting hyperplanes (the hyperplanes parallel to the dividing hyperplane at the edges of the margin)
All of the support vectors lie exactly on the margin. Regardless of the number of dimensions or size of data set, the number of support vectors could be as little as 2.
Soft-Margin Linear SVM
But what if our dataset isn't linearly separable? We introduce soft margin SVM. We no longer require that our datapoints lie outside the margin, we allow some amount of them to stray over the line into the margin. We use the slack parameter C to control this. (nu in nu-SVM) This gives us a wider margin and greater error on the training dataset, but improves generalization and/or allows us to find a linear separation of data that is not linearly separable.
Now, the number of support vectors depends on how much slack we allow and the distribution of the data. If we allow a large amount of slack, we will have a large number of support vectors. If we allow very little slack, we will have very few support vectors. The accuracy depends on finding the right level of slack for the data being analyzed. Some data it will not be possible to get a high level of accuracy, we must simply find the best fit we can.
Non-Linear SVM
This brings us to non-linear SVM. We are still trying to linearly divide the data, but we are now trying to do it in a higher dimensional space. This is done via a kernel function, which of course has its own set of parameters. When we translate this back to the original feature space, the result is non-linear:
Now, the number of support vectors still depends on how much slack we allow, but it also depends on the complexity of our model. Each twist and turn in the final model in our input space requires one or more support vectors to define. Ultimately, the output of an SVM is the support vectors and an alpha, which in essence is defining how much influence that specific support vector has on the final decision.
Here, accuracy depends on the trade-off between a high-complexity model which may over-fit the data and a large-margin which will incorrectly classify some of the training data in the interest of better generalization. The number of support vectors can range from very few to every single data point if you completely over-fit your data. This tradeoff is controlled via C and through the choice of kernel and kernel parameters.
I assume when you said performance you were referring to accuracy, but I thought I would also speak to performance in terms of computational complexity. In order to test a data point using an SVM model, you need to compute the dot product of each support vector with the test point. Therefore the computational complexity of the model is linear in the number of support vectors. Fewer support vectors means faster classification of test points.
A good resource:
A Tutorial on Support Vector Machines for Pattern Recognition
800 out of 1000 basically tells you that the SVM needs to use almost every single training sample to encode the training set. That basically tells you that there isn't much regularity in your data.
Sounds like you have major issues with not enough training data. Also, maybe think about some specific features that separate this data better.
Both number of samples and number of attributes may influence the number of support vectors, making model more complex. I believe you use words or even ngrams as attributes, so there are quite many of them, and natural language models are very complex themselves. So, 800 support vectors of 1000 samples seem to be ok. (Also pay attention to #karenu's comments about C/nu parameters that also have large effect on SVs number).
To get intuition about this recall SVM main idea. SVM works in a multidimensional feature space and tries to find hyperplane that separates all given samples. If you have a lot of samples and only 2 features (2 dimensions), the data and hyperplane may look like this:
Here there are only 3 support vectors, all the others are behind them and thus don't play any role. Note, that these support vectors are defined by only 2 coordinates.
Now imagine that you have 3 dimensional space and thus support vectors are defined by 3 coordinates.
This means that there's one more parameter (coordinate) to be adjusted, and this adjustment may need more samples to find optimal hyperplane. In other words, in worst case SVM finds only 1 hyperplane coordinate per sample.
When the data is well-structured (i.e. holds patterns quite well) only several support vectors may be needed - all the others will stay behind those. But text is very, very bad structured data. SVM does its best, trying to fit sample as well as possible, and thus takes as support vectors even more samples than drops. With increasing number of samples this "anomaly" is reduced (more insignificant samples appear), but absolute number of support vectors stays very high.
SVM classification is linear in the number of support vectors (SVs). The number of SVs is in the worst case equal to the number of training samples, so 800/1000 is not yet the worst case, but it's still pretty bad.
Then again, 1000 training documents is a small training set. You should check what happens when you scale up to 10000s or more documents. If things don't improve, consider using linear SVMs, trained with LibLinear, for document classification; those scale up much better (model size and classification time are linear in the number of features and independent of the number of training samples).
There is some confusion between sources. In the textbook ISLR 6th Ed, for instance, C is described as a "boundary violation budget" from where it follows that higher C will allow for more boundary violations and more support vectors.
But in svm implementations in R and python the parameter C is implemented as "violation penalty" which is the opposite and then you will observe that for higher values of C there are fewer support vectors.