Perhaps this is an easy question, but I want to make sure I understand the conceptual basis of the LibSVM implementation of one-class SVMs and if what I am doing is permissible.
I am using one class SVMs in this case for outlier detection and removal. This is used in the context of a greater time series prediction model as a data preprocessing step. That said, I have a Y vector (which is the quantity we are trying to predict and is continuous, not class labels) and an X matrix (continuous features used to predict). Since I want to detect outliers in the data early in the preprocessing step, I have yet to normalize or lag the X matrix for use in prediction, or for that matter detrend/remove noise/or otherwise process the Y vector (which is already scaled to within [-1,1]). My main question is whether it is correct to model the one class SVM like so (using libSVM):
svmod = svmtrain(ones(size(Y,1),1),Y,'-s 2 -t 2 -g 0.00001 -n 0.01');
[od,~,~] = svmpredict(ones(size(Y,1),1),Y,svmod);
The resulting model does yield performance somewhat in line with what I would expect (99% or so prediction accuracy, meaning 1% of the observations are outliers). But why I ask is because in other questions regarding one class SVMs, people appear to be using their X matrices where I use Y. Thanks for your help.
What you are doing here is nothing more than a fancy range check. If you are not willing to use X to find outliers in Y (even though you really should), it would be a lot simpler and better to just check the distribution of Y to find outliers instead of this improvised SVM solution (for example remove the upper and lower 0.5-percentiles from Y).
In reality, this is probably not even close to what you really want to do. With this setup you are rejecting Y values as outliers without considering any context (e.g. X). Why are you using RBF and how did you come up with that specific value for gamma? A kernel is total overkill for one-dimensional data.
Secondly, you are training and testing on the same data (Y). A kitten dies every time this happens. One-class SVM attempts to build a model which recognizes the training data, it should not be used on the same data it was built with. Please, think of the kittens.
Additionally, note that the nu parameter of one-class SVM controls the amount of outliers the classifier will accept. This is explained in the LIBSVM implementation document (page 4): It is proved that nu is an upper bound on the fraction of training errors and
a lower bound of the fraction of support vectors. In other words: your training options specifically state that up to 1% of the data can be rejected. For one-class SVM, replace can by should.
So when you say that the resulting model does yield performance somewhat in line with what I would expect ... ofcourse it does, by definition. Since you have set nu=0.01, 1% of the data is rejected by the model and thus flagged as an outlier.
Related
I am facing a peculiar problem and I was wondering if there is an explanation. I am trying to run a linear regression problem and test different optimization methods and two of them have a strange outcome when comparing to each other. I build a data set that satisfies y=2x+5 and I add a random noise to that.
xtrain=np.range(0,50,1).reshape(50,1)
ytrain=2*train+5+np.random.normal(0,2,(50,1))
opt1=torch.optim.SGD(model.parameters(),lr=1e-5,momentum=0.8))
opt2=torch.optim.Rprop(model.parameters(),lr=1e-5)
F_loss=F.mse_loss
from torch.utils.data import TensorDataset,DataLoader
train_d=TensorDataset(xtrain,ytrain)
train=DataLoader(train_d,50,shuffle=True)
model1=nn.Linear(1,1)
loss=F_loss(model1(xtrain),ytrain)
def fit(nepoch, model1, F_loss, opt):
for epoch in range(nepoch):
for i,j in train:
predict = model1(i)
loss = F_loss(predict, j)
loss.backward()
opt.step()
opt.zero_grad()
When i compare the results of the following commands:
fit(500000, model1, F_loss, opt1)
fit(500000, model1, F_loss, opt2)
In the last epoch for opt1:loss=2.86,weight=2.02,bias=4.46
In the last epoch for opt2:loss=3.47,weight=2.02,bias=4.68
These results do not make sense to me, shouldn't opt2 have a smaller loss than opt1 since the weight and bias it finds is closer to the real value? opt2's method finds weights and biases to be closer to the real value (they are respectively 2 and 5). Am i doing something wrong?
This has to do with the fact that you are drawing the training samples themselves from a random distribution.
By doing so, you inherently randomized the ground truth to some extent. Sure, you will get values that are inherently distributed around 2x+5, but you do not guarantee that 2x+5 will also be the best fit to this data distribution.
It could thus happen that you accidentally end up with values that deviate quite significantly from the original function, and, since you use a mean squared error, these values get weighted quite significantly.
In expectation (i.e., for the number of samples going towards infinity), you will likely get closer and closer to the expected parameters.
A way to verify this would be to plot your training samples against the parameter set, as well as the (ideal) underlying function.
Also note that Linear Regression does have a direct solution - something that is very uncommon in Machine Learning - meaning you can directly calculate an optimal solution, e.g., with sklearn's function
Im new with NN and i have this problem:
I have a dataset with 300 rows and 33 columns. Each row has 3 more columns for the results.
Im trying to use MLP for trainning a model so that when i have a new row, it estimates those 3 result columns.
I can easily reduce the error during trainning to 0.001 but when i use cross validation it keep estimating very poorly.
It estimates correctly if i use the same entry it used to train, but if i use another values that werent used for trainning the results are very wrong
Im using two hidden layers with 20 neurons each, so my architecture is [33 20 20 3]
For activation function im using biporlarsigmoid function.
Do you guys have some suggestion on where i could try to change to improve this?
Overfitting
As mentioned in the comments, this perfectly describes overfitting.
I strongly suggest reading the wikipedia article on overfitting, as it well describes causes, but I'll summarize some key points here.
Model complexity
Overfitting often happens when you model is needlessly complex for the problem. I don't know anything about your dataset, but I'm guessing [33 20 20 3] is more parameters than necessary for predicting.
Try running your cross-validation methods again, this time with either fewer layers, or fewer nodes per layer. Right now you are using 33*20 + 20*20 + 20*3 = 1120 parameters (weights) to make your prediction, is this necessary?
Regularization
A common solution to overfitting is regularization. The driving principle is KISS (keep it simple, stupid).
By applying an L1 regularizer to your weights, you keep preference for the smallest number of weights to solve your problem. The network will pull many weights to 0 as they aren't need.
By applying an L2 regularizer to your weights, you keep preference for lower rank solutions to your problem. This means that your network will prefer weights matrices that span lower dimensions. Practically this means your weights will be smaller numbers, and are less likely to be able to "memorize" the data.
What is L1 and L2? These are types of vector norms. L1 is the sum of the absolute value of your weights. L2 is the sqrt of the sum of squares of your weights. (L3 is the cubed root of the sum of cubes of weights, L4 ...).
Distortions
Another commonly used technique is to augment your training data with distorted versions of your training samples. This only makes sense with certain types of data. For instance images can be rotated, scaled, shifted, add gaussian noise, etc. without dramatically changing the content of the image.
By adding distortions, your network will no longer memorize your data, but will also learn when things look similar to your data. The number 1 rotated 2 degrees still looks like a 1, so the network should be able to learn from both of these.
Only you know your data. If this is something that can be done with your data (even just adding a little gaussian noise to each feature), then maybe this is worth looking into. But do not use this blindly without considering the implications it may have on your dataset.
Careful analysis of data
I put this last because it is an indirect response to the overfitting problem. Check your data before pumping it through a black-box algorithm (like a neural network). Here are a few questions worth answering if your network doesn't work:
Are any of my features strongly correlated with each other?
How do baseline algorithms perform? (Linear regression, logistic regression, etc.)
How are my training samples distributed among classes? Do I have 298 samples of one class and 1 sample of the other two?
How similar are my samples within a class? Maybe I have 100 samples for this class, but all of them are the same (or nearly the same).
My goal is to classify an image using multi class linear SVM (with out kernel). I would like to write my own SVM classifier
I am using MATLAB and have trained linear SVM using image sets provided.
I have around 20 classes, 5 images in each class (total of 100 images) and I am using one-versus-all strategy.
Each image is a (112,92) matrix. That means 112*92=10304 values.
I am using quadprog(H,f,A,C) to solve the quadratic equation (y=w'x+b) in the SVM. One call to quadprog returns w vector of size 10304 for one image. That means I have to call quadprog for 100 times.
The problem is one quadprog call takes 35 secs to get executed. That means for 100 images it will take 3500 secs. This might be due to large size of vectors and matrices involved.
I want to reduce the execution time of quadprog. Is there any way to do it?
First of all, when you do classification using SVM, you usually extract a feature (like HOG) of an image, so that the dimensionality of space on which SVM has to operate gets reduced. You are using raw pixel values, which generates a 10304-D vector. That is not good. Use some standard feature.
Secondly, you do not call quadprog 100 times. You call only once. The concept behind the optimization is, you want to find a weight vector w and a bias b such that it satisfies w'x_i+b=y_i for all images (i.e. all x_i). Note that i will go from 1 to the number of examples in your training set, but w and b stay the same. If you wanted a (w,b) that will satisfy only one x, you do not need any fancy optimization. So stack your x in a big matrix of dimension NxD, y will be a vector of Nx1, and then call quadprog. It will take a longer time than 35 seconds, but you do it only once. This is called training an SVM. While testing for a new image, you just extract its feature, and do w*x+b to get its class.
Third, unless you are doing this as an exercise to understand SVMs, quadprog is not the best way to solve this problem. You need to use some other techniques which work well with large data, for example, Sequential Minimal Optimization.
How can one linear hyperplane classify more than 2 classes: For multi-class classification, SVMs usually use two popular approaches:
One-vs-one SVM: For a C class problem, you train C(C-1)/2 classifiers and at test time, you test that many classifiers and choose the class which receives most votes.
One-vs-all SVM: As name suggests, you train one classifier per class with positive samples from that class and negative samples from all other classes.
From LIBSVM FAQs:
It is one-against-one. We chose it after doing the following comparison: C.-W. Hsu and C.-J. Lin. A comparison of methods for multi-class support vector machines , IEEE Transactions on Neural Networks, 13(2002), 415-425.
"1-against-the rest" is a good method whose performance is comparable to "1-against-1." We do the latter simply because its training time is shorter.
However, also note that a naive implementation of one-vs-one may not be practical for large-scale problems. LIBSVM website also lists this shortcoming and provides an extension.
LIBLINEAR does not support one-versus-one multi-classification, so we provide an extension here. If k is the number of classes, we generate k(k-1)/2 models, each of which involves only two classes of training data. According to Yuan et al. (2012), one-versus-one is not practical for large-scale linear classification because of the huge space needed to store k(k-1)/2 models. However, this approach may still be viable if model vectors (i.e., weight vectors) are very sparse. Our implementation stores models in a sparse form and can effectively handle some large-scale data.
I have a 500x1000 feature vector and principal component analysis says that over 99% of total variance is covered by the first component. So I replace 1000 dimension point by 1 dimension point giving 500x1 feature vector(using Matlab's pca function). But, my classifier accuracy which was initially around 80% with 1000 features now drops to 30% with 1 feature even though more than 99% of the variance is accounted by this feature. What could be the explanation to this or are my methods wrong?
(This question partly arises from my earlier question Significance of 99% of variance covered by the first component in PCA)
Edit:
I used weka's principal components method to perform the dimensionality reduction and support vector machines(SVM) classifier.
Principal Components do not necessarily have any correlation to classification accuracy. There could be a 2-variable situation where 99% of the variance corresponds to the first PC but that PC has no relation to the underlying classes in the data. Whereas the second PC (which only contributes to 1% of the variance) is the one that can separate the classes. If you only keep the first PC, then you lose the feature that actually provides the ability to classify the data.
In practice, smaller (lower variance) PCs often are associated with noise so there can be benefit in removing them but there is no guarantee of this.
Consider a case where you have two variables: a person's mass (in grams) and body temperature (in degrees Celsius). You want to predict which people have the flu and which do not. In this case, weight has a much greater variance but probably no correlation to the flu, whereas temperature, which has low variance, has a strong correlation to the flu. After the Principal Components transformation, the first PC will be strongly aligned with mass (since it has much greater variance) so if you dropped the second PC, would be losing almost all of your classification accuracy.
It is important to remember that Principal Components is an unsupervised transformation of the data. It does not consider labels of your training data when calculating the transformation (as opposed to something like Fisher's linear discriminant).
I have to write a classificator (gaussian mixture model) that I use for human action recognition.
I have 4 dataset of video. I choose 3 of them as training set and 1 of them as testing set.
Before I apply the gm model on the training set I run the pca on it.
pca_coeff=princomp(trainig_data);
score = training_data * pca_coeff;
training_data = score(:,1:min(size(score,2),numDimension));
During the testing step what should I do? Should I execute a new princomp on testing data
new_pca_coeff=princomp(testing_data);
score = testing_data * new_pca_coeff;
testing_data = score(:,1:min(size(score,2),numDimension));
or I should use the pca_coeff that I compute for the training data?
score = testing_data * pca_coeff;
testing_data = score(:,1:min(size(score,2),numDimension));
The classifier is being trained on data in the space defined by the principle components of the training data. It doesn't make sense to evaluate it in a different space - therefore, you should apply the same transformation to testing data as you did to training data, so don't compute a different pca_coef.
Incidently, if your testing data is drawn independently from the same distribution as the training data, then for large enough training and test sets, the principle components should be approximately the same.
One method for choosing how many principle components to use involves examining the eigenvalues from the PCA decomposition. You can get these from the princomp function like this:
[pca_coeff score eigenvalues] = princomp(data);
The eigenvalues variable will then be an array where each element describes the amount of variance accounted for by the corresponding principle component. If you do:
plot(eigenvalues);
you should see that the first eigenvalue will be the largest, and they will rapidly decrease (this is called a "Scree Plot", and should look like this: http://www.ats.ucla.edu/stat/SPSS/output/spss_output_pca_5.gif, though your one may have up to 800 points instead of 12).
Principle components with small corresponding eigenvalues are unlikely to be useful, since the variance of the data in those dimensions is so small. Many people choose a threshold value, and then select all principle components where the eigenvalue is above that threshold. An informal way of picking the threshold is to look at the Scree plot and choose the threshold to be just after the line 'levels out' - in the image I linked earlier, a good value might be ~0.8, selecting 3 or 4 principle components.
IIRC, you could do something like:
proportion_of_variance = sum(eigenvalues(1:k)) ./ sum(eigenvalues);
to calculate "the proportion of variance described by the low dimensional data".
However, since you are using the principle components for a classification task, you can't really be sure that any particular number of PCs is optimal; the variance of a feature doesn't necessarily tell you anything about how useful it will be for classification. An alternative to choosing PCs with the Scree plot is just to try classification with various numbers of principle components and see what the best number is empirically.