On page 352, "An Introduction to Statistical Learning" by James et al. explains the effect of using the polynomial kernel as follows:
It essentially amounts to fitting a support vector classifier in a
higher-dimensional space involving polynomials of degree d, rather
than in the original feature space.
However, it then proceeds to state that, when using a non-linear kernel, predictions are made using:
Now, the number of parameters in this equation is 1 + |S|, where |S| is the number of support vectors. If I were to use additional features explicitly instead, I would have an individual weight for each feature, which may amount to many more than 1 + |S| parameters. So, it seems that using the kernel gives less expressive power than using explicit features. Is this correct?
You're right in the sense that you will fit more parameters if the number of features (|F|) is larger than |S|. Fitting an SVM with |S| < |F| will often be more accurate than a linear fit on F, since you're performing a non-linear fit to F. Think of the example in the left image
You can't make a linear combination of the given features (x and y axis) that separates the black and white labels perfectly, however you can make some non-linear transformation of F which does (right image). You could attempt to add transformations of your features as additional features (such as [f_1, f_2, f_1*f_1, f_1*f_2, f_2*f_2], where f_1, f_2 are your original features), or you can instead fit an SVM. In an SVM you don't make a linear fit of some transformed features, but you instead make a linear fit to how each data point relates to the support vectors via the chosen kernel.
A second consideration is that adding additional features that are unrelated to your labels introduce noise to your fit and might have non-zero weights since they pick up on some random pattern in your data. As such, fitting more parameters doesn't always help the model.
Related
In deep learning, I saw many papers apply the pre-processing step as normalization step. It normalizes the input as zero mean and unit variance before feeding to the convolutional network (has BatchNorm). Why not use original intensity? What is the benefit of the normalization step? If I used histogram matching among images, should I still use the normalization step? Thanks
Normalization is important to bring features onto the same scale for the network to behave much better. Let's assume there are two features where one is measured on a scale of 1 to 10 and the second on a scale from 1 to 10,000. In terms of squared error function the network will be busy optimizing the weights according to the larger error on the second feature.
Therefore it is better to normalize.
The answer to this can be found in Andrew Ng's tutorial: https://youtu.be/UIp2CMI0748?t=133.
TLDR: If you do not normalize input features, some features can have a very different scale and will slow down Gradient Descent.
Long explanation: Let us consider a model that uses two features Feature1 and Feature2 with the following ranges:
Feature1: [10,10000]
Feature2: [0.00001, 0.001]
The Contour plot of these will look something like this (scaled for easier visibility).
Contour plot of Feature1 and Feature2
When you perform Gradient Descent, you will calculate d(Feature1) and d(Feature2) where "d" denotes differential in order to move the model weights closer to minimizing the loss. As evident from the contour plot above, d(Feature1) is going to be significantly smaller compared to d(Feature2), so even if you choose a reasonably medium value of learning rate, then you will be zig-zagging around because of relatively large values of d(Feature2) and may even miss the global minima.
Medium value of learning rate
In order to avoid this, if you choose a very small value of learning rate, Gradient Descent will take a very long time to converge and you may stop training even before reaching the global minima.
Very small Gradient Descent
So as you can see from the above examples, not scaling your features lead to an inefficient Gradient Descent which results in not finding the most optimal model
I know that feature selection helps me remove features that may have low contribution. I know that PCA helps reduce possibly correlated features into one, reducing the dimensions. I know that normalization transforms features to the same scale.
But is there a recommended order to do these three steps? Logically I would think that I should weed out bad features by feature selection first, followed by normalizing them, and finally use PCA to reduce dimensions and make the features as independent from each other as possible.
Is this logic correct?
Bonus question - are there any more things to do (preprocess or transform)
to the features before feeding them into the estimator?
If I were doing a classifier of some sort I would personally use this order
Normalization
PCA
Feature Selection
Normalization: You would do normalization first to get data into reasonable bounds. If you have data (x,y) and the range of x is from -1000 to +1000 and y is from -1 to +1 You can see any distance metric would automatically say a change in y is less significant than a change in X. we don't know that is the case yet. So we want to normalize our data.
PCA: Uses the eigenvalue decomposition of data to find an orthogonal basis set that describes the variance in data points. If you have 4 characteristics, PCA can show you that only 2 characteristics really differentiate data points which brings us to the last step
Feature Selection: once you have a coordinate space that better describes your data you can select which features are salient.Typically you'd use the largest eigenvalues(EVs) and their corresponding eigenvectors from PCA for your representation. Since larger EVs mean there is more variance in that data direction, you can get more granularity in isolating features. This is a good method to reduce number of dimensions of your problem.
of course this could change from problem to problem, but that is simply a generic guide.
Generally speaking, Normalization is needed before PCA.
The key to the problem is the order of feature selection, and it's depends on the method of feature selection.
A simple feature selection is to see whether the variance or standard deviation of the feature is small. If these values are relatively small, this feature may not help the classifier. But if you do normalization before you do this, the standard deviation and variance will become smaller (generally less than 1), which will result in very small differences in std or var between the different features.If you use zero-mean normalization, the mean of all the features will equal 0 and std equals 1.At this point, it might be bad to do normalization before feature selection
Feature selection is flexible, and there are many ways to select features. The order of feature selection should be chosen according to the actual situation
Good answers here. One point needs to be highlighted. PCA is a form of dimensionality reduction. It will find a lower dimensional linear subspace that approximates the data well. When the axes of this subspace align with the features that one started with, it will lead to interpretable feature selection as well. Otherwise, feature selection after PCA, will lead to features that are linear combinations of the original set of features and they are difficult to interpret based on the original set of features.
I am using Linear regression to predict data. But, I am getting totally contrasting results when I Normalize (Vs) Standardize variables.
Normalization = x -xmin/ xmax – xmin
Zero Score Standardization = x - xmean/ xstd
a) Also, when to Normalize (Vs) Standardize ?
b) How Normalization affects Linear Regression?
c) Is it okay if I don't normalize all the attributes/lables in the linear regression?
Thanks,
Santosh
Note that the results might not necessarily be so different. You might simply need different hyperparameters for the two options to give similar results.
The ideal thing is to test what works best for your problem. If you can't afford this for some reason, most algorithms will probably benefit from standardization more so than from normalization.
See here for some examples of when one should be preferred over the other:
For example, in clustering analyses, standardization may be especially crucial in order to compare similarities between features based on certain distance measures. Another prominent example is the Principal Component Analysis, where we usually prefer standardization over Min-Max scaling, since we are interested in the components that maximize the variance (depending on the question and if the PCA computes the components via the correlation matrix instead of the covariance matrix; but more about PCA in my previous article).
However, this doesn’t mean that Min-Max scaling is not useful at all! A popular application is image processing, where pixel intensities have to be normalized to fit within a certain range (i.e., 0 to 255 for the RGB color range). Also, typical neural network algorithm require data that on a 0-1 scale.
One disadvantage of normalization over standardization is that it loses some information in the data, especially about outliers.
Also on the linked page, there is this picture:
As you can see, scaling clusters all the data very close together, which may not be what you want. It might cause algorithms such as gradient descent to take longer to converge to the same solution they would on a standardized data set, or it might even make it impossible.
"Normalizing variables" doesn't really make sense. The correct terminology is "normalizing / scaling the features". If you're going to normalize or scale one feature, you should do the same for the rest.
That makes sense because normalization and standardization do different things.
Normalization transforms your data into a range between 0 and 1
Standardization transforms your data such that the resulting distribution has a mean of 0 and a standard deviation of 1
Normalization/standardization are designed to achieve a similar goal, which is to create features that have similar ranges to each other. We want that so we can be sure we are capturing the true information in a feature, and that we dont over weigh a particular feature just because its values are much larger than other features.
If all of your features are within a similar range of each other then theres no real need to standardize/normalize. If, however, some features naturally take on values that are much larger/smaller than others then normalization/standardization is called for
If you're going to be normalizing at least one variable/feature, I would do the same thing to all of the others as well
First question is why we need Normalisation/Standardisation?
=> We take a example of dataset where we have salary variable and age variable.
Age can take range from 0 to 90 where salary can be from 25thousand to 2.5lakh.
We compare difference for 2 person then age difference will be in range of below 100 where salary difference will in range of thousands.
So if we don't want one variable to dominate other then we use either Normalisation or Standardization. Now both age and salary will be in same scale
but when we use standardiztion or normalisation, we lose original values and it is transformed to some values. So loss of interpretation but extremely important when we want to draw inference from our data.
Normalization rescales the values into a range of [0,1]. also called min-max scaled.
Standardization rescales data to have a mean (μ) of 0 and standard deviation (σ) of 1.So it gives a normal graph.
Example below:
Another example:
In above image, you can see that our actual data(in green) is spread b/w 1 to 6, standardised data(in red) is spread around -1 to 3 whereas normalised data(in blue) is spread around 0 to 1.
Normally many algorithm required you to first standardise/normalise data before passing as parameter. Like in PCA, where we do dimension reduction by plotting our 3D data into 1D(say).Here we required standardisation.
But in Image processing, it is required to normalise pixels before processing.
But during normalisation, we lose outliers(extreme datapoints-either too low or too high) which is slight disadvantage.
So it depends on our preference what we chose but standardisation is most recommended as it gives a normal curve.
None of the mentioned transformations shall matter for linear regression as these are all affine transformations.
Found coefficients would change but explained variance will ultimately remain the same. So, from linear regression perspective, Outliers remain as outliers (leverage points).
And these transformations also will not change the distribution. Shape of the distribution remains the same.
lot of people use Normalisation and Standardisation interchangeably. The purpose remains the same is to bring features into the same scale. The approach is to subtract each value from min value or mean and divide by max value minus min value or SD respectively. The difference you can observe that when using min value u will get all value + ve and mean value u will get bot + ve and -ve values. This is also one of the factors to decide which approach to use.
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
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.
Could somebody explain to me what do I miss?
Thanks,
Daniel
The dual formulation of the linear SVM depends only on scalar products of all training vectors. Scalar product essentially measures similarity of two vectors. We can then generalize it by replacing with any other "well-behaved" (it should be positive-definite, it's needed to preserve convexity, as well as enables Mercer's theorem) similarity measure. And RBF is just one of them.
If you take a look at the formula here you'll see that RBF is basically a scalar product in a certain infinitely dimensional space
Thus RBF is kind of a union of polynomial kernels of all possible degrees.
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.
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.
The short answer is that this business about infinite dimensional spaces is only part of the theoretical justification, and of no practical importance. You never actually touch an infinite-dimensional space in any sense. It's part of the proof that the radial basis function works.
Basically, SVMs are proved to work the way they do by relying on properties of dot products over vector spaces. You can't just swap in the radial basis function and expect it necessarily works. To prove that it does, however, you show that the radial basis function is actually like a dot product over a different vector space, and it's as if we're doing regular SVMs in a transformed space, which works. And it happens that infinite dimensioal-ness is OK, and that the radial basis function does correspond to a dot product in such a space. So you can say SVMs still work when you use this particular kernel.
How exactly is an U-matrix constructed in order to visualise a self-organizing-map? More specifically, suppose that I have an output grid of 3x3 nodes (that have already been trained), how do I construct a U-matrix from this? You can e.g. assume that the neurons (and inputs) have dimension 4.
I have found several resources on the web, but they are not clear or they are contradictory. For example, the original paper is full of typos.
A U-matrix is a visual representation of the distances between neurons in the input data dimension space. Namely you calculate the distance between adjacent neurons, using their trained vector. If your input dimension was 4, then each neuron in the trained map also corresponds to a 4-dimensional vector. Let's say you have a 3x3 hexagonal map.
The U-matrix will be a 5x5 matrix with interpolated elements for each connection between two neurons like this
The {x,y} elements are the distance between neuron x and y, and the values in {x} elements are the mean of the surrounding values. For example, {4,5} = distance(4,5) and {4} = mean({1,4}, {2,4}, {4,5}, {4,7}). For the calculation of the distance you use the trained 4-dimensional vector of each neuron and the distance formula that you used for the training of the map (usually Euclidian distance). So, the values of the U-matrix are only numbers (not vectors). Then you can assign a light gray colour to the largest of these values and a dark gray to the smallest and the other values to corresponding shades of gray. You can use these colours to paint the cells of the U-matrix and have a visualized representation of the distances between neurons.
Have also a look at this web article.
The original paper cited in the question states:
A naive application of Kohonen's algorithm, although preserving the topology of the input data is not able to show clusters inherent in the input data.
Firstly, that's true, secondly, it is a deep mis-understanding of the SOM, thirdly it is also a mis-understanding of the purpose of calculating the SOM.
Just take the RGB color space as an example: are there 3 colors (RGB), or 6 (RGBCMY), or 8 (+BW), or more? How would you define that independent of the purpose, ie inherent in the data itself?
My recommendation would be not to use maximum likelihood estimators of cluster boundaries at all - not even such primitive ones as the U-Matrix -, because the underlying argument is already flawed. No matter which method you then use to determine the cluster, you would inherit that flaw. More precisely, the determination of cluster boundaries is not interesting at all, and it is loosing information regarding the true intention of building a SOM. So, why do we build SOM's from data?
Let us start with some basics:
Any SOM is a representative model of a data space, for it reduces the dimensionality of the latter. For it is a model it can be used as a diagnostic as well as a predictive tool. Yet, both cases are not justified by some universal objectivity. Instead, models are deeply dependent on the purpose and the accepted associated risk for errors.
Let us assume for a moment the U-Matrix (or similar) would be reasonable. So we determine some clusters on the map. It is not only an issue how to justify the criterion for it (outside of the purpose itself), it is also problematic because any further calculation destroys some information (it is a model about a model).
The only interesting thing on a SOM is the accuracy itself viz the classification error, not some estimation of it. Thus, the estimation of the model in terms of validation and robustness is the only thing that is interesting.
Any prediction has a purpose and the acceptance of the prediction is a function of the accuracy, which in turn can be expressed by the classification error. Note that the classification error can be determined for 2-class models as well as for multi-class models. If you don't have a purpose, you should not do anything with your data.
Inversely, the concept of "number of clusters" is completely dependent on the criterion "allowed divergence within clusters", so it is masking the most important thing of the structure of the data. It is also dependent on the risk and the risk structure (in terms of type I/II errors) you are willing to take.
So, how could we determine the number classes on a SOM? If there is no exterior apriori reasoning available, the only feasible way would be an a-posteriori check of the goodness-of-fit. On a given SOM, impose different numbers of classes and measure the deviations in terms of mis-classification cost, then choose (subjectively) the most pleasing one (using some fancy heuristics, like Occam's razor)
Taken together, the U-matrix is pretending objectivity where no objectivity can be. It is a serious misunderstanding of modeling altogether.
IMHO it is one of the greatest advantages of the SOM that all the parameters implied by it are accessible and open for being parameterized. Approaches like the U-matrix destroy just that, by disregarding this transparency and closing it again with opaque statistical reasoning.