I'm looking to set up a linear regression using 2D Gaussian basis functions. My input training variables cover a two dimensional space. Before applying the machine learning (Bayesian linear regression), I need to select parameters for the Gaussians - mean and variance and also decide how many basis functions to use.
I am currently spacing the means (of a preallocated number of basis Gaussians) evenly over a grid, and just assuming constant variance. This is obviously not the best approach.
Any ideas on how to calculate these variables?
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'm trying to understand SVR model.
To do it I looked at SVM and it's pretty clear for me. But there is no much explications about SVR.
The first question is why it's called Support Vector Regression or how we use vectors to predict numerical values?
Also I don't understand some parameters such as epsilon and gamma. How they influence predicted result?
A SVM learns a so called decision function from your features, such that features from you positive class produce positive real numbers, and features from the negative class produce negative numbers (at least most of the time, depending on your data).
For two features you can visualize this in a 2D plane. The function assigns a real value to each point in the plane, this value can be depicted as color. This plot shows the values as different blue colors.
The feature values resulting in zero form the so called decision boundary.
This function itself has two kind of parameters:
kernel dependend parameters. In your case for the radial basis functions, these parameters are epsilon and gamma, which you set before learning.
And the so called support-vectors which are determined during learning. support-vectors are just parameters of your decision function.
Learning is nothing than determining good support-vectors (parameters !).
In this 2d example video the colors don't show the actual function value, but only the sign. You can see how gamma influences the smoothness of the decision function.
To answer you question:
SVR builds such a function but with a different goal. The function does not try to assign positive outcomes to your postive examples, and negative outcomes to the negative examples.
Instead the function is built to approximate the given numeric outcomes.
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.
If we have K classes, do I have to plot K learning curves?
Because it seems impossible to me to calculate the train/validation error against all K theta vectors at once.
To clarify, the learning curve is a plot of the training & cross validation/test set error/cost vs training set size. This plot should allow you to see if increasing the training set size improves performance. More generally, the learning curve allows you to identify whether your algorithm suffers from a bias (under fitting) or variance (over fitting) problem.
It depends. Learning curves do not concern themselves with the number of classes. Like you said, it is a plot of training set and test set error, where that error is a numerical value. This is all learning curves are.
That error can be anything you want: accuracy, precision, recall, F1 score etc. (even MAE, MSE and others for regression).
However, the error you choose to use is the one that does or does not apply to your specific problem, which in turn indirectly affects how you should use learning curves.
Accuracy is well defined for any number of classes, so if you use this, a single plot should suffice.
Precision and recall, however, are defined only for binary problems. You can somewhat generalize them (see here for example) by considering the binary problem with classes x and not x for each class x. In that case, you will probably want to plot learning curves for each class. This will also help you identify problems relating to certain classes better.
If you want to read more about performance metrics, I like this paper a lot.
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.