I have a good (or at least a self-consistent) calibration set and have applied PCA and recently PLS regression on n.i.r. spectrum of known mixtures of water and additive to predict the percentage of additive by volume. I thus far have done self-calibration and now want to predict the concentration from the n.i.r.spectrum blindly. Octave returns XLOADINGS, YLOADINGS, XSCORES, YSCORES, COEFFICIENTS, and FITTED with the plsregress command. The "fitted" is the estimate of concentration. Octave uses the SIMPLS approach.
How do I use these returned variables to predict concentration give a new samples spectrum?
Scores are usually denoted by T and loadings by P and X=TP'+E where E is the residual. I am stuck.
Note that T and P are X scores and loadings, respectively. Unlike PCA, PLS has scores and loadings for Y as well (usually denoted U and Q).
While the documentation of plsregress is sketchy at best, the paper it refers to Sijmen de Jong: SIMPLS: an alternativ approach to partial least squares regression Chemom Intell Lab Syst, 1993, 18, 251-263, DOI: 10.1016/0169-7439(93)85002-X
discusses prediction with equations (36) and (37), which give:
Yhat0 = X0 B
Note that this uses centered data X0 to predict centered y-values. B are the COEFFICIENTS.
I recommend that as a first step you predict your training spectra and make sure you get the correct results (FITTED).
Related
I am going through Introduction to Statistical Learning in R by Hastie and Tibshirani. I came across two concepts: RSE and MSE. My understanding is like this:
RSE = sqrt(RSS/N-2)
MSE = RSS/N
Now I am building 3 models for a problem and need to compare them. While MSE come intuitively to me, I was also wondering if calculating RSS/N-2 will make any use which is according to above is RSE^2
I think I am not sure which to use where?
RSE is an estimate of the standard deviation of the residuals, and therefore also of the observations. Which is why it's equal to RSS/df. And in your case, as a simple linear model df = 2.
MSE is mean squared error observed in your models, and it's usually calculated using a test set to compare the predictive accuracy of your fitted models. Since we're concerned with the mean error of the model, we divide by n.
I think RSE ⊂ MSE (i.e. RSE is part of MSE).
And MSE = RSS/ degree of freedom
MSE for a single set of data (X1,X2,....Xn) would be RSS over N
or more accurately is RSS/N-1
(since your freedom to vary will be reduced by one when U have used up all the freedom)
But in linear regression concerning X and Y with binomial term, the degree of freedom is affected by both X and Y thus N-2
thus yr MSE = RSS/N-2
and one can also call this RSE
And in over parameterized model, meaning one have a collection of many ßs (more than 2 terms| y~ ß0 + ß1*X + ß2*X..), one can even penalize the model by reducing the denominator by including the number of parameters:
MSE= RSS/N-p (p is the number of fitted parameters)
I am getting started with deep learning and have a basic question on CNN's.
I understand how gradients are adjusted using backpropagation according to a loss function.
But I thought the values of the convolving filter matrices (in CNN's) needs to be determined by us.
I'm using Keras and this is how (from a tutorial) the convolution layer was defined:
classifier = Sequential()
classifier.add(Conv2D(32, (3, 3), input_shape = (64, 64, 3), activation = 'relu'))
There are 32 filter matrices with dimensions 3x3 is used.
But, how are the values for these 32x3x3 matrices are determined?
It's not the gradients that are adjusted, the gradient calculated with the backpropagation algorithm is just the group of partial derivatives with respect to each weight in the network, and these components are in turn used to adjust the network weights in order to minimize the loss.
Take a look at this introductive guide.
The weights in the convolution layer in your example will be initialized to random values (according to a specific method), and then tweaked during training, using the gradient at each iteration to adjust each individual weight. Same goes for weights in a fully connected layer, or any other layer with weights.
EDIT: I'm adding some more details about the answer above.
Let's say you have a neural network with a single layer, which has some weights W. Now, during the forward pass, you calculate your output yHat for your network, compare it with your expected output y for your training samples, and compute some cost C (for example, using the quadratic cost function).
Now, you're interested in making the network more accurate, ie. you'd like to minimize C as much as possible. Imagine you want to find the minimum value for simple function like f(x)=x^2. You can start at some random point (as you did with your network), then compute the slope of the function at that point (ie, the derivative) and move down that direction, until you reach a minimum value (a local minimum at least).
With a neural network it's the same idea, with the difference that your inputs are fixed (the training samples), and you can see your cost function C as having n variables, where n is the number of weights in your network. To minimize C, you need the slope of the cost function C in each direction (ie. with respect to each variable, each weight w), and that vector of partial derivatives is the gradient.
Once you have the gradient, the part where you "move a bit following the slope" is the weights update part, where you update each network weight according to its partial derivative (in general, you subtract some learning rate multiplied by the partial derivative with respect to that weight).
A trained network is just a network whose weights have been adjusted over many iterations in such a way that the value of the cost function C over the training dataset is as small as possible.
This is the same for a convolutional layer too: you first initialize the weights at random (ie. you place yourself on a random position on the plot for the cost function C), then compute the gradients, then "move downhill", ie. you adjust each weight following the gradient in order to minimize C.
The only difference between a fully connected layer and a convolutional layer is how they calculate their outputs, and how the gradient is in turn computed, but the part where you update each weight with the gradient is the same for every weight in the network.
So, to answer your question, those filters in the convolutional kernels are initially random and are later adjusted with the backpropagation algorithm, as described above.
Hope this helps!
Sergio0694 states ,"The weights in the convolution layer in your example will be initialized to random values". So if they are random and say I want 10 filters. Every execution algorithm could find different filter. Also say I have Mnist data set. Numbers are formed of edges and curves. Is it guaranteed that there will be a edge filter or curve filter in 10?
I mean is first 10 filters most meaningful most distinctive filters we can find.
best
So, I have a vector that corresponds to a given feature (same dimensionality). Is there a package in Julia that would provide a mathematical function that fits these data points, in relation to the original feature? In other words, I have x and y (both vectors) and need to find a decent mapping between the two, even if it's a highly complex one. The output of this process should be a symbolic formula that connects x and y, e.g. (:x)^3 + log(:x) - 4.2454. It's fine if it's just a polynomial approximation.
I imagine this is a walk in the park if you employ Genetic Programming, but I'd rather opt for a simpler (and faster) approach, if it's available. Thanks
Turns out the Polynomials.jl package includes the function polyfit which does Lagrange interpolation. A usage example would go:
using Polynomials # install with Pkg.add("Polynomials")
x = [1,2,3] # demo x
y = [10,12,4] # demo y
polyfit(x,y)
The last line returns:
Poly(-2.0 + 17.0x - 5.0x^2)`
which evaluates to the correct values.
The polyfit function accepts a maximal degree for the output polynomial, but defaults to using the length of the input vectors x and y minus 1. This is the same degree as the polynomial from the Lagrange formula, and since polynomials of such degree agree on the inputs only if they are identical (this is a basic theorem) - it can be certain this is the same Lagrange polynomial and in fact the only one of such a degree to have this property.
Thanks to the developers of Polynomial.jl for leaving me just to google my way to an Answer.
Take a look to MARS regression. Multi adaptive regression splines.
I would like to make a prediction by using Least Squares Support Vector Machine for Regression, which is proposed by Suykens et al. I am using LS-SVMlab, which you can find the MATLAB toolbox here. Let's consider I have an independent variable X and a dependent variable Y, that both are simulated. I am following the instructions in the tutorial.
>>X = linspace(-1,1,50)’;
>>Y = (15*(X.^2-1).^2.*X.^4).*exp(-X)+normrnd(0,0.1,length(X),1);
>>type = ’function estimation’;
>>[gam,sig2] = tunelssvm({X,Y,type,[], [],’RBF_kernel’},’simplex’,...’leaveoneoutlssvm’,’mse’});
>>[alpha,b] = trainlssvm({X,Y,type,gam,sig2,’RBF_kernel’});
>>plotlssvm({X,Y,type,gam,sig2,’RBF_kernel’},{alpha,b});
The code above finds the best parameters using simplex method and leave-one-out cross validation and trains the model and give me alphas (support vector values for all the data points in the training set) and b coefficients. However, it does not give me the predictions of the variable Y. It only draws the plot. In some articles, I saw plots like the one below,
As I said before, the LS-SVM toolbox does not give me the predicted values of Y, it only draws the plot but no values in the workspace. How can I get these values and draw a graph of predicted values together with actual values?
There is one solution that I think of. By using X values in the training set, I re-run the model and get the prediction of values Y by using simlssvm command but it does not seem reasonable to me. Any solution that you can offer? Thanks in advance.
I am afraid you have answered your own question. The only way to obtain the prediction for the training points in LS-SVMLab is by simulating the training points after training your model.
[yp,alpha,b,gam,sig2,model] = lssvm(x,y,'f')
when u use this function yp is the predicted value
I have a function f(x) = 1/(x + a+ b*I*sign(x)) and I want to calculate the
integral of
dx dy dz f(x) f(y) f(z) f(x+y+z) f(x-y - z)
over the entire R^3 (b>0 and a,- b are of order unity). This is just a representative example -- in practice I have n<7 variables and 2n-1 instances of f(), n of them involving the n integration variables and n-1 of them involving some linear combintation of the integration variables. At this stage I'm only interested in a rough estimate with relative error of 1e-3 or so.
I have tried the following libraries :
Steven Johnson's cubature code: the hcubature algorithm works but is abysmally slow, taking hundreds of millions of integrand evaluations for even n=2.
HintLib: I tried adaptive integration with a Genz-Malik rule, the cubature routines, VEGAS and MISER with the Mersenne twister RNG. For n=3 only the first seems to be somewhat viable option but it again takes hundreds of millions of integrand evaluations for n=3 and relerr = 1e-2, which is not encouraging.
For the region of integration I have tried both approaches: Integrating over [-200, 200]^n (i.e. a region so large that it essentially captures most of the integral) and the substitution x = sinh(t) which seems to be a standard trick.
I do not have much experience with numerical analysis but presumably the difficulty lies in the discontinuities from the sign() term. For n=2 and f(x)f(y)f(x-y) there are discontinuities along x=0, y=0, x=y. These create a very sharp peak around the origin (with a different sign in the various quadrants) and sort of 'ridges' at x=0,y=0,x=y along which the integrand is large in absolute value and changes sign as you cross them. So at least I know which regions are important. I was thinking that maybe I could do Monte Carlo but somehow "tell" the algorithm in advance where to focus. But I'm not quite sure how to do that.
I would be very grateful if you had any advice on how to evaluate the integral with a reasonable amount of computing power or how to make my Monte Carlo "idea" work. I've been stuck on this for a while so any input would be welcome. Thanks in advance.
One thing you can do is to use a guiding function for your Monte Carlo integration: given an integral (am writing it in 1D for simplicity) of ∫ f(x) dx, write it as ∫ f(x)/g(x) g(x) dx, and use g(x) as a distribution from which you sample x.
Since g(x) is arbitrary, construct it such that (1) it has peaks where you expect them to be in f(x), and (2) such that you can sample x from g(x) (e.g., a gaussian, or 1/(1+x^2)).
Alternatively, you can use a Metropolis-type Markov chain MC. It will find the relevant regions of the integrand (almost) by itself.
Here are a couple of trivial examples.