How would I approach learning changes in speed using RNNs/LSTMs given x,y coordinates of continuous data? (I have to use a recurrent layer as this is a sub-problem of a bigger end-to-end model that does other things too)
Training data example:
x,y,speed_changed
0,0,0
0,0.1,0
0,0.2,0
0,0.3,0
0,0.5,1
0,0.6,0
0,0.7,0
...
So far I constructed stateful LSTM and train it on one item per batch. After, I reset the state of the LSTM every time there is a change in speed, so I learn that a segment had the same speed (segments can have different lengths).
How do I use such model in production then since the segments have different lengths?
Or is there a better way to train the recurrent net on such data? Perhaps an anomaly detection? (I want to avoid having a fixed batch size (e.g. window of 3 frames))
The structure of RNNs and LSTMs will not let you do it directly, and this is the reason why - The activation function for an RNN is:
h(t) = Tanh(W * h(t-1) + U * x(t) + Bias)
Note that W, U and the Bias are all the same - no matter how many time frames you use for the RNN. So given some X vector, the output will be a function of p1*X1+p2*X2 and so forth, where X1 is X in your example and X2 is Y.
However - to detect a change in speed - you need a different calculation. A change in speed indicates that a different distance was traveled between time frames 1 and 2, and between time frames 2 and 3.
The traveled distance is SQRT((X1(t)-X1(t-1))^2 + (X2(t)-X2(t-1))^2).
This means that you need an activation function that takes into consideration X1*X1 in some way - and this is not possible within an RNN or LSTM.
However, you could achieve what you need indirectly, by using a custom activation function that calculates the distance passed on the latest time frame. Take a look at this link. By using your custom activation function, you can insert the vector of X1(t), X2(t), X1(t-1), X2(t-1) and calculate the distance D. At t=1 you may use 0's as X1(t=0) and X2(t=0).
Your custom activation function should look like D = (X1(t) - X1(t-1))^2 + (X2(t)-X2(t-1))^2. This way - if the speed is the same between time frames you will feed the RNN with constant D values, so you expect the RNN to achieve weights that will simulate a function of D(t) - D(t-1).
Quick question, when I am backpropagating the loss function to my parameters and I used a scaled output (ex. tanh(x) * 2), do I need to include the derivative of the scaled output w.r.t the original output? Thank you!
Before we can backprop the errors, we've to compute the gradient of the loss function with respect to each of the parameters. This computation involves computing the gradients of the outputs first and then use chain rule repeatedly. So, when you do this, the scaling constant remains as is. So, yes, you've to scale the errors accordingly.
As an example, you might have observed the following L2 regularized loss - a.k.a Ridge regression:
Loss = 1/2 * |T - Y|^2 + \lambda * ||w||^2
Here, we are scaling down the squared error. So, when we compute the gradient 1/2 & 2 would cancel out. If we would not have multiplied this by 0.5 in the first place, then we would have to scale up our gradient by 2. Else the gradient vector would point in some other direction instead of the direction which minimizes the loss.
I tried to implement GMMs but I have a few problems during the em-algorithm.
Let's say I've got 3D Samples (stat1, stat2, stat3) which I use to train the GMMs.
One of my training sets for one of the GMMs has in nearly every sample a "0" for stat1. During training I get really small Numbers (like "1.4456539880060609E-124") in the first row and column of the covariance matrix which leads in the next iteration of the EM-Algorithm to 0.0 in the first row and column.
I get something like this:
0.0 0.0 0.0
0.0 5.0 6.0
0.0 2.0 1.0
I need the inverse covariance matrix to calculate the density but since one column is zero I can't do this.
I thought about falling back to the old covariance matrix (and mean) or to replace every 0 with a really small number.
Or is there a another simple solution to this problem?
Simply your data lies in degenerated subspace of your actual input space, and GMM is not well suited in most generic form for such setting. THe problem is that empirical covariance estimator that you use simply fail for such data (as you said - you cannot inverse it). What you usually do? You chenge covariance estimator to the constrained/regularized ones, which contain:
Constant-based shrinking, thus instead of using Sigma = Cov(X) you do Sigma = Cov(X) + eps * I, where eps is prefedefined small constant, and I is identity matrix. Consequently you never have a zero values on the diagonal, and it is easy to prove that for reasonable epsilon, this will be inversible
Nicely fitted shrinking, like Oracle Covariance Estimator or Ledoit-Wolf Covariance Estimator which find best epsilon based on the data itself.
Constrain your gaussians to for example spherical family, thus N(m, sigma I), where sigma = avg_i( cov( X[:, i] ) is the mean covariance per dimension. This limits you to spherical gaussians, and also solves the above issue
There are many more solutions possible, but all based on the same thing - chenge covariance estimator in such a way, that you have a guarantee of invertability.
I am taking this course on Neural networks in Coursera by Geoffrey Hinton (not current).
I have a very basic doubt on weight spaces.
https://d396qusza40orc.cloudfront.net/neuralnets/lecture_slides%2Flec2.pdf
Page 18.
If I have a weight vector (bias is 0) as [w1=1,w2=2] and training case as {1,2,-1} and {2,1,1}
where I guess {1,2} and {2,1} are the input vectors. How can it be represented geometrically?
I am unable to visualize it? Why is training case giving a plane which divides the weight space into 2? Could somebody explain this in a coordinate axes of 3 dimensions?
The following is the text from the ppt:
1.Weight-space has one dimension per weight.
2.A point in the space has particular setting for all the weights.
3.Assuming that we have eliminated the threshold each hyperplane could be represented as a hyperplane through the origin.
My doubt is in the third point above. Kindly help me understand.
It's probably easier to explain if you look deeper into the math. Basically what a single layer of a neural net is performing some function on your input vector transforming it into a different vector space.
You don't want to jump right into thinking of this in 3-dimensions. Start smaller, it's easy to make diagrams in 1-2 dimensions, and nearly impossible to draw anything worthwhile in 3 dimensions (unless you're a brilliant artist), and being able to sketch this stuff out is invaluable.
Let's take the simplest case, where you're taking in an input vector of length 2, you have a weight vector of dimension 2x1, which implies an output vector of length one (effectively a scalar)
In this case it's pretty easy to imagine that you've got something of the form:
input = [x, y]
weight = [a, b]
output = ax + by
If we assume that weight = [1, 3], we can see, and hopefully intuit that the response of our perceptron will be something like this:
With the behavior being largely unchanged for different values of the weight vector.
It's easy to imagine then, that if you're constraining your output to a binary space, there is a plane, maybe 0.5 units above the one shown above that constitutes your "decision boundary".
As you move into higher dimensions this becomes harder and harder to visualize, but if you imagine that that plane shown isn't merely a 2-d plane, but an n-d plane or a hyperplane, you can imagine that this same process happens.
Since actually creating the hyperplane requires either the input or output to be fixed, you can think of giving your perceptron a single training value as creating a "fixed" [x,y] value. This can be used to create a hyperplane. Sadly, this cannot be effectively be visualized as 4-d drawings are not really feasible in browser.
Hope that clears things up, let me know if you have more questions.
I have encountered this question on SO while preparing a large article on linear combinations (it's in Russian, https://habrahabr.ru/post/324736/). It has a section on the weight space and I would like to share some thoughts from it.
Let's take a simple case of linearly separable dataset with two classes, red and green:
The illustration above is in the dataspace X, where samples are represented by points and weight coefficients constitutes a line. It could be conveyed by the following formula:
w^T * x + b = 0
But we can rewrite it vice-versa making x component a vector-coefficient and w a vector-variable:
x^T * w + b = 0
because dot product is symmetrical. Now it could be visualized in the weight space the following way:
where red and green lines are the samples and blue point is the weight.
More possible weights are limited to the area below (shown in magenta):
which could be visualized in dataspace X as:
Hope it clarifies dataspace/weightspace correlation a bit. Feel free to ask questions, will be glad to explain in more detail.
The "decision boundary" for a single layer perceptron is a plane (hyper plane)
where n in the image is the weight vector w, in your case w={w1=1,w2=2}=(1,2) and the direction specifies which side is the right side. n is orthogonal (90 degrees) to the plane)
A plane always splits a space into 2 naturally (extend the plane to infinity in each direction)
you can also try to input different value into the perceptron and try to find where the response is zero (only on the decision boundary).
Recommend you read up on linear algebra to understand it better:
https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces
For a perceptron with 1 input & 1 output layer, there can only be 1 LINEAR hyperplane. And since there is no bias, the hyperplane won't be able to shift in an axis and so it will always share the same origin point. However, if there is a bias, they may not share a same point anymore.
I think the reason why a training case can be represented as a hyperplane because...
Let's say
[j,k] is the weight vector and
[m,n] is the training-input
training-output = jm + kn
Given that a training case in this perspective is fixed and the weights varies, the training-input (m, n) becomes the coefficient and the weights (j, k) become the variables.
Just as in any text book where z = ax + by is a plane,
training-output = jm + kn is also a plane defined by training-output, m, and n.
Equation of a plane passing through origin is written in the form:
ax+by+cz=0
If a=1,b=2,c=3;Equation of the plane can be written as:
x+2y+3z=0
So,in the XYZ plane,Equation: x+2y+3z=0
Now,in the weight space;every dimension will represent a weight.So,if the perceptron has 10 weights,Weight space will be 10 dimensional.
Equation of the perceptron: ax+by+cz<=0 ==> Class 0
ax+by+cz>0 ==> Class 1
In this case;a,b & c are the weights.x,y & z are the input features.
In the weight space;a,b & c are the variables(axis).
So,for every training example;for eg: (x,y,z)=(2,3,4);a hyperplane would be formed in the weight space whose equation would be:
2a+3b+4c=0
passing through the origin.
I hope,now,you understand it.
Consider we have 2 weights. So w = [w1, w2]. Suppose we have input x = [x1, x2] = [1, 2]. If you use the weight to do a prediction, you have z = w1*x1 + w2*x2 and prediction y = z > 0 ? 1 : 0.
Suppose the label for the input x is 1. Thus, we hope y = 1, and thus we want z = w1*x1 + w2*x2 > 0. Consider vector multiplication, z = (w ^ T)x. So we want (w ^ T)x > 0. The geometric interpretation of this expression is that the angle between w and x is less than 90 degree. For example, the green vector is a candidate for w that would give the correct prediction of 1 in this case. Actually, any vector that lies on the same side, with respect to the line of w1 + 2 * w2 = 0, as the green vector would give the correct solution. However, if it lies on the other side as the red vector does, then it would give the wrong answer.
However, suppose the label is 0. Then the case would just be the reverse.
The above case gives the intuition understand and just illustrates the 3 points in the lecture slide. The testing case x determines the plane, and depending on the label, the weight vector must lie on one particular side of the plane to give the correct answer.
Seems like a basic question, but I need to use feature scaling (take each feature value, subtract the mean then divide by the standard deviation) in my implementation of linear regression with gradient descent. After I'm finished, I'd like the weights and regression line rescaled to the original data. I'm only using one feature, plus the y-intercept term. How would I change the weights, after I get them using the scaled data, so that they apply to the original unscaled data?
Suppose your regression is y = W*x + b with x the scaled data, with the original data it is
y = W/std * x0 + b - u/std * W
where u and std are mean value and standard deviation of x0. Yet I don't think you need to transform back the data. Just use the same u and std to scale the new test data.