了解支持向量回归 (SVR)

Question

I'm working with SVR, and using this resource . Erverything is super clear, with epsilon intensive loss function (from figure). Prediction comes with tube, to cover most training sample, and generalize bounds, using support vectors.

Then we have this explanation. This can be described by introducing (non-negative) slack variables , to measure the deviation of training samples outside -insensitive zone. I understand this error, outside tube, but don't know, how we can use this in optimization. Could somebody explain this?

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In local source. I'm trying to achieve very simple optimization solution, without libraries. This what I have for loss function.

import numpy as np

# Kernel func, linear by default
def hypothesis(x, weight, k=None):
    k = k if k else lambda z : z
    k_x = np.vectorize(k)(x)
    return np.dot(k_x, np.transpose(weight))

.......

import math

def boundary_loss(x, y, weight, epsilon):
    prediction = hypothesis(x, weight)

    scatter = np.absolute(
        np.transpose(y) - prediction)
    bound = lambda z: z \
        if z >= epsilon else 0

    return np.sum(np.vectorize(bound)(scatter))

Answer 1

First, let's look at the objective function. The first term, 1/2 * w^2 (wish this site had LaTeX support but this will suffice) correlates with the margin of the SVM. The article you linked doesn't, in my opinion, explain this very well and calls this term describing "the model's complexity", but perhaps this is not the best way of explaining it. Minimizing this term maximizes the margin (while still representing the data well), which is the predominant goal of using SVM's doing regression.

Warning, Math Heavy Explanation: The reason this is the case is that when maximizing the margin, you want to find the "farthest" non-outlier points right on the margin and minimize its distance. Let this farthest point be x_n . We want to find its Euclidean distance d from the plane f(w, x) = 0 , which I will rewrite as w^T * x + b = 0 (where w^T is just the transpose of the weights matrix so that we can multiply the two). To find the distance, let us first normalize the plane such that |w^T * x_n + b| = epsilon |w^T * x_n + b| = epsilon , which we can do WLOG as w is still able to form all possible planes of the form w^T * x + b= 0 . Then, let's note that w is perpendicular to the plane. This is obvious if you have dealt a lot with planes (particularly in vector calculus), but can be proven by choosing two points on the plane x_1 and x_2 , then noticing that w^T * x_1 + b = 0 , and w^T * x_2 + b = 0 . Subtracting the two equations we get w^T(x_1 - x_2) = 0 . Since x_1 - x_2 is just any vector strictly on the plane, and its dot product with w is 0, then we know that w is perpendicular to the plane. Finally, to actually calculate the distance between x_n and the plane, we take the vector formed by x_n' and some point on the plane x' (The vectors would then be x_n - x' , and projecting it onto the vector w . Doing this, we get d = |w * (x_n - x') / |w|| , which we can rewrite as d = (1 / |w|) * | w^T * x_n - w^T x'| , and then add and subtract b to the inside to get d = (1 / |w|) * | w^T * x_n + b - w^T * x' - b| . Notice that w^T * x_n + b is epsilon (from our normalization above), and that w^T * x' + b is 0, as this is just a point on our plane. Thus, d = epsilon / |w| . Notice that maximizing this distance subject to our constraint of finding the x_n and having |w^T * x_n + b| = epsilon is a difficult optimization problem. What we can do is restructure this optimization problem as minimizing 1/2 * w^T * w subject to the first two constraints in the picture you attached, that is, |y_i - f(x_i, w)| <= epsilon . You may think that I have forgotten the slack variables, and this is t rue, but when just focusing on this term and ignoring the second term, we ignore the slack variables for now, I will bring them back later. The reason these two optimizations are equivalent is not obvious, but the underlying reason lies in discrimination boundaries, which you are free to read more about (it's a lot more math that frankly I don't think this answer needs more of). Then, note that minimizing 1/2 * w^T * w is the same as minimizing 1/2 * |w|^2 , which is the desired result we were hoping for. End of the Heavy Math

Now, notice that we want to make the margin big, but not so big that includes noisy outliers like the one in the picture you provided.

Thus, we introduce a second term. To motivate the margin down to a reasonable size the slack variables are introduced, (I will call them p and p* because I don't want to type out "psi" every time). These slack variables will ignore everything in the margin, ie those are the points that do not harm the objective and the ones that are "correct" in terms of their regression status. However, the points outside the margin are outliers, they do not reflect well on the regression, so we penalize them simply for existing. The slack error function that is given there is relatively easy to understand, it just adds up the slack error of every point ( p_i + p*_i ) for i = 1,...,N , and then multiplies by a modulating constant C which determines the relative importance of the two terms. A low value of C means that we are okay with having outliers, so the margin will be thinned and more outliers will be produced. A high value of C indicates that we care a lot about not having slack, so the margin will be made bigger to accommodate these outliers at the expense of representing the overall data less well.

A few things to note about p and p* . First, note that they are both always >= 0 . The constraint in your picture shows this, but it also intuitively makes sense as slack should always add to the error, so it is positive. Second, notice that if p > 0 , then p* = 0 and vice versa as an outlier can only be on one side of the margin. Last, all points inside the margin will have p and p* be 0, since they are fine where they are and thus do not contribute to the loss.

Notice that with the introduction of the slack variables, if you have any outliers then you won't want the condition from the first term, that is, |w^T * x_n + b| = epsilon |w^T * x_n + b| = epsilon as the x_n would be this outlier, and your whole model would be screwed up. What we allow for, then, is to change the constraint to be |w^T * x_n + b| = epsilon + (p + p*) |w^T * x_n + b| = epsilon + (p + p*) . When translated to the new optimization's constraint, we get the full constraint from the picture you attached, that is, |y_i - f(x_i, w)| <= epsilon + p + p* |y_i - f(x_i, w)| <= epsilon + p + p* . (I combined the two equations into one here, but you could rewrite them as the picture is and that would be the same thing).

Hopefully after covering all this up, the motivation for the objective function and the corresponding slack variables makes sense to you.

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If I understand the question correctly, you also want code to calculate this objective/loss function, which I think isn't too bad. I have not tested this (yet), but I think this should be what you want.

# Function for calculating the error/loss for a SVM. I assume that:
#  - 'x' is 2d array representing the vectors of the data points
#  - 'y' is an array representing the values each vector actually gives
#  - 'weights' is an array of weights that we tune for the regression
#  - 'epsilon' is a scalar representing the breadth of our margin.
def optimization_objective(x, y, weights, epsilon):
    # Calculates first term of objective (note that norm^2 = dot product)
    margin_term = np.dot(weight, weight) / 2

    # Now calculate second term of objective. First get the sum of slacks.
    slack_sum = 0
    for i in range(len(x)): # For each observation
        # First find the absolute distance between expected and observed.
        diff = abs(hypothesis(x[i]) - y[i])
        # Now subtract epsilon
        diff -= epsilon
        # If diff is still more than 0, then it is an 'outlier' and will have slack.
        slack = max(0, diff)
        # Add it to the slack sum
        slack_sum += slack

    # Now we have the slack_sum, so then multiply by C (I picked this as 1 aribtrarily)
    C = 1
    slack_term = C * slack_sum

    # Now, simply return the sum of the two terms, and we are done.
    return margin_term + slack_term

I got this function working on my computer with small data, and you may have to change it a little to work with your data if, for example, the arrays are structured differently, but the idea is there. Also, I am not the most proficient with python, so this may not be the most efficient implementation, but my intent was to make it understandable.

Now, note that this just calculates the error/loss (whatever you want to call it). To actually minimize it requires going into Lagrangians and intense quadratic programming which is a much more daunting task. There are libraries available for doing this but if you want to do this library free as you are doing with this, I wish you good luck because doing that is not a walk in the park.

Finally, I would like to note that most of this information I got from notes I took in my ML class I took last year, and the professor (Dr. Abu-Mostafa) was a great help to have me learn the material. The lectures for this class are online (by the same prof), and the pertinent ones for this topic are here and here (although in my very biased opinion you should watch all the lectures, they were a great help). Leave a comment/question if you need anything cleared up or if you think I made a mistake somewhere. If you still don't understand, I can try to edit my answer to make more sense. Hope this helps!