机器学习第三周

To attempt classification, one method is to use linear regression and map all predictions greater than 0.5 as a 1 and all less than 0.5 as a 0. However, this method doesn’t work well because classification is not actually a linear function.

The classification problem is just like the regression problem, except that the values we now want to predict take on only a small number of discrete values. For now, we will focus on the binary classification problem in which y can take on only two values, 0 and 1. (Most of what we say here will also generalize to the multiple-class case.) For instance, if we are trying to build a spam classifier for email, then x^(i) may be some features of a piece of email, and y may be 1 if it is a piece of spam mail, and 0 otherwise. Hence, y∈{0,1}. 0 is also called the negative class, and 1 the positive class, and they are sometimes also denoted by the symbols - and +. Given x^(i), the corresponding y^(i) is also called the label for the training example.

We could approach the classification problem ignoring the fact that y is discrete-valued, and use our old linear regression algorithm to try to predict y given x. However, it is easy to construct examples where this method performs very poorly. Intuitively, it also doesn’t make sense for h_θ(x) to take values larger than 1 or smaller than 0 when we know that y∈{0, 1}. To fix this, let’s change the form for our hypotheses h_θ(x) to satisfy 0 ≤ h_θ(x) ≤ 10. This is accomplished by plugging θ^Tx into the Logistic Function.

Our new form uses the “Sigmoid Function,” also called the “Logistic Function”:

The following image shows us what the sigmoid function looks like:

The function g(z), shown here, maps any real number to the (0, 1) interval, making it useful for transforming an arbitrary-valued function into a function better suited for classification.

h_θ(x) will give us the probability that our output is 1. For example, h_θ(x)=0.7 gives us a probability of 70% that our output is 1. Our probability that our prediction is 0 is just the complement of our probability that it is 1 (e.g. if probability that it is 1 is 70%, then the probability that it is 0 is 30%).

In order to get our discrete 0 or 1 classification, we can translate the output of the hypothesis function as follows:

The way our logistic function g behaves is that when its input is greater than or equal to zero, its output is greater than or equal to 0.5:

Remember.

So if our input to g is θ^TX, then that means:

From these statements we can now say:

The decision boundary is the line that separates the area where y = 0 and where y = 1. It is created by our hypothesis function.

Example:

In this case, our decision boundary is a straight vertical line placed on the graph where x1 = 5, and everything to the left of that denotes y = 1, while everything to the right denotes y = 0.

Again, the input to the sigmoid function g(z) (e.g. θ^TX doesn’t need to be linear, and could be a function that describes a circle (e.g. z = θ_0 + θ_1x_1^2 + θ_2x_2^2) or any shape to fit our data.

We cannot use the same cost function that we use for linear regression because the Logistic Function will cause the output to be wavy, causing many local optima. In other words, it will not be a convex function.

Instead, our cost function for logistic regression looks like:

Similarly, when y = 0, we get the following plot for J(θ) vs h_θ(x):

If our correct answer ‘y’ is 0, then the cost function will be 0 if our hypothesis function also outputs 0. If our hypothesis approaches 1, then the cost function will approach infinity.

If our correct answer ‘y’ is 1, then the cost function will be 0 if our hypothesis function outputs 1. If our hypothesis approaches 0, then the cost function will approach infinity.

Note that writing the cost function in this way guarantees that J(θ) is convex for logistic regression.

We can compress our cost function’s two conditional cases into one case:

Notice that when y is equal to 1, then the second term (1-y)\log(1-h_\theta(x)) will be zero and will not affect the result. If y is equal to 0, then the first term -y \log(h_\theta(x)) will be zero and will not affect the result.

We can fully write out our entire cost function as follows:

A vectorized implementation is:

Gradient Descent

Remember that the general form of gradient descent is:

We can work out the derivative part using calculus to get:

Notice that this algorithm is identical to the one we used in linear regression. We still have to simultaneously update all values in \theta.

A vectorized implementation is:

Conjugate gradient, BFGS, and L-BFGS are more sophisticated, faster ways to optimize θ that can be used instead of gradient descent. We suggest that you should not write these more sophisticated algorithms yourself (unless you are an expert in numerical computing) but use the libraries instead, as they’re already tested and highly optimized. Octave provides them.

We first need to provide a function that evaluates the following two functions for a given input value θ:

We can write a single function that returns both of these:

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function [jVal, gradient] = costFunction(theta)
  jVal = [...code to compute J(theta)...];
  gradient = [...code to compute derivative of J(theta)...];
end

Then we can use octave’s fminunc() optimization algorithm along with the optimset() function that creates an object containing the options we want to send to fminunc(). (Note: the value for MaxIter should be an integer, not a character string - errata in the video at 7:30)

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options = optimset('GradObj', 'on', 'MaxIter', 100);
initialTheta = zeros(2,1);
   [optTheta, functionVal, exitFlag] = fminunc(@costFunction, initialTheta, options);

We give to the function fminunc() our cost function, our initial vector of theta values, and the options object that we created beforehand.

Now we will approach the classification of data when we have more than two categories. Instead of y = {0,1} we will expand our definition so that y = {0,1…n}.

Since y = {0,1…n}, we divide our problem into n+1 (+1 because the index starts at 0) binary classification problems; in each one, we predict the probability that ‘y’ is a member of one of our classes.

We are basically choosing one class and then lumping all the others into a single second class. We do this repeatedly, applying binary logistic regression to each case, and then use the hypothesis that returned the highest value as our prediction.

The following image shows how one could classify 3 classes:

To summarize:

Train a logistic regression classifier h_θ(x) for each class to predict the probability that y = i.

To make a prediction on a new x, pick the class that maximizes h_θ(x)

1. Suppose that you have trained a logistic regression classifier, and it outputs on a new example x a prediction h_θ(x) = 0.2. This means (check all that apply):

• a. Our estimate for P(y=1|x;\theta) is 0.2.
• b. Our estimate for P(y=0|x;\theta) is 0.2.
• a. Our estimate for P(y=1|x;\theta) is 0.8.
• a. Our estimate for P(y=0|x;\theta) is 0.8.

2. Suppose you have the following training set, and fit a logistic regression classifier h_\theta(x) = g(\theta_0 + \theta_1x_1 + \theta_2 x_2).

Which of the following are true? Check all that apply.

• a. J(\theta) will be a convex function, so gradient descent should converge to the global minimum.
• b. Adding polynomial features (e.g., instead using h_\theta(x) = g(\theta_0 + \theta_1x_1 + \theta_2 x_2 + \theta_3 x_1^2 + \theta_4 x_1 x_2 + \theta_5 x_2^2) ) could increase how well we can fit the training data.
• c. The positive and negative examples cannot be separated using a straight line. So, gradient descent will fail to converge.
• d. Because the positive and negative examples cannot be separated using a straight line, linear regression will perform as well as logistic regression on this data.

3. For logistic regression, the gradient is given by \frac{\partial}{\partial \theta_j} J(\theta) =\frac{1}{m}\sum_{i=1}^m{ (h_\theta(x^{(i)}) - y^{(i)}) x_j^{(i)}}. Which of these is a correct gradient descent update for logistic regression with a learning rate of α? Check all that apply.

• a. \theta := \theta - \alpha \frac{1}{m} \sum_{i=1}^m{ \left(\theta^T x - y^{(i)}\right) x^{(i)}}.
• b. \theta_j := \theta_j - \alpha \frac{1}{m} \sum_{i=1}^m{ (h_\theta(x^{(i)}) - y^{(i)}) x_j^{(i)}} (simultaneously update for all j).
• c. \theta_j := \theta_j - \alpha \frac{1}{m} \sum_{i=1}^m{ (h_\theta(x^{(i)}) - y^{(i)}) x^{(i)}} (simultaneously update for all j).
• d. \theta_j := \theta_j - \alpha \frac{1}{m} \sum_{i=1}^m{ \left(\frac{1}{1 + e^{-\theta^T x^{(i)}}} - y^{(i)}\right) x_j^{(i)}} (simultaneously update for all j).

4. Which of the following statements are true? Check all that apply.

• a. Since we train one classifier when there are two classes, we train two classifiers when there are three classes (and we do one-vs-all classification).
• b. The one-vs-all technique allows you to use logistic regression for problems in which each y^{(i)} comes from a fixed, discrete set of values.
• c. The cost function J(\theta) for logistic regression trained with m \geq 1 examples is always greater than or equal to zero.
• d. For logistic regression, sometimes gradient descent will converge to a local minimum (and fail to find the global minimum). This is the reason we prefer more advanced optimization algorithms such as fminunc (conjugate gradient/BFGS/L-BFGS/etc).

5. Suppose you train a logistic classifier h_\theta(x) = g(\theta_0 + \theta_1x_1 + \theta_2 x_2). Suppose \theta_0 = - 6, \theta_1 = 1, \theta_2 = 0. Which of the following figures represents the decision boundary found by your classifier?

• a. Figure:
• b. Figure:
• c. Figure:
• d. Figure:

1. a, d
2. a, b 样本数量少所以肯定只有一个最优解，使用多项式特征能够更好的拟合数据集。
3. a, b, d b 是标准算式，a 是矩阵算式。
4. b, c 对于逻辑分类 y 的种类大于 2 需要的分类器数量为 y。使用更高级的算法并不能避免出现局部最优解的问题。
5. b 如果是 \theta_0 = - 6, \theta_1 = 0, \theta_2 = 1 需要选 d。

1. You are training a classification model with logistic regression. Which of the following statements are true? Check all that apply.

• a. Adding a new feature to the model always results in equal or better performance on the training set.
• b. Introducing regularization to the model always results in equal or better performance on the training set.
• c. Introducing regularization to the model always results in equal or better performance on examples not in the training set.
• d. Adding many new features to the model helps prevent overfitting on the training set.

2. Suppose you ran logistic regression twice, once with \lambda = 0, and once with \lambda = 1 .
   One of the times, you got parameters \theta = \begin{bmatrix} 81.47 \newline 12.69 \end{bmatrix}, and the other time you got \theta = \begin{bmatrix} 13.01 \newline 0.91 \end{bmatrix}.
   However, you forgot which value of \lambda corresponds to which value of \theta. Which one do you think corresponds to \lambda = 1?

• a. \theta = \begin{bmatrix} 81.47 \newline 12.69 \end{bmatrix}
• b. \theta = \begin{bmatrix} 13.01 \newline 0.91 \end{bmatrix}

3. Which of the following statements about regularization are true? Check all that apply.

• a. Because logistic regression outputs values 0 \leq h_\theta(x) \leq 1, its range of output values can only be “shrunk” slightly by regularization anyway, so regularization is generally not helpful for it.
• b. Using too large a value of \lambda can cause your hypothesis to overfit the data; this can be avoided by reducing \lambda.
• c. Using a very large value of \lambda cannot hurt the performance of your hypothesis; the only reason we do not set \lambda to be too large is to avoid numerical problems.
• d. Consider a classification problem. Adding regularization may cause your classifier to incorrectly classify some training examples (which it had correctly classified when not using regularization, i.e. when \lambda = 0).

4. In which one of the following figures do you think the hypothesis has overfit the training set?

• a. Figure:
• b. Figure:
• c. Figure:
• d. Figure:

5. In which one of the following figures do you think the hypothesis has underfit the training set?

• a. Figure:
• b. Figure:
• c. Figure:
• d. Figure: