Monday, July 21, 2014

Machine Learning in Clojure - part 2

I am trying to implement the material from the Machine Learning course on Coursera in Clojure.

My last post was about doing linear regression with 1 variable. This post will show that the same process works for multiple variables, and then explain why we represent the problem with matrices.

The only code in this post is calling the functions introduced in the last one. I also use the same examples, so post this will make a lot more sense if you read that one first.

For reference, here is the linear regression function:

(defn linear-regression [x Y a i]
  (let [m (first (cl/size Y))
        X (add-ones x)]
    (loop [Theta (cl/zeros 1 (second (cl/size X))) i i]
      (if (zero? i)
        (let [ans (cl/* X (cl/t Theta))
              diffs (cl/- ans Y)
              dx (cl/* (cl/t diffs) X)
              adjust-x (cl/* dx (/ a m))]
          (recur (cl/- Theta adjust-x)
                   (dec i)))))))

Because the regression function works with matrices, it does not need any changes to run a regression over multiple variables.

Some Examples

In the English Premier League, a team gets 3 points for a win, and 1 point for a draw. Trying to find a relationship between wins and points gets close to the answer.

(->> (get-matrices [:win] :pts)
    (print-results "wins->points"))

** wins->points **
 A 1x2 matrix
 1.24e+01  2.82e+00

When we add a second variable, the number of draws, we get close enough to ascribe the difference to rounding error.

(->> (get-matrices [:win :draw] :pts)
     (print-results "wins+draws->points"))

** wins+draws->points **
 A 1x3 matrix
-2.72e-01  3.01e+00  1.01e+00 

In the last post, I asserted that scoring goals was the key to success in soccer.

(->> (get-matrices [:for] :pts)
     (print-results "for->points"))

** for->points **
 A 1x2 matrix
 2.73e+00  9.81e-01 

If you saw Costa Rica in the World Cup, you know that defense counts for a lot too. Looking at both goals for and against can give a broader picture.

(->> (get-matrices [:for :against] :pts)
     (print-results "for-against->pts"))

** for-against->pts **
 A 1x3 matrix
 3.83e+01  7.66e-01 -4.97e-01

The league tables contain 20 fields of data, and the code works for any number of variables. Will adding more features (variables) make for a better model?

We can expand the model to include whether the goals were scored at home or away.

(->> (get-matrices [:for-h :for-a :against-h :against-a] :pts)
     (print-results "forh-fora-againsth-againsta->pts")) 

** forh-fora-againsth-againsta->pts **
 A 1x5 matrix
 3.81e+01  7.22e-01  8.26e-01 -5.99e-01 -4.17e-01 

The statistical relationship we have found suggests that that goals scored on the road are with .1 points more than those scored at home. The difference in goals allowed is even greater; they cost .6 points at home and only .4 on the road.

Wins and draws are worth the same number of points, no matter where the game takes place, so what is going on?

In many sports there is a “home field advantage”, and this is certainly true in soccer. A team that is strong on the road is probably a really strong team, so the relationship we have found may indeed be accurate.

Adding more features indiscriminately can lead to confusion.

(->> (get-matrices [:for :against :played :gd :for-h :for-a] :pts)
     (map *)
     (print-results "kitchen sink”))

** kitchen sink **
(0.03515239958218979 0.17500425607459014 -0.22696465757628984 1.3357911841232217 0.4019689136508527 0.014497060396707949 0.1605071956778842)

When I printed out this result the first time, the parameter representing the number of games played displayed as a decimal point with no digit before or after. Multiplying each term by 1 got the numbers to appear. Weird.

The :gd stands for “goal difference” it is the difference between the number of goals that a team scores and the number they give up. Because we are also pulling for and against, this is a redundant piece of information. Pulling home and away goals for makes the combined goals-for column redundant as well.

All of the teams in the sample played the same number of games, so that variable should not have influenced the model. Looking at the values, our model says that playing a game is worth 1.3 points, and this is more important than all of the other factors combined. Adding that piece of data removed information.

Let’s look at one more model with redundant data. Lets look at goals for, against and the goal difference, which is just the difference of the two.

(->> (get-matrices [:for :against :gd] :pts)
     (print-results "for-against-gd->pts"))

** for-against-gd->pts **
 A 1x4 matrix
 3.83e+01  3.45e-01 -7.57e-02  4.21e-01 

points = 38.3 + 0.345 * goals-for - 0.0757 * goals-against + 0.421 * goal-difference

The first term, Theta[0] is right around 38. If a team neither scores nor allows any goals during a season, they will draw all of their matches, earning 38 points. I didn’t notice that the leading term was 38 in all of the cases that included both goals for and against until I wrote this model without the exponents.

Is this model better or worse than the one that looks at goals for and goals against, without goal difference. I can’t decide.

Why Matrices?

Each of our training examples have a series of X values, and one corresponding Y value. Our dataset contains 380 examples (20 teams * 19 seasons).
Our process is to make a guess as to the proper value for each parameter to multiply the X values by and compare the results in each case to the Y value. We use the differences between the product of our guesses, and the real life values to improve our guesses.

This could be done with a loop. With m examples and n features we could do something like

for i = 1 to m 
     guess = 0 
     for j = 1 to n 
          guess = guess + X[i, j] * Theta[j] 
     end for j
 difference[i] = guess - Y
end for i

We would need another loop to calculate the new values for Theta.

Matrices have operations defined that replace the above loops. When we multiply the X matrix by the Theta vector, for each row of X, we multiply each element by the corresponding element in Theta, and add the products together to get the first element of the result.

Matrix subtraction requires two matrices that are the same size. The result of subtraction is a new matrix that is the same size, where each element is the difference of the corresponding elements in the original matrices.

Using these two operations, we can replace the loops above with

Guess = X * Theta
Difference = Guess - Y

Clearly the notation is shorter. The other advantage is that there are matrix libraries that are able to do these operations much more efficiently than can be done with loops.

There are two more operations that our needed in the linear regression calculations. One is multiplying matrices by a single number, called a scalar. When multiplying a matrix by a number, multiply each element by that number. [1 2 3] * 3 = [3 6 9].

The other operation we perform is called a transpose. Transposing a matrix turns all of its rows into columns, and its columns into rows. In our examples, the size of X is m by n, and the size of Theta is 1 x n. We don’t have any way to multiply an m by n matrix and a 1 by n matrix, but we can multiply a m by n matrix and an n by 1 matrix. The product will be an m by 1 matrix.

In the regression function there are a couple of transposes to make the dimensions line up. That is the meaning of the cl/t expression. cl is an alias for the Clatrix matrix library.

Even though we replaced a couple of calculations that could have been done in loops with matrix calculations, we are still performing these calculations in a series of iterations. There is a technique for calculating linear regression without the iterative process called Normal Equation.

I am not going to discuss normal equation for two reasons. First, I don’t understand the mathematics. Second the process we use, Gradient Descent, can be used with other types of machine learning techniques, and normal equation cannot.

Saturday, July 5, 2014

Linear Regression in Clojure, Part I

Several months ago I recommended the Machine Learning course from Coursera. At the time, I intended to retake the course and try to implement the solutions to the homework in Clojure. Unfortunately, I got involved in some other things, and wasn’t able to spend time on the class. 

Recently, a new book has come out, Clojure for Machine Learning. I am only a couple of chapters in, but it has already been a good help to me. I do agree with this review that the book is neither a good first Clojure book, or a good first machine learning resource, but it does join the two topics well.

Linear Regression
The place to start with machine learning is Linear Regression with one variable. The goal is to come up with an equation in the familiar form of y = mx + b, where x is the value you know and y is the value you are trying to predict. 

Linear regression is a supervised learning technique. This means that for each of the examples used to create the model the correct answer is known. 

We will use slightly different notation to represent the function we are trying to find. In place of b we will put Theta[0] and in place of m we will put Theta[1]. The reason for this, is that we are going to be using a generalized technique that will work for any number of variables, and the result of our model will be a vector called Theta. 

Even though our technique will work for multiple variables, we will focus on predicting based on a single variable. This is conceptually a little simpler, but more importantly it allows us to plot the input data and our results, so we can see what we are doing.

The Question
A number of years ago I read the book Moneyball, which is about the application of statistics to baseball. One of the claims in the book is that the best predictor for the number of games a baseball team wins in a season is the number of runs they score that season. To improve their results, teams should focus on strategies that maximize runs.

The question I want to answer is whether the same is true in soccer: Are the number of points a team earns in a season correlated with the number of goals that they score. For any that don’t know, a soccer team is awarded 3 points for a win and 1 point for a tie.

The importance of goals is a relevant question for a Manchester United fan. At the end of the 2012-13 season, head coach Sir Alex Ferguson retired after winning his 13th Premier League title. He was replaced by David Moyes. Under Moyes the offense which had been so potent the year before looked clumsy. Also, the team seemed unlucky, giving up goals late in games, turning wins into draws and draws into defeats. The team that finished 1st the year before finished 7th in 2013-14. Was the problem a bad strategy, or bad luck?

The Data
I have downloaded the league tables for the last 19 years of the English Premier League from There have actually been 22 seasons in the Premier League, but in the first 3 seasons each team played 42 games, vs 38 games for the last 19 seasons, and I opted for consistency over quantity.

I actually want to run 3 regressions, first one on a case where I am sure there is a correlation, then on a case where I am sure there is not, and then finally to determine whether a correlation exists between goals and points. 

There should be a high correlation between the number of wins a team has and their number of points. Since every team plays the same number of games, there should be no correlation between the number of games played and a teams position in the standings.

The Process
We will use a technique called gradient descent to find the equation we want to use for our predictions. We will start with an arbitrary value for Theta[0] and Theta[1]; setting both to 0. We will multiply each x value by Theta[1] and add Theta[0], and compare that result to the corresponding value of Y. We will use the differences between Y and the results of Theata * X to calculate new values for Theta, and repeat the process.

One way of measuring the quality of the prediction is with a cost function that measures the mean square error of the predictions. 

1/2m * sum(h(x[i]) - y[i])^2

Where m is the number of test cases we are evaluating, and h(x[i]) is the predicted value for a test case i. We will not use the cost function directly, but its derivative is used in improving our predictions of Theta as follows:

Theta[0] = Theta[0] - alpha * 1/m * sum(h(x[i]) - y([i])
Theta[1] = Theta[1] - alpha * 1/m * sum((h(x[i]) - y([i])  * x[i]) 

We have added one more symbol here. alpha is called the learning rate. The learning rate determines how much we modify Theta each iteration. If alpha is set too high, the process will oscillate between Thetas that are too low and two high and the process will never converge. When alpha is set lower than necessary, extra iterations are necessary to converge.

I need to mention again that this methodology and these equations come directly from Professor Ng’s machine learning course on Coursera that I linked above. He spends over an hour on linear regression with one variable, and if you want more information that is the place to go.

The Code
The actual calculations we are going to do are operations on matrices. When we multiply the matrix X by the matrix Theta, we obtain a matrix of predictions that can be compared element by element with the matrix Y. The same results could be obtained by looping over each test case, but expressing the computations as matrix operations yields simpler equations, shorter code and better performance.

I used the clatrix matrix library for the calculations.

One other thing to note, in the equations above, Theta[0] is treated differently than Theta[1], it is not multiplied by any x terms, either in the predictions or in the adjustments after the predictions. If we add an additional column to our X matrix, an X[0], and make all of the values in this column 1, we then no longer have to make a distinction between Theta[0] and Theta[1].

(defn add-ones "Add an X[0] column of all 1's to use with Theta[0]"
  (let [width (first (cl/size x))
        new-row (vec (repeat width 1))
        new-mat (cl/matrix new-row)]
    (cl/hstack new-mat x)))

(defn linear-regression [x Y a i]
  (let [m (first (cl/size Y))
        X (add-ones x)]
    (loop [Theta (cl/zeros 1 (second (cl/size X))) i i]
      (if (zero? i)
        (let [ans (cl/* X (cl/t Theta))
              diffs (cl/- ans Y)
              dx (cl/* (cl/t diffs) X)
              adjust-x (cl/* dx (/ a m))]
          (recur (cl/- Theta adjust-x)
                   (dec i)))))))

The linear-regression function takes as parameters the X and Y values that we use for training, the learning rate and the number of iterations to perform. We add a column of ones to the passed in X values. We initialize the Theta vector, setting all the values to 0. 

At this point X is a matrix of 380 rows and 2 columns. Theta is a matrix of 1 row and 2 columns. If we take the transpose of Theta (turn the rows into columns, and columns into rows) we get a new matrix, Theta’ which has 2 rows and 1 columns. Multiplying the matrix X with Theta’ yields a matrix of 380x1 containing all of the predictions, and the same size as Y.  

Taking the difference between the calculated answers and our known values yields a 380x1 matrix. We transpose this matrix, making it 1x380, and multiply it by our 380x2 X matrix, yielding a 1x2 matrix. We multiply each element in this matrix by a and divide by m, ending up with a 1x2 matrix which has the amounts we want to subtract from Theta, which is also a 1x2 matrix. All that is left to do is recur with the new values for Theta.

The Results
Since I am going to be performing the same operations on three different data sets, I wrote a couple of helper functions. plot-it uses Incanter to display a scatter plot of the data. reg-epl calls the linear-regression function specifying a learning rate of .0001 and 1000000 iterations. I also have a get-matrices function, which downloads the data and creates the X and Y matrices for the specified fields.

(def wins (get-matrices [:win] :pts))
(plot-it wins)
(def win-theta (reg-epl wins))
(println "Wins-points: " win-theta)

Yields this graph

and these results

Wins-points:   A 1x2 matrix
 1.24e+01  2.82e+00

The relationship between wins and points is obvious in the graph. The equation we developed estimates wins as being worth 2.82 points, instead of the correct 3. This is because it had no way to account for draws, and use a high intercept to get those extra points in there.

A team with 0 wins would be expected to have 12.4 points. A team with 10 wins would have 12.4 + 2.82 * 10 = 40.6 points. A team with 20 wins would have 12.4 + 2.82 * 25 = 
82.9 points.

(def played (get-matrices [:played] :rank))
(plot-it played)
(def played-theta (reg-epl played))
(println "played-rank: " played-theta)
(println "expected finish:" (+ (first played-theta)
                               (* 38 (second played-theta))))

Playing 38 games gives you an equal chance of having a finishing position anywhere between 1 and 20. The graph gives a good illustration of what no-correlation looks like.

If we use the terms in Theta to find the expected finishing position for a team playing 38 games, we find exactly what we expect, 10.5.

played-rank:   A 1x2 matrix
 7.27e-03  2.76e-01

expected finish: 10.499999999999996

Ok, now that we have seen what it looks like when we have a strong correlation, and no correlation, is there a correlation between goals and points?

(def goals (get-matrices [:for] :pts))
(plot-it goals)
(def goal-theta (reg-epl goals))
(def goal-lm (find-lm goals))
(println "goals-points: " goal-theta)
(println "goals-points (incanter): " goal-lm)

Looking at the graph, while not quite as sharp as the goals-points graph, it definitely looks like scoring more goals earns you more points.

To double check my function, I also used Incanter’s linear-model function to also generate an intercept and slope. (And yes, I am relieved that they match).

goals-points:   A 1x2 matrix
 2.73e+00  9.81e-01

goals-points (incanter):  [2.7320304686089685 0.9806635460888629]

We can superimpose the line from our regression formula on the graph, to see how they fit together.

(def goal-plot (scatter-plot (first goals) (second goals)))
(defn plot-fn [x]
  (+ (* (second goal-theta) x) (first goal-theta)))
(def plot-with-regression (add-function goal-plot plot-fn 0 100))

(view plot-with-regression)

The Answer
We can calculate how many points we would expect the team to earn based on their 86 goals in 2012-13 and 64 goals in 2013-14.

(println "86 goals = " (+ (first goal-theta)
                          (* (second goal-theta) 86)))

(println "64 goals = " (+ (first goal-theta)
                          (* (second goal-theta) 64)))

86 goals =  87.07011197597255
64 goals =  65.49481001604704

In the last year under Sir Alex, Manchester United earned 89 points, 2 more than the formula predicts. In their year under David Moyes, they earned 64 points, 1.5 less than the formula predicts. 

Of the 25 point decline in Manchester United’s results, 21.5 points can be attributed to the failure of the offense under Moyes, and 3.5 points can be attributed to bad luck or other factors. 

Manchester United’s attacking style isn’t just fun to watch, it is also the reason they win so much. Hopefully the team’s owners have learned that lesson, and will stick to attack minded managers in the future.

You can find all of the code for the project on github.