April 25, 2016

This article will be of interest to you if you want to learn about recommender systems and predicting movie ratings (or book ratings, or product ratings, or any other kind of rating).

Contests like the $1 million Netflix Challenge are an example of what collaborative filtering can be used for.

## Problem Setup

Let’s use the “users rating movies” example for this tutorial. After some Internet searching, we can determine that there are approximately 500, 000 movies in existence. Let’s also suppose that your very popular movie website has 1 billion users (Facebook has 1.6 billion users as of 2015, so this number is plausible).

How many possible user-movie ratings can you have? That is \( 10^9 \times 5 \times 10^5 = 5 \times 10^{14} \). That’s a lot of ratings! Way too much to fit into your RAM, in fact.

But that’s just one problem.

How many movies have you seen in your life? Of those movies, what percentage of them have you rated? The number is miniscule. In fact, *most* users have not rated *most* movies.

This is why recommender systems exist in the first place – so we can recommend you movies that you haven’t seen yet, that we know you’ll like.

So if you were to create a user-movie matrix of movie ratings, most of it would just have missing values.

However, that’s not to say there isn’t a pattern to be found.

Suppose we look at a subset of movie ratings, and we find the following:

Where we’ve used N/A to show that a movie has not yet been rated by a user.

If we used the “cosine distance” ( \( \frac{u^T v}{ |u||v| } \) ) on the vectors created by looking at only the common movies, we could see that Guy A and Guy B have similar tastes. We could then surmise, based on this closeness, that Guy A might rate the Batman movie a “4”, and Guy B might rate Batman Returns a “4”. And since this is a pretty high rating, we might want to recommend these movies to these users.

This is the idea behind collaborative filtering.

## Enter Matrix Factorization

Matrix factorization solves the above problems by reducing the number of free parameters (so the total number of parameters is much smaller than #users times #movies), and by fitting these parameters to the data (ratings) that do exist.

What is matrix factorization?

Think of factorization in general:

15 = 3 x 5 (15 is made up of the factors 3 and 5)

\( x^2 + x = x(x + 1) \)

We can do the same thing with matrices:

$$\left( \begin{matrix}3 & 4 & 5 \\ 6 & 8 & 10 \end{matrix} \right) = \left( \begin{matrix}1 \\ 2 \end{matrix} \right) \left( \begin{matrix}3 & 4 & 5 \end{matrix} \right) $$

In fact, this is exactly what we do in matrix factorization. We “pretend” the big ratings matrix (the one that can’t fit into our RAM) is actually made up of 2 smaller matrices multiplied together.

Remember that to do a valid matrix multiply, the inner dimensions must match. What is the size of this dimension? We call it “K”. It is unknown, but we can choose it via possibly cross-validation so that our model generalizes well.

If we have \( M \) users and \( N \) ratings, then the total number of parameters in our model is \( MK + NK \). If we set \( K = 10 \), the total number of parameters we’d have for the user-movie problem would be \( 10^{10} + 5 \times 10^6 \), which is still approximately \( 10^{10} \), which is a factor of \( 10^4 \) smaller than before.

This is a big improvement!

So now we have:

$$ A \simeq \hat{ A } = UV $$

If you were to picture the matrices themselves, they would look like this:

Because I am lazy and took this image from elsewhere on the Internet, the “d” here is what I am calling “K”. And their “R” is my “A”.

You know that with any machine learning algorithm we have 2 procedures – the fitting procedure and the prediction procedure.

For the fitting procedure, we want every known \( A_{ij} \) to be as close to \( \hat{A}_{ij} = u_i^Tv_j \) as possible. \( u_i \) is the ith row of \( U \). \( v_j \) is the jth column of \( V \).

For the prediction procedure, we won’t have an \( A_{ij} \), but we can use \( \hat{A}_{ij} = u_i^Tv_j \) to tell us what user i *might* rate movie j given the existing patterns.

## The Cost Function

A natural cost function for this problem is the squared error. Think of it as a regression. This is just:

$$ J = \sum_{(i, j) \in \Omega} (A_{ij} – \hat{A}_{ij})^2 $$

Where \( \Omega \) is the set of all pairs \( (i, j) \) where user i has rated movie j.

Later, we will use \( \Omega_i \) to be the set of all j’s (movies) that user i has rated, and we will use \( \Omega_j \) to be the set of all i’s (users) that have rated movie j.

## Coordinate Descent

What do you do when you want to minimize a function? Take the derivative and set it to 0, of course. No need to use anything more complicated if the simple approach is solvable and performs well. It is also possible to use gradient descent on this problem by taking the derivative and then taking small steps in that direction.

You will notice that there are 2 derivatives to take here. The first is \( \partial{J} / \partial{u} \).

The other is \( \partial{J} / \partial{v} \). After calculating the derivatives and solving for \( u \) and \( v \), you get:

$$ u_i = ( \sum_{j \in \Omega_i} v_j v_j^T )^{-1} \sum_{j \in \Omega_i} A_{ij} v_j $$

$$ v_j = ( \sum_{i \in \Omega_j} u_i u_i^T )^{-1} \sum_{i \in \Omega_j} A_{ij} u_i $$

So you take both derivatives. You set both to 0. You solve for the optimal u and v. Now what?

The answer is: coordinate descent.

You first update \( u \) using the current setting of \( v \), then you update \( v \) using the current setting of \( u \). The order doesn’t matter, just that you alternate between the two.

There is a mathematical guarantee that J will improve on each iteration.

This technique is also known as **alternating least squares**. (This makes sense because we’re minimizing the squared error and updating \( u \) and \( v \) in an alternating fashion.)

## Bias Parameters

As with other methods like linear regression and logistic regression, we can add bias parameters to our model to improve accuracy. In this case our model becomes:

$$ \hat{A}_{ij} = u_i^T v_j + b_i + c_j + \mu $$

Where \( \mu \) is the global mean (average of all known ratings).

You can interpret \( b_i \) as the bias of a user. A negative bias means this user just hates movies more than the average person. A positive bias would mean the opposite. Similarly, \( c_j \) is the bias of a movie. A positive bias would mean, “Wow, this movie is good, regardless of who is watching it!” A negative bias would be a movie like* Avatar: The Last Airbender*.

We can re-calculate the optimal settings for each parameter (again by taking the derivatives and setting them to 0) to get:

$$ u_i = ( \sum_{j \in \Omega_i} v_j v_j^T )^{-1} \sum_{j \in \Omega_i} (A_{ij} – b_i – c_j – \mu )v_j $$

$$ v_j = ( \sum_{i \in \Omega_j} u_i u_i^T )^{-1} \sum_{i \in \Omega_j}(A_{ij} – b_i – c_j – \mu )u_i $$

$$ b_i = \frac{1}{| \Omega_i |}\sum_{j \in \Omega_i} A_{ij} – u_i^Tv_j – c_j – \mu $$

$$ c_j= \frac{1}{| \Omega_j |}\sum_{i \in \Omega_j} A_{ij} – u_i^Tv_j – b_i – \mu $$

## Regularization

With the above model, you may encounter what is called the “singular covariance” problem. This is what happens when you can’t invert the matrix that appears in the updates for \( u \) and \( v \).

The solution is again, similar to what you would do in linear regression or logistic regression: Add a squared error term with a weight \( \lambda \) that keeps the parameters small.

In terms of the likelihood, the previous formulation assumes that the difference between \( A_{ij} \) and \( \hat{A}_{ij} \) is normally distributed, while the cost function with regularization is like adding a normally-distributed prior on each parameter centered at 0.

i.e. \( u_i, v_j, b_i, c_j \sim N(0, 1/\lambda) \).

So the cost function becomes:

$$ J = \sum_{(i, j) \in \Omega} (A_{ij} – \hat{A}_{ij})^2 + \lambda(||U||_F + ||V||_F + ||b||^2 + ||c||^2) $$

Where \( ||X||_F \) is the Frobenius norm of \( X \).

For each parameter, setting the derivative with respect to that parameter, setting it to 0 and solving for the optimal value yields:

$$ u_i = ( \sum_{j \in \Omega_i} v_j v_j^T + \lambda{I})^{-1} \sum_{j \in \Omega_i} (A_{ij} – b_i – c_j – \mu )v_j $$

$$ v_j = ( \sum_{i \in \Omega_j} u_i u_i^T + \lambda{I})^{-1} \sum_{i \in \Omega_j}(A_{ij} – b_i – c_j – \mu )u_i $$

$$ b_i = \frac{1}{| \Omega_i | +\lambda}\sum_{j \in \Omega_i} A_{ij} – u_i^Tv_j – c_j – \mu $$

$$ c_j= \frac{1}{| \Omega_j | +\lambda}\sum_{i \in \Omega_j} A_{ij} – u_i^Tv_j – b_i – \mu $$

## Python Code

The simplest way to implement the above formulas would be to just code them directly.

Initialize your parameters as follows:

U = np.random.randn(M, K) / K V = np.random.randn(K, N) / K B = np.zeros(M) C = np.zeros(N)

Next, you want \( \Omega_i \) and \( \Omega_j \) to be easily accessible, so create dictionaries “ratings_by_i” where “i” is the key, and the value is an array of all the (j, r) pairs that user i has rated (r is the rating). Do the same for “ratings_by_j”.

Then, your updates would be as follows:

for t in xrange(T): # update B for i in xrange(M): if i in ratings_by_i: accum = 0 for j, r in ratings_by_i[i]: accum += (r - U[i,:].dot(V[:,j]) - C[j] - mu) B[i] = accum / (len(ratings_by_i[i]) + reg) # update U for i in xrange(M): if i in ratings_by_i: matrix = np.zeros((K, K)) + reg*np.eye(K) vector = np.zeros(K) for j, r in ratings_by_i[i]: matrix += np.outer(V[:,j], V[:,j]) vector += (r - B[i] - C[j] - mu)*V[:,j] U[i,:] = np.linalg.solve(matrix, vector) # update C for j in xrange(N): if j in ratings_by_j: accum = 0 for i, r in ratings_by_j[j]: accum += (r - U[i,:].dot(V[:,j]) - B[i] - mu) C[j] = accum / (len(ratings_by_j[j]) + reg) # update V for j in xrange(N): if j in ratings_by_j: matrix = np.zeros((K, K)) + reg*np.eye(K) vector = np.zeros(K) for i, r in ratings_by_j[j]: matrix += np.outer(U[i,:], U[i,:]) vector += (r - B[i] - C[j] - mu)*U[i,:] V[:,j] = np.linalg.solve(matrix, vector)

And that’s all there is to it!

For more free machine learning and data science tutorials, sign up for my newsletter.