# Understanding matrix factorization for recommendation (part 3) - SVD for recommendation

Foreword: this is the third part of a 4 parts series. Here are parts 1, 2 and 4. This series is an extended version of a talk I gave at PyParis 17.

# SVD for recommendation

Now that we have a good understanding of what SVD is and how it models the ratings, we can get to the heart of the matter: using SVD for recommendation purpose. Or rather, using SVD for predicting missing ratings. Let’s go back to our actual matrix $R$, which is sparse:

$R= \begin{pmatrix} 1 & \color{#e74c3c}{?} & 2 & \color{#e74c3c}{?} & \color{#e74c3c}{?}\\ \color{#e74c3c}{?} & \color{#e74c3c}{?} & \color{#e74c3c}{?} & \color{#e74c3c}{?} & 4\\ 2 & \color{#e74c3c}{?} & 4 & 5 & \color{#e74c3c}{?}\\ \color{#e74c3c}{?} & \color{#e74c3c}{?} & 3 & \color{#e74c3c}{?} & \color{#e74c3c}{?}\\ \color{#e74c3c}{?} & 1 & \color{#e74c3c}{?} & 3 & \color{#e74c3c}{?}\\ 5 & \color{#e74c3c}{?} & \color{#e74c3c}{?} & \color{#e74c3c}{?} & 2\\ \end{pmatrix}$

By sparse, we here mean “with missing entries”, not “containing a lot of zeros”. We don’t replace the missing values with zeros. Remember that our goal is to predict the $\color{#ff2c2d}{?}$.

#### Ooooops

Now guess what: The SVD of $R$ is not defined. It does not exist. Yup, it is impossible to compute, it is not defined, it does not exist :). But don’t worry, all your efforts to read this article that far were not futile.

If $R$ was dense, we could compute $M$ and $U$ easily: the columns of $M$ are the eigenvectors of $RR^T$, and the columns of $U$ are the eigenvectors of $R^TR$. The associated eigenvalues make up the diagonal matrix $\Sigma$. There are very efficient algorithms that can do that.

But as $R$ is sparse, the matrices $RR^T$ and $R^TR$ do not exist, so their eigenvectors do not exist either and we can’t factorize $R$ as the product $M\Sigma U^T$. However, there is a way around. A first option that was used for some time is to fill the missing entries of $R$ with some simple heuristic, e.g. the mean of the columns (or the rows). Once the matrix is dense, we can compute its SVD using the traditional algorithms. This works OK, but results are usually highly biased. We will rather use another way, based on a minimization problem.

#### The alternative

Computing the eigenvectors of $RR^T$ and $R^TR$ is not the only way of computing the SVD of a dense matrix $R$. We can actually find the matrices $M$ and $U$ if we can find all the vectors $p_u$ and $q_i$ such that (the $p_u$ make up the rows of $M$ and the $q_i$ make up the columns of $U^T$):

• $r_{ui} = p_u \cdot q_i$ for all $u$ and $i$
• All the vectors $p_u$ are mutually orthogonal, as well as the vectors $q_i$.

Finding such vectors $p_u$ and $q_i$ for all users and items can be done by solving the following optimization problem (while respecting the orthogonality constraints):

$\min_{p_u, q_i\\p_u \perp p_v\\ q_i \perp q_j}\sum_{r_{ui} \in R} (r_{ui} - p_u \cdot q_i)^2$

The above one reads as find vectors $p_u$ and $q_i$ that makes the sum minimal. In other words: we’re trying to match as well as possible the values $r_{ui}$ with what they are supposed to be: $p_u \cdot q_i$.

I’m abusing notation here and referring to $R$ as both a matrix and a set of ratings. Once we know the values of the vectors $p_u$ and $q_i$ that make this sum minimal (and here the minimum is zero), we can reconstruct $M$ and $U$ and we get our SVD.

So what do we do when $R$ is sparse, i.e. when some ratings are missing from the matrix? Simon Funk’s answer is that we should just not give a crap. We still solve the same optimization problem:

$\min_{p_u, q_i}\sum_{r_{ui} \in R} (r_{ui} - p_u \cdot q_i)^2.$

The only difference is that this time, some ratings are missing, i.e. $R$ is incomplete. Note that we are not treating the missing entries as zeros: we are purely and simply ignoring them. Also, we will forget about the orthogonality constraints, because even if they are useful for interpretation purpose, constraining the vectors usually does not help us to obtain more accurate predictions.

Thanks to his solution, Simon Funk ended up in the top 3 of the Netflix Prize for some time. His algorithm was heavily used, studied and improved by the other teams.

#### The algorithm

This optimization problem is not convex. That is, it will be very difficult to find the values of the vectors $p_u$ and $q_i$ that make this sum minimal (and the optimal solution may not even be unique). However, there are tons of techniques that can find approximate solutions. We will here use SGD (Stochastic Gradient Descent).

Gradient descent is a very classical technique for finding the (sometimes local) minimum of a function. If you have ever heard of back-propagation for training neural networks, well backprop is just a technique to compute gradients, which are later used for gradient descent. SGD is one of the zillions variants of gradient descent. We won’t detail too much how SGD works (there are tons of good resources on the web) but the general pitch is as follows.

When you have a function $f$ with a parameter $\theta$ that looks like this:

$f(\theta) = \sum_k f_k(\theta),$

the SGD procedue minimizes $f$ (i.e. finds the value of $\theta$ such that $f(\theta)$ is as small as possible), with the following steps:

1. Randomly initialize $\theta$
2. for a given number of times, repeat:
• for all $k$, repeat:
• compute $\frac{\partial f_k}{\partial \theta}$
• update $\theta$ with the following rule: $$\theta \leftarrow \theta - \alpha \frac{\partial f_k}{\partial \theta},$$, where $\alpha$ is the learning rate (a small value).

In our case, the parameter $\theta$ corresponds to all the vectors $p_u$ and all the vectors $q_i$ (which we will denote by $$(p_*, q_*)$$), and the function $f$ we want to minimize is

$f(p_*, q_*) = \sum_{r_{ui} \in R} (r_{ui} - p_u \cdot q_i)^2 =\sum_{r_{ui} \in R} f_{ui}(p_u, q_i),$

where $f_{ui}$ is defined by $f_{ui}(p_u, q_i) = (r_{ui} - p_u \cdot q_i)^2$.

So, in order to apply SGD, what we are looking for is the value of the derivative of $f_{ui}$ with respect to any $p_u$ and any $q_i$.

• The derivative of $f_{ui}$ with respect to a given vector $p_u$ is given by:

$\frac{\partial f_{ui}}{\partial p_u} = \frac{\partial}{\partial p_u} (r_{ui} - p_u \cdot q_i)^2 = - 2 q_i (r_{ui} - p_u \cdot q_i)$
• Symmetrically, the derivative of $f_{ui}$ with respect to a given vector $q_i$ is given by:

$\frac{\partial f_{ui}}{\partial q_i} = \frac{\partial}{\partial q_i} (r_{ui} - p_u \cdot q_i)^2 = - 2 p_u (r_{ui} - p_u \cdot q_i)$

Don’t be scared, honestly this is highschool-level calculus.

The SGD procedure then becomes:

1. Randomly initialize all vectors $p_u$ and $q_i$
2. for a given number of times, repeat:
• for all known ratings $r_{ui}$, repeat:
• compute $\frac{\partial f_{ui}}{\partial p_u}$ and $\frac{\partial f_{ui}}{\partial q_i}$ (we just did)
• update $p_u$ and $q_i$ with the following rule: $$p_u \leftarrow p_u + \alpha \cdot q_i (r_{ui} - p_u \cdot q_i)$$, and $$q_i \leftarrow q_i + \alpha \cdot p_u (r_{ui} - p_u \cdot q_i)$$. We avoided the multiplicative constant $2$ and merged it into the learning rate $\alpha$.

Notice how in this algorithm, the different factors in $p_u$ (and $q_i$) are all updated at the same time. Funk’s original algorithm was a bit different: he actually trained the first factor, then the second, then the third, etc. This gave his algorithm a more SVDesque flavor. A nice discussion about this can be found in Aggarwal’s Textbook on recommender systems.

Once all the vectors $p_u$ and $q_i$ have been computed, we can estimate all the ratings we want using the formula:

$\hat{r}_{ui} = p_u \cdot q_i.$

There’s a hat on $\hat{r}_{ui}$ to indicate that it’s an estimation, not its real value.

#### Dimensionality reduction

Now before we can jump to the Python implementation of this algorithm, there’s one thing we need to decide: what should be the size of the vectors $p_u$ and $q_i$? One thing we know for sure though, is that it has to be the same for all vectors $p_u$ and $q_i$, else we could not compute dot products.

To answer this question, let’s go back briefly to PCA and to our creepy guys:          As you remember, these creepy guys can reconstruct all of the original faces:          \begin{align*} \text{Face 1}~=~&\alpha_1 \cdot \color{#048BA8}{\text{Creepy guy #1}}\\ +~ &\alpha_2 \cdot \color{#048BA8}{\text{Creepy guy #2}}\\ +~ &\cdots\\ +~ &\alpha_{400} \cdot \color{#048BA8}{\text{Creepy guy #400}} \end{align*} ~~~ \begin{align*} \text{Face 2}~=~&\beta_1 \cdot \color{#048BA8}{\text{Creepy guy #1}}\\ +~ &\beta_2 \cdot \color{#048BA8}{\text{Creepy guy #2}}\\ +~ &\cdots\\ +~ &\beta_{400} \cdot \color{#048BA8}{\text{Creepy guy #400}} \end{align*} $$~~~\cdots$$

But in fact, we don’t need to use all the creepy guys to get a good approximation of each original face. I actually lied to you: the gifs you see above only use the first 200 creepy guys (instead of 400)! And you couldn’t see the difference, could you? To further illustrate this point, here is the reconstruction of the first original face, using from 1 to 400 creepy guys, each time adding 40 creepy guys into the reconstruction.           The last picture is the perfect reconstruction, and as you can see even using only 80 creepy guys (third picture) is enough to recognize the original guy.

As a side note, you may wonder why the first picture does not look like the first creepy guy, and why the creepy guys have a much higher contrast than the original pictures: it’s because PCA first subtracts the mean of all images. The first image you are seeing corresponds in fact to the contribution of the first creepy guy plus the average face. If that doesn’t make sense to you, don’t worry. We don’t care when we’re doing recommendation.

So the take-away message here is this: you don’t need all the creepy/typical guys to have a good approximation of your initial matrix. The same goes for SVD and the recommendation problem: you don’t need all the typical movies or all the typical users to get a good approximation.

This means that when we are representing our users and items like this (SVD does it for us):

$\begin{array}{ll} \text{Alice} & = 10\% \color{#048BA8}{\text{ Action fan}} &+ 10\% \color{#048BA8}{\text{ Comedy fan}} &+ 50\% \color{#048BA8}{\text{ Romance fan}} &+\cdots\\ \text{Bob} &= 50\% \color{#048BA8}{\text{ Action fan}}& + 30\% \color{#048BA8}{\text{ Comedy fan}} &+ 10\% \color{#048BA8}{\text{ Romance fan}} &+\cdots\\ \text{Titanic} &= 20\% \color{#048BA8}{\text{ Action}}& + 00\% \color{#048BA8}{\text{ Comedy}} &+ 70\% \color{#048BA8}{\text{ Romance}} &+\cdots\\ \text{Toy Story} &= 30\% \color{#048BA8}{\text{ Action }} &+ 60\% \color{#048BA8}{\text{ Comedy}}&+ 00\% \color{#048BA8}{\text{ Romance}} &+\cdots\\ \end{array}$

we can just use a small number of typical movies/users and get a good solution. In our case, we will restrict the size of the $p_u$ and the $q_i$ to 10. That is, we will only consider 10 latent factors.

You have the right to be skeptical about this, but we have in fact good theoretical guaranties about this approximation. A fantastic result about SVD (and PCA) is that when we’re using only $k$ factors, we obtain the best low-rank approximation (understand low number of factors) of the original matrix. Details are a bit technical and outside the scope of this article (although very interesting), so I refer you to this Stanford course notes (Fact 4.2). (Quick note: in section 5 of the course notes, the author proposes a way to recover missing entries from SVD. This heuristic technique is the one that we first suggested above, but it’s not what works best. What works best in recommendation is to optimize on the known ratings, as we are doing with SGD ;)).

We now have all it takes to write a matrix factorization algorithm in the next (and last!) part of this series. We’ll do that in Python (obviously ;)), using the Surprise library. As you’ll see, the algortihm is surprisingly simple to write, yet fairly efficient.