April Sagan

The How and Why of Matrix Completion

What is Matrix Completion?

Simply put, the goal of matrix completion is fill in missing entries of a matrix (or dataset) given the fact that the matrix is low rank, or low dimensional. Essentially, it’s like a game of Sudoku with a different set of rules. Lets say I have a matrix that I know is supposed to be rank 2. That means that every column can be written as a linear combination (weighted sum) of two vectors. Lets look at an example of what this puzzle might look like.

\[\begin{bmatrix} 1 & 1 &2 & 2\\2&1&3&-\\1&2&-&1 \end{bmatrix}\]

The first two columns are completly filled in, so we can use those to figure out the rest of the columns. Based on the few entries in the third column that are given, we can see that the third column should probably be the first column plus the second column. Likewise, the fourth column is two times the first column minus the second column.

\[\begin{bmatrix} 1 & 1 &2 & 2\\2&1&3&5\\1&2&3&1 \end{bmatrix}\]

That was a particularly easy example since we knew the first two columns completely.

To see why we should care about this, here’s a claim that shouldn’t be too hard to believe: Datasets are inherently low rank . In the example we just did, the columns could be movies, the rows could be people, and the numbers could be how each person rated each movie. Obviously, this is going to be sparse since not everyone has seen every movie. That’s where matrix completions comes in. When we filled in the missing entries, we gave our guess as to what movies people are going to enjoy. After explaining an algorithm to do matrix completion, we’re going to try this for a data set with a million ratings people gave movies and see how well we recommend movies to people.

How do we do it?

There’s two paradigms for matrix completion. One is to minimize the rank of a matrix that fits our measurements, and the other is to find a matrix of a given rank that matches up with our known entries. In this blog post, I’ll just be talking about the second.

Before we explain the algorithm, we need to introduce a little more notation. We are going to let \(\Omega\) be the set of indices where we know the entry. For example, if we have the partially observed matrix

\[\begin{matrix}\color{blue}1\\\color{blue}2\\\color{blue}3\end{matrix}\begin{bmatrix} & 1 & \\ & & 1\\1 & & \end{bmatrix}\] \[\begin{matrix} &\color{red}1 & \color{red}2 & \color{red}3 \end{matrix}\]

then, \(\Omega\) would be \(\{ (\color{blue} 1, \color{red}2), (\color{blue}2 , \color{red}3),(\color{blue} 3, \color{red}1)\}\) We can now pose the problem of finding a matrix with rank \(r\) that best fits the entries we’ve observe as an optimization problem.

\[\begin{aligned}&\underset{X}{\text{minimize}}& \sum_{(i,j)\text{ in }\Omega} (X_{ij}-M_{ij})^2 \\& \text{such that} & \text{rank}(X)=r\end{aligned}\]

The first line specifies objective function (the function we want to minimize), which is the sum of the square of the difference between \(X_{ij}\) and \(M_{ij}\) for every \((i,j)\) that we have a measurement for. The second line is our constraint, which says that the matrix has to be rank \(r\) .

While minimizing a function like that isn’t too hard, forcing the matrix to be rank \(r\) can be tricky. One property of a low rank matrix that has \(m\) rows and \(n\) columns is that we can factor it into two smaller matrices like such:


where \(U\) is \(n\) by \(r\) and \(V\) is \(r\) by \(m\) . So now, if we can find matrices \(U\) and \(V\) such that the matrix \(UV\) fits our data, we know its going to be rank \(r\) and that will be the solution to our problem.

If \(u_i\) is the \(i^{th}\) column of \(U\) and \(v_j\) is the \(j^{th}\) column of \(V\) , then \(X_{ij}\) is the inner product of \(u_i\) and \(v_j\) , \(X_{ij}= \langle u_i, v_i \rangle\) . We can rewrite the optimization problem we want to solve as

\[\begin{aligned} \underset{U, V}{\text{minimize}}& \sum_{(i,j)\in \Omega} (\langle u_i, v_i \rangle-M_{ij})^2 \end{aligned}\]

In order to solve this, we can alternate between optimizing for \(U\) while letting \(V\) be a constant, and optimizing over \(V\) while letting \(U\) be a constant. If \(t\) is the iteration number, then the algorithm is simply

\[\begin{aligned} \text{for } t=1,2,\ldots:& \\ U^{t}=&\underset{U}{\text{minimize}}& \sum_{(i,j)\in \Omega} (\langle u_i, v^{t-1} i \rangle-M_{ij})^2 \\ V^{t}=&\underset{ V}{\text{minimize}}& \sum_{(i,j)\in \Omega} (\langle u^t_i, v_i \rangle-M_{ij})^2 \\ \end{aligned}\]

At each iteration, we just need to solve a least squares problem which is easy enough.

import numpy as np
from scipy.optimize import minimize

def alt_min(m,n,r, Omega, known):

  for i in range(0,100):  
    objU=lambda x: np.linalg.norm(np.reshape(x, [m,r]).dot(V)[Omega]-known)**2
    U = np.reshape(minimize(objU, U).x, [m,r])
    objV=lambda x: np.linalg.norm(U.dot(np.reshape(x, [r,n]))[Omega]-known)**2
    V = np.reshape(minimize(objV, V).x, [r,n])

    if res < 0.0001:
  return (U,V)

Lets test our algorithm with the simple example given earlier.

X=([0,0,0,0,1,1,1,2,2,2], [0,1,2,3,0,1,2,0,1,3])
(U,V)=alt_min(3,4,2,X, y)
  [[1.00007939 0.99998808 1.9999993 2.00000032]
   [1.99997349 1.00000394 3.00000041 4.99908422]
   [0.99997352 2.00000397 2.99974998 0.99999984]]

Thats the same matrix we came up with!

SpaLoR: A python package for Sparse and Low Rank models

While matrix completion is a form of machine learning, I always see it talked about and coded up as just an optimization algorithm. To help bridge this gap, I’ve been working on a python package for called SpaLoR, which should be easy to pick up for scikit-learn users.

In it, there is a model called MatrixCompletion, which you can use to fit your data to a low rank matrix and then make predictions. No parameters are required when you initialize, but a few you might want to specify are

How do we use it for movie recomendations?

Now that we have a good understanding of what matrix completion is and how to do it, we can get to the fun part. Theres a ton of applications of matrix completion, from reconstructing the molecular structure of protiens from limited measurements to image classification, but by far the most commonly cited example is the Netflix problem. The state of the art dataset for movie recommendations comes from MovieLens, and though they have datasets with 25 million ratings, we’re going to stick with 1 million for simplicity.

Before we get into the data, we should justify to ourslelves that this is going to be a low-rank matrix. Let’s take the movies Breakfast Club and Pretty in Pink as an example. I would bet that the way individuals rate these two movies is pretty much the same way, and so they columns associated with each of them should be very close to eachother. Now lets throw Titanic into the mix. While I wouldn’t expect it to be the same, it might be similiar. It might also be similiar to other period pieces featuring forbidden love, like Pride and Prejudice, or movies with Leonardo DeCaprio, like Wolf of Wallstreet. So, I would expect that the ratings for Titanic might look like an average of all of these movies. The point is that the ratings for a specific movie should be pretty close to a linear combination of ratings of just a few other movies.

First, lets load the data set and see what it looks like.

data = np.loadtxt( 'movieLens/ratings.dat',delimiter='::' )
  [[1.00000000e+00 1.19300000e+03 5.00000000e+00 9.78300760e+08]
   [1.00000000e+00 6.61000000e+02 3.00000000e+00 9.78302109e+08]
   [1.00000000e+00 9.14000000e+02 3.00000000e+00 9.78301968e+08]]

The first column is the user ID, the second is the movie ID, the third is the rating (1,2,3,4, or 5), and the last is a time stamp (which we don’t need to worry about). We want the rows of the matrix to be users, and the columns should be movies.

X=data[:, [0,1]].astype(int)-1


  (6040, 3952)

So, we have 6040 users and 3952 movies. That’s a total of about 23 million potential ratings, of which we know 1 million. We’re going to reserve 200,000 of the ratings to test our results.

from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)

Now to train the model and make some predictions!

from spalor.MC import MC
from statistics import mean

mc_model.fit(X_train, y_train)

print("Percent of predictions off my less than 1: ",np.sum(abs(y_test-y_predict)<1)/len(y_test))
  MAE: 0.6910439339771605
  Percent of predictions off my less than 1: 0.7603603243318903

I would say 0.691 is pretty good for the mean absolute error. Plus, 76% of the predictions are off by less than 1. This might not seem great, but its pretty close to the state of the art.

These numbers can most definetly get better. Here’s a few ideas I might write about in the future to get better results: