# 24PCA I

Learning Outcomes
• Discuss the dimensionality of a dataset and strategies for dimensionality reduction
• Derive and carry out the procedure of PCA

So far in this course, we’ve focused on supervised learning techniques that create a function to map inputs (features) to labelled outputs. Regression and classification are two main examples, where the output value of regression is quantitative while the output value of classification is categorical.

Today, we’ll introduce an unsupervised learning technique called PCA. Unlike supervised learning, unsupervised learning is applied to unlabeled data. Because we have features but no labels, we aim to identify patterns in those features.

## 24.1 Visualization (Revisited)

Visualization can help us identify clusters or patterns in our dataset, and it can give us an intuition about our data and how to clean it for the model. For this demo, we’ll return to the MPG dataset from Lecture 19 and see how far we can push visualization for multiple features.

Code
import pandas as pd
import numpy as np
import scipy as sp
import plotly.express as px
import seaborn as sns
Code
mpg = sns.load_dataset("mpg").dropna()
mpg.head()
mpg cylinders displacement horsepower weight acceleration model_year origin name
0 18.0 8 307.0 130.0 3504 12.0 70 usa chevrolet chevelle malibu
1 15.0 8 350.0 165.0 3693 11.5 70 usa buick skylark 320
2 18.0 8 318.0 150.0 3436 11.0 70 usa plymouth satellite
3 16.0 8 304.0 150.0 3433 12.0 70 usa amc rebel sst
4 17.0 8 302.0 140.0 3449 10.5 70 usa ford torino

We can plot one feature as a histogram to see it’s distribution. Since we only plot one feature, we consider this a 1-dimensional plot.

Code
px.histogram(mpg, x="displacement")

We can also visualize two features (2-dimensional scatter plot):

Code
px.scatter(mpg, x="displacement", y="horsepower")

Three features (3-dimensional scatter plot):

Code
fig = px.scatter_3d(mpg, x="displacement", y="horsepower", z="weight",
width=800, height=800)
fig.update_traces(marker=dict(size=3))

We can even push to 4 features using a 3D scatter plot and a colorbar:

Code
fig = px.scatter_3d(mpg, x="displacement",
y="horsepower",
z="weight",
color="model_year",
width=800, height=800,
opacity=.7)
fig.update_traces(marker=dict(size=5))

Visualizing 5 features is also possible if we make the scatter dots unique to the datapoint’s origin.

Code
fig = px.scatter_3d(mpg, x="displacement",
y="horsepower",
z="weight",
color="model_year",
size="mpg",
symbol="origin",
width=900, height=800,
opacity=.7)
# hide color scale legend on the plotly fig
fig.update_layout(coloraxis_showscale=False)

However, adding more features to our visualization can make our plot look messy and uninformative, and it can also be near impossible if we have a large number of features. The problem is that many datasets come with more than 5 features —— hundreds, even. Is it still possible to visualize all those features?

## 24.2 Dimensionality

Suppose we have a dataset of:

• $$N$$ observations (datapoints/rows)
• $$d$$ attributes (features/columns)

Let’s “rename” this in terms of linear algebra so that we can be more clear with our wording. Using linear algebra, we can view our matrix as:

• $$N$$ row vectors in a $$d$$-Dimensional space, OR
• $$d$$ column vectors in an $$N$$-Dimensions space

The intrinsic dimension of a dataset is the minimal set of dimensions needed to approximately represent the data. In linear algebra terms, it is the dimension of the column space of a matrix, or the number of linearly independent columns in a matrix; this is equivalently called the rank of a matrix.

In the examples below, Dataset 1 has 2 dimensions because it has 2 linearly independent columns. Similarly, Dataset 2 has 3 dimensions because it has 3 linearly independent columns.

It may be tempting to say that it has 4 dimensions, but the Weight (lbs) column is actually just a linear transformation of the Weight (kg) column. Thus, no new information is captured, and the matrix of our dataset has a (column) rank of 3! Therefore, despite having 4 columns, we still say that this data is 3-dimensional.

Plotting the weight columns together reveals the key visual intuition. While the two columns visually span a 2D space as a line, the data does not deviate at all from that singular line. This means that one of the weight columns is redundant! Even given the option to cover the whole 2D space, the data below does not. It might as well not have this dimension, which is why we still do not consider the data below to span more than 1 dimension.

What happens when there are outliers? Below, we’ve added one outlier point to the dataset above, and just that one point is enough to change the rank of the matrix from 1 to 2 dimensions. However, the data is still approximately 1-dimensional.

Dimensionality reduction is generally an approximation of the original data that’s achieved by projecting the data onto a desired dimension. In the example below, our original datapoints (blue dots) are 2-dimensional. We have a few choices if we want to project them down to 1-dimension: project them onto the $$x$$-axis (left), project them onto the $$y$$-axis (middle), or project them to a line $$mx + b$$ (right). The resulting datapoints after the projection is shown in red. Which projection do you think is better? How can we calculate that?

In general, we want the projection which is the best approximation for the original data (the graph on the right). In other words, we want the projection that captures the most variance of the original data. In the next section, we’ll see how this is calculated.

## 24.3 Matrix Decomposition (Factorization)

One linear technique for dimensionality reduction is matrix decomposition, which is closely tied to matrix multiplication. In this section, we will decompose our data matrix $$X$$ into a lower-dimensional matrix $$Z$$ that approximately recovers the original data when multiplied by $$W$$.

First, consider the matrix multiplication example below:

• For table 1, each row of the fruits matrix represents one bowl of fruit; for example, the first bowl/row has 2 apples, 2 lemons, and 2 melons.
• For table 2, each column of the dollars matrix represents the cost of fruit at a store; for example, the first store/column charges 2 dollars for an apple, 1 dollar for a lemon, and 4 dollars for a melon.
• The output is the cost of each bowl at each store.
Linear Algebra Review: Matrix Multiplication

In general, there are two ways to interpret matrix multiplication:

1. Each datapoint in our output is a dot product between a row in the data matrix and a column in the operations matrix. In this view, we perform multiple linear operations on the data
2. Each column in our output is a linear transformation of the original columns based on a column in the transformation matrix

We will use the second interpretation to link matrix multiplication with matrix decomposition, where we receive a lower dimensional representation of data along with a transformation matrix.

Matrix decomposition (a.k.a matrix factorization) is the opposite of matrix multiplication. Instead of multiplying two matrices, we want to decompose a single matrix into 2 separate matrices. Just like with real numbers, there are infinite ways to decompose a matrix into a product of two matrices. For example, $$9.9$$ can be decomposed as $$1.1 * 9$$, $$3.3 * 3.3$$, $$1 * 9.9$$, etc. Additionally, the sizes of the 2 decomposed matrices can vary drastically. In the example below, the first factorization (top) multiplies a $$3x2$$ matrix by a $$2x3$$ matrix while the second factorization (bottom) multiplies a $$3x3$$ matrix by a $$3x3$$ matrix; both result in the original matrix on the right.

We can even expand the $$3x3$$ matrices to $$3x4$$ and $$4x3$$ (shown below as the factorization on top), but this defeats the point of dimensionality reduction since we’re adding more “useless” dimensions. On the flip side, we also can’t reduce the dimension to $$3x1$$ and $$1x3$$ (shown below as the factorization on the bottom); since the rank of the original matrix is greater than 1, this decomposition will not result in the original matrix.

In practice, we often work with datasets containing many features, so we usually want to construct decompositions where the dimensionality is below the rank of the original matrix. While this does not recover the data exactly, we can still provide approximate reconstructions of the matrix.

In the next section, we will discuss a method to automatically and approximately factorize data. This avoids redundant features and makes computation easier because we can train on less data. Since some approximations are better than others, we will also discuss how the method helps us capture a lot of information in a low number of dimensions.

## 24.4 Principal Component Analysis (PCA)

In PCA, our goal is to transform observations from high-dimensional data down to low dimensions (often 2, as most visualizations are 2D) through linear transformations. In other words, we want to find a linear transformation that creates a low-dimension representation that captures as much of the original data’s total variance as possible.

We often perform PCA during the Exploratory Data Analysis (EDA) stage of our data science lifecycle when we don’t know what model to use. It helps us with:

• Visually identifying clusters of similar observations in high dimensions.
• Removing irrelevant dimensions if we suspect that the dataset is inherently low rank. For example, if the columns are collinear, there are many attributes, but only a few mostly determine the rest through linear associations.
• Creating a transformed dataset of decorrelated features.

There are two equivalent ways of framing PCA:

1. Finding directions of maximum variability in the data.
2. Finding the low dimensional (rank) matrix factorization that best approximates the data.

To execute the first approach of variance maximization framing (more common), we can find the variances of each attribute with np.var and then keep the $$k$$ attributes with the highest variance. However, this approach limits us to work with attributes individually; it cannot resolve collinearity, and we cannot combine features.

The second approach uses PCA to construct principal components with the most variance in the data (even higher than the first approach) using linear combinations of features. We’ll describe the procedure in the next section.

### 24.4.1 PCA Procedure (Overview)

To perform PCA on a matrix:

1. Center the data matrix by subtracting the mean of each attribute column.
2. To find the $$i$$-th principal component, $$v_i$$:
1. $$v$$ is a unit vector that linearly combines the attributes.
2. $$v$$ gives a one-dimensional projection of the data.
3. $$v$$ is chosen to maximize the variance along the projection onto $$v$$. This is equivalent to minimizing the sum of squared distances between each point and its projection onto $$v$$.
4. Choose $$v$$ such that it is orthogonal to all previous principal components.

The $$k$$ principal components capture the most variance of any $$k$$-dimensional reduction of the data matrix.

In practice, however, we don’t carry out the procedures in step 2 because they take too long to compute. Instead, we use singular value decomposition (SVD) to find all principal components efficiently.

### 24.4.2 Deriving PCA as Error Minimization

In this section, we will derive PCA keeping the following goal in mind: minimize the reconstruction loss for our matrix factorization model. You are not expected to be able to be able to redo this derivation, but understanding the derivation may help with future assignments.

Given a matrix $$X$$ with $$n$$ rows and $$d$$ columns, our goal is to find its best decomposition such that $X \approx Z W$ Z has $$n$$ rows and $$k$$ columns; W has $$k$$ rows and $$d$$ columns.

To measure the accuracy of our reconstruction, we define the reconstruction loss below, where $$X_i$$ is the row vector of $$X$$, and $$Z_i$$ is the row vector of $$Z$$:

There are many solutions to the above, so let’s constrain our model such that $$W$$ is a row-orthonormal matrix (i.e. $$WW^T=I$$) where the rows of $$W$$ are our principal components.

In our derivation, let’s first work with the case where $$k=1$$. Here Z will be an $$n \times 1$$ vector and W will be a $$1 \times d$$ vector.

\begin{aligned} L(z,w) &= \frac{1}{n}\sum_{i=1}^{n}(X_i - z_{i}w)(X_i - z_{i}w)^T \\ &= \frac{1}{n}\sum_{i=1}^{n}(X_{i}X_{i}^T - 2z_{i}X_{i}w^T + z_{i}^{2}ww^T) & \text{(expand the loss)} \\ = \frac{1}{n}\sum_{i=1}^{n}(-2z_{i}X_{i}w^T + z_{i}^{2}) & \text{(First term is constant and }ww^T=1\text{ by orthonormality)} \\ \end{aligned}

Now, we can take the derivative with respect to $$Z_i$$. \begin{aligned} \frac{\partial{L(Z,W)}}{\partial{z_i}} &= \frac{1}{n}(-2X_{i}w^T + 2z_{i}) \\ z_i &= X_iw^T & \text{(Setting derivative equal to 0 and solving for }z_i\text{)}\end{aligned}

We can now substitute our solution for $$z_i$$ in our loss function:

\begin{aligned} L(z,w) &= \frac{1}{n}\sum_{i=1}^{n}(-2z_{i}X_{i}w^T + z_{i}^{2}) \\ L(z=X_iw^T,w) &= \frac{1}{n}\sum_{i=1}^{n}(-2X_iw^TX_{i}w^T + (X_iw^T)^{2}) \\ &= \frac{1}{n}\sum_{i=1}^{n}(-X_iw^TX_{i}w^T) \\ &= \frac{1}{n}\sum_{i=1}^{n}(-wX_{i}^TX_{i}w^T) \\ &= -w\frac{1}{n}\sum_{i=1}^{n}(X_i^TX_{i})w^T \\ &= -w\Sigma w^T \end{aligned}

Now, we need to minimize our loss with respect to $$w$$. Since we have a negative sign, one way we can do this is by making $$w$$ really big. However, we also have the orthonormality constraint $$ww^T=1$$. To incorporate this constraint into the equation, we can add a Lagrange multiplier, $$\lambda$$. Note that lagrangian multipliers are out of scope for Data 100.

$L(w,\lambda) = -w\Sigma w^T + \lambda(ww^T-1)$

Taking the derivative with respect to $$w$$, \begin{aligned} \frac{\partial{L(w,\lambda)}}{w} &= -2\Sigma w^T + 2\lambda w^T \\ 2\Sigma w^T - 2\lambda w^T &= 0 & \text{(Setting derivative equal to 0)} \\ \Sigma w^T &= \lambda w^T \\ \end{aligned}

This result implies that:

• $$w$$ is a unitary eigenvector of the covariance matrix. This means that $$||w||^2 = ww^T = 1$$
• The error is minimized when $$w$$ is the eigenvector with the largest eigenvalue $$\lambda$$.

This derivation can inductively be used for the next (second) principal component (not shown).

The final takeaway from this derivation is that the principal components are the eigenvectors with the largest eigenvalues of the covariance matrix. These are the directions of the maximum variance of the data. We can construct the latent factors (the Z matrix) by projecting the centered data X onto the principal component vectors: