Lecture 12 – Data 100, Spring 2024¶

Data 100, Spring 2024

Acknowledgments Page

In [1]:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns

Multiple Linear Regression¶

Let's load in a new dataset. This is aggregate per-player data from the 2018-19 NBA season, downloaded from Kaggle.

In [2]:
nba = pd.read_csv('data/nba18-19.csv', index_col=0)
nba.index.name = None # Drops name of index (players are ordered by rank)
In [3]:
nba.head(5)
Out[3]:
Player Pos Age Tm G GS MP FG FGA FG% ... FT% ORB DRB TRB AST STL BLK TOV PF PTS
1 Álex Abrines\abrinal01 SG 25 OKC 31 2 19.0 1.8 5.1 0.357 ... 0.923 0.2 1.4 1.5 0.6 0.5 0.2 0.5 1.7 5.3
2 Quincy Acy\acyqu01 PF 28 PHO 10 0 12.3 0.4 1.8 0.222 ... 0.700 0.3 2.2 2.5 0.8 0.1 0.4 0.4 2.4 1.7
3 Jaylen Adams\adamsja01 PG 22 ATL 34 1 12.6 1.1 3.2 0.345 ... 0.778 0.3 1.4 1.8 1.9 0.4 0.1 0.8 1.3 3.2
4 Steven Adams\adamsst01 C 25 OKC 80 80 33.4 6.0 10.1 0.595 ... 0.500 4.9 4.6 9.5 1.6 1.5 1.0 1.7 2.6 13.9
5 Bam Adebayo\adebaba01 C 21 MIA 82 28 23.3 3.4 5.9 0.576 ... 0.735 2.0 5.3 7.3 2.2 0.9 0.8 1.5 2.5 8.9

5 rows × 29 columns


We are interested in predicting the number of points (`PTS`) an athlete will score in a basketball game this season.

Suppose we want to fit a linear model by using some characteristics, or features of a player. Specifically, we'll focus on field goals, assists, and 3-point attempts.

  • FG, the number of (2-point) field goals per game
  • AST, the average number of assists per game, and
  • 3PA, the number of 3-point field goals attempted per game

This is how we express that model:

$$ \hat{y} = \theta_0 + \theta_1 \cdot \textbf{FG} + \theta_2 \cdot \textbf{AST} + \theta_3 \cdot \textbf{3PA} $$
In [4]:
nba[['FG', 'AST', '3PA', 'PTS']]
Out[4]:
FG AST 3PA PTS
1 1.8 0.6 4.1 5.3
2 0.4 0.8 1.5 1.7
3 1.1 1.9 2.2 3.2
4 6.0 1.6 0.0 13.9
5 3.4 2.2 0.2 8.9
... ... ... ... ...
528 4.0 0.8 0.0 11.5
529 3.1 0.9 0.0 7.8
530 3.6 1.1 0.0 8.9
530 3.4 0.8 0.0 8.5
530 3.8 1.5 0.0 9.4

708 rows × 4 columns



In lecture, we saw that the predictions for the entire data set $\hat{\mathbb{Y}}$ can be computed as:

$$ \Large \hat{\mathbb{Y}} = \mathbb{X} \theta $$

The design matrix $\mathbb{X} \in \mathbb{R}^{n,d+1}$ has $n$ rows corresponding to each record in the dataset and $d+1$ columns corresponding to the original $d$ columns in the data plus an additional 1s column vector (bias/intercept feature).

Let's build this design matrix using Pandas:

In [5]:
X = nba[['FG', 'AST', '3PA']]
X.insert(0, 'Bias', 1)
X
Out[5]:
Bias FG AST 3PA
1 1 1.8 0.6 4.1
2 1 0.4 0.8 1.5
3 1 1.1 1.9 2.2
4 1 6.0 1.6 0.0
5 1 3.4 2.2 0.2
... ... ... ... ...
528 1 4.0 0.8 0.0
529 1 3.1 0.9 0.0
530 1 3.6 1.1 0.0
530 1 3.4 0.8 0.0
530 1 3.8 1.5 0.0

708 rows × 4 columns

In [6]:
X = X.to_numpy()
X.shape
Out[6]:
(708, 4)

While we're at it, let's build the $\mathbb{Y}$ vector of our true PTS values.

In [7]:
# For nba data
Y = nba[["PTS"]].to_numpy()
n = len(Y)
print("number datapoints", n)
Y[:5]

number datapoints 708
Out[7]:
array([[ 5.3],
       [ 1.7],
       [ 3.2],
       [13.9],
       [ 8.9]])

Example prediction¶

Suppose we decide to pick an arbitrary parameter $\theta$:

$$\theta = \begin{bmatrix}0.50 \\ -0.14 \\ 0.65 \\ 1.52 \end{bmatrix}$$

(For those interested: I drew these from random values simulated from a standard normal distribution using np.random.randn.)

In [8]:
theta_arbitrary = np.array([[0.5], [-1.14], [0.65], [1.52]])
theta_arbitrary
Out[8]:
array([[ 0.5 ],
       [-1.14],
       [ 0.65],
       [ 1.52]])

For this value of $\theta$ we can make a prediction using our model with matrix multiplication.

The @ symbol is the matrix multiply operation and is equivalent to writing xt.dot(theta).

In [9]:
display((X @ theta_arbitrary)[:5])

display(Y[:5])

array([[ 5.07 ],
       [ 2.844],
       [ 3.825],
       [-5.3  ],
       [-1.642]])
array([[ 5.3],
       [ 1.7],
       [ 3.2],
       [13.9],
       [ 8.9]])

Computing MSE¶

For Ordinary Least Squares, the average loss is MSE:

$$ \Large R(\theta) = \frac{1}{n} || \mathbb{Y} - \mathbb{X}\theta||^2_2 $$

NumPy has a handy function np.linalg.norm that computes the norm of a matrix (default is L2 norm).

In [10]:
theta_arbitrary = np.array([0.5, -1.14, 0.65, 1.52])
def mse_nba(theta):
    """
    Y is PTS
    X is intercept, FG, AST, 3PA
    """
    return (1/n) * (np.linalg.norm(Y - X @ theta) ** 2)

mse_nba(theta_arbitrary)
Out[10]:
57637.9018454746

Is this good? Is this bad? Let's compute the optimal theta and compare!

Implementing Least Squares¶

From lecture, the Least Squares Estimate $\hat{\theta}$ is: $$ \Large \hat{\theta} = \left( \mathbb{X}^{\top} \mathbb{X} \right)^{-1} \mathbb{X}^{\top} \mathbb{Y} $$

While it is not as numerically stable or efficient. We can compute $\hat{\theta}$ by direction using matrix inversion. To do this, we import the inv function for the numpy linear algebra library:

In [11]:
from numpy.linalg import inv
In [12]:
def least_squares_estimate(X, Y):
    return inv(X.T @ X) @ X.T @ Y

theta_hat = least_squares_estimate(X, Y)
theta_hat
Out[12]:
array([[-0.29253798],
       [ 2.51705703],
       [ 0.05075571],
       [ 0.31307653]])
In [13]:
print("Arbitrary theta MSE: ", mse_nba(theta_arbitrary))
print("Optimal theta MSE:"  , mse_nba(theta_hat))

Arbitrary theta MSE:  57637.9018454746
Optimal theta MSE: 0.3963133329080335

Nice!!!

Making Least Squares Predictions¶

Now that we have estimated the model parameters $\hat{\theta}$ we can now also predict the points scored $\hat{\mathbb{Y}}$ for each of our players.

In [14]:
Y_hat = X @ theta_hat

Model Performance/Diagnosing the Model¶

In previous lectures, we have plotted the residual vs. our single input feature $x$.

For higher dimensional regression models, we often graph the residual with respect to the fitted values $\hat{y}$.

In [15]:
sns.scatterplot(x = Y_hat.flatten(), y = (Y - Y_hat).flatten())
sns.lineplot(x=[-1, 35], y=[0,0], label="$\hat{Y}=Y$", color = 'orange', linewidth = 3)
plt.xlabel("$\hat{Y}$")
plt.ylabel("$\hat{Y} - Y$")
plt.xlim([0, 32])
plt.show()

Overall while the residuals are roughly centered around 0 ($\hat{y} = y$), we see heteroskedasticity: Our regression spread is uneven, particularly as predictions get big.

Multiple $R^2$¶

Let's compute the coefficient of determination, or multiple $R^2$, for our model.

In [16]:
r2_ast_fg_3pa = np.var(Y_hat) / np.var(Y)
r2_ast_fg_3pa
Out[16]:
0.9883162128703274

That's super high!!! Wait, what's up with that?

(Hint: Basketball facts)

Let's try reducing the number of features to see how this Multiple $R^2$ metric changes.

In [17]:
# Use intercept, AST, 3PA
X_3d = nba[['AST', '3PA']]
X_3d.insert(0, 'Bias', 1)
X_3d = X_3d.to_numpy()

theta_ast_3pa = least_squares_estimate(X_3d, Y)
Y_hat_ast_3pa = X_3d @ theta_ast_3pa

r2_ast_3pa = np.var(Y_hat_ast_3pa) / np.var(Y)
r2_ast_3pa
Out[17]:
0.608786276366571
In [18]:
# Use intercept, AST only (SLR)
X_slr = nba[['AST']]
X_slr.insert(0, 'Bias', 1)
X_slr = X_slr.to_numpy()

theta_ast_only = least_squares_estimate(X_slr, Y)
Y_hat_ast_only = X_slr @ theta_ast_only

r2_ast_only = np.var(Y_hat_ast_only) / np.var(Y)
r2_ast_only
Out[18]:
0.4570055507968593
In [19]:
theta_ast_only
Out[19]:
array([[3.98332315],
       [2.39888152]])
In [20]:
theta_hat
Out[20]:
array([[-0.29253798],
       [ 2.51705703],
       [ 0.05075571],
       [ 0.31307653]])

Comparing these Multiple $R^2$ together:

In [21]:
print("(SLR) intercept, AST:    ", r2_ast_only)
print("intercept, 3PA, AST:     ", r2_ast_3pa)
print("intercept, FG, 3PA, AST: ", r2_ast_fg_3pa)

(SLR) intercept, AST:     0.4570055507968593
intercept, 3PA, AST:      0.608786276366571
intercept, FG, 3PA, AST:  0.9883162128703274

Because of how basketball is scored, knowing Field Goals FG and 3 point goal attempts 3PA will reliably tell you how many total points a player scored in that game. This is assuming NBA players make a good number of their 3 pointers.

Side note, if you wanted to check that Multiple R^2 for Simple Linear Regression is indeed correlation coefficient $r^2$:

In [22]:
r = np.corrcoef(nba['AST'], nba['PTS'])[0,1]
r ** 2
Out[22]:
0.4570055507968595