📈 Lecture 11 – Data 100, Spring 2025

Data 100, Spring 2025

Acknowledgments Page

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
import itertools
from mpl_toolkits.mplot3d import Axes3D

# Big font helper
def adjust_fontsize(size=None):
    SMALL_SIZE = 8
    MEDIUM_SIZE = 10
    BIGGER_SIZE = 12
    if size != None:
        SMALL_SIZE = MEDIUM_SIZE = BIGGER_SIZE = size

    plt.rc('font', size=SMALL_SIZE)          # controls default text sizes
    plt.rc('axes', titlesize=SMALL_SIZE)     # fontsize of the axes title
    plt.rc('axes', labelsize=MEDIUM_SIZE)    # fontsize of the x and y labels
    plt.rc('xtick', labelsize=SMALL_SIZE)    # fontsize of the tick labels
    plt.rc('ytick', labelsize=SMALL_SIZE)    # fontsize of the tick labels
    plt.rc('legend', fontsize=SMALL_SIZE)    # legend fontsize
    plt.rc('figure', titlesize=BIGGER_SIZE)  # fontsize of the figure title

🦭 Dugongs Part 1: Comparing Two Different Models, Both Fit with MSE

dugongs = pd.read_csv("data/dugongs.csv")
display(dugongs.head())

data_constant = dugongs["Age"]
data_linear = dugongs[["Length", "Age"]]

	Length	Age
0	1.77	1.5
1	1.80	1.0
2	1.85	1.5
3	1.87	1.5
4	2.02	2.5

🎢 Loss Surfaces

The average loss of the constant model using L2 loss (i.e., squared loss) is

$$ \Large \hat{R}(\theta_0) = \frac{1}{n}\sum_{i=1}^n (y_i - \theta_0)^2 $$

plt.style.use('default') # Revert style to default mpl
adjust_fontsize(size=16)
%matplotlib inline

# Calculates the MSE for a constant model given some value of theta
def mse_constant(theta, data):
    return np.mean(np.array([(y_obs - theta) ** 2 for y_obs in data]), axis=0)

# Set up a range of theta values and calculate the MSE for each
thetas = np.linspace(-20, 42, 1000)
l2_loss_thetas = mse_constant(thetas, data_constant)

# Plot the MSE as a function of the thetas
plt.plot(thetas, l2_loss_thetas)
plt.xlabel(r'$\theta_0$')
plt.ylabel(r'MSE')

# Optimal point
thetahat = np.mean(data_constant)
plt.scatter([thetahat], [mse_constant(thetahat, data_constant)], s=50, label = r"$\hat{\theta}_0$")
plt.legend()
# plt.savefig('mse_constant_loss.png', bbox_inches = 'tight');
plt.show()

The average loss for the SLR model using L2 loss is the following:

$$ \Large \hat{R}(\theta_0, \theta_1) = \frac{1}{n}\sum_{i=1}^n (y_i - (\theta_0 + \theta_1x))^2 $$

Note: Unlike the average loss of the constant model, which has one input, the average loss of SLR has two inputs!

# Run this cell to see a 3D visualization of the loss surface
import itertools
import plotly.graph_objects as go

# Calculates the MSE for a linear model given some values of theta_0 and theta_1
def mse_linear(theta_0, theta_1, data_linear):
    data_x, data_y = data_linear.iloc[:,0], data_linear.iloc[:,1]
    return np.mean(np.array([(y - (theta_0+theta_1*x)) ** 2 for x, y in zip(data_x, data_y)]), axis=0)

# Set up a grid of theta values and calculate the MSE for each
theta_0_values = np.linspace(-80, 20, 80)
theta_1_values = np.linspace(-10,30, 80)
mse_values = [mse_linear(theta_0, theta_1, data_linear) \
              for theta_0, theta_1 in itertools.product(theta_0_values, theta_1_values)]
mse_values = np.reshape(mse_values, (len(theta_0_values), len(theta_1_values)), order='F')

# Optimal point
data_x, data_y = data_linear.iloc[:,0], data_linear.iloc[:,1]
theta_1_hat = np.corrcoef(data_x, data_y)[0, 1] * np.std(data_y) / np.std(data_x)
theta_0_hat = np.mean(data_y) - theta_1_hat * np.mean(data_x)
fig_lin = go.Figure(data=[go.Surface(x=theta_0_values, y=theta_1_values, z=mse_values,name='z=MSE')])
fig_lin.add_trace(go.Scatter3d(x=[theta_0_hat], y=[theta_1_hat], \
                               z=[mse_linear(theta_0_hat, theta_1_hat, data_linear)],\
                               mode='markers', marker=dict(size=12),name='theta hat'))

fig_lin.update_layout(
    autosize=False,
    scene = dict(
                    xaxis_title='theta_0',
                    yaxis_title='theta_1',
                    zaxis_title='z=MSE'),
                    width=500,
                    margin=dict(r=20, b=10, l=10, t=10))

fig_lin.show()

🧮 Computing the RMSE of SLR and constant model fit with MSE

print("Least Squares Constant Model RMSE:",
          round(np.sqrt(mse_constant(thetahat, data_constant)), 2)
      )
print("Least Squares Linear Model RMSE:  ",
          round(np.sqrt(mse_linear(theta_0_hat, theta_1_hat, data_linear)),2)
      )

Least Squares Constant Model RMSE: 7.72
Least Squares Linear Model RMSE:   4.31

Interpreting the RMSE (Root Mean Square Error):

• Constant model error is HIGHER than linear model error
• Linear model fit is BETTER than constant model fit (if using RMSE as our metric)

🔮 Predictions

This plotting code is just for your reference.

This code generates the figures used in the lecture.

yobs = data_linear["Age"]      # The true observations y
xs = data_linear["Length"]     # Needed for linear predictions
n = len(yobs)                  # Predictions

yhats_constant = [thetahat for i in range(n)]    # Not used, but food for thought
yhats_linear = [theta_0_hat + theta_1_hat * x for x in xs]

sns.set_theme()
adjust_fontsize(size=16)
%matplotlib inline

fig = plt.figure(figsize=(8, 1.5))
sns.rugplot(yobs, height=0.25, lw=2) ;
plt.axvline(thetahat, color='red', lw=4, label=r"$\hat{\theta}_0$");
plt.legend()
plt.yticks([])
# plt.savefig('dugong_rug.png', bbox_inches = 'tight');
plt.show()

# In case we're in a weird style state
sns.set_theme()
adjust_fontsize(size=16)
%matplotlib inline

sns.scatterplot(x=xs, y=yobs)
plt.axhline(y=thetahat, color='red', lw=4)
# plt.savefig('dugong_constant_line.png', bbox_inches = 'tight');
plt.show()

# In case we're in a weird style state
sns.set_theme()
adjust_fontsize(size=16)
%matplotlib inline

sns.scatterplot(x=xs, y=yobs)
plt.plot(xs, yhats_linear, color='red', lw=4)
# plt.savefig('dugong_line.png', bbox_inches = 'tight');
plt.show()

---

🧋 Boba Tea: Two Constant Models, Fit to Different Losses

Exploring MAE

boba = np.array([20, 21, 22, 29, 33])

Let's plot the $L_1$ loss for a single observation. We'll plot the $L_1$ loss for the first observation; since $y_1 = 20$, we'll be plotting

$$ \Large L_1(20, \theta_0) = |20 - \theta_0| $$

thetas = np.linspace(10, 30, 1000)
l1_loss_single_obvs = np.abs(boba[0] - thetas)

plt.style.use('default') # Revert style to default mpl
adjust_fontsize(size=16)
%matplotlib inline

plt.plot(thetas, l1_loss_single_obvs,  'g--', );
plt.xlabel(r'$\theta_0$');
plt.ylabel(r'L1 Loss for $y = 20$');
# plt.savefig('l1_loss_single_obs.png', bbox_inches = 'tight');

def mae_constant(theta_0, data):
    return np.mean(np.array([np.abs(y_obs - theta_0) for y_obs in data]), axis=0)

thetas = np.linspace(10, 40, 1000)
l1_loss_thetas = mae_constant(thetas, boba)
plt.plot(thetas, l1_loss_thetas, color = 'green', lw=3);
plt.xlabel(r'$\theta_0$');
plt.ylabel(r'MAE across all data');
# plt.savefig('l1_loss_average.png', bbox_inches = 'tight');

🥣 Median Minimizes MAE for the Constant Model

# In case we're in a weird style state
sns.set_theme()
adjust_fontsize(size=16)
%matplotlib inline

yobs = boba
thetahat = np.median(yobs)

fig = plt.figure(figsize=(8, 1.5))
sns.rugplot(yobs, height=0.25, lw=2) ;
plt.scatter([thetahat], [-.1], color='green', lw=4, label=r"$\hat{\theta}_0$");
plt.xlabel("Sales")
plt.legend()
plt.yticks([])
# plt.savefig('boba_rug.png', bbox_inches = 'tight');
plt.show()

2️⃣ Two Constant Models, Fit to Different Losses

plt.style.use('default') # Revert style to default mpl
adjust_fontsize(size=16)
%matplotlib inline

nplots = 2
def plot_losses(data, title=None, theta_range=(10, 40)):
    thetas = np.linspace(theta_range[0], theta_range[1], 1000)
    l2_loss_thetas = mse_constant(thetas, data)
    thetahat_mse = np.mean(data)

    l1_loss_thetas = mae_constant(thetas, data)
    thetahat_mae = np.median(data)

    fig, axs = plt.subplots(1, nplots, figsize=(5*2+0.5, 3.5))
    axs[0].plot(thetas, l2_loss_thetas, lw=3);
    axs[0].scatter([thetahat_mse], [mse_constant(thetahat_mse, data)], s=100)
    axs[0].annotate(r"$\hat{\theta}_0$ = " + f"{thetahat_mse:.1f}",
                    xy=(thetahat_mse, np.average(axs[0].get_ylim())),
                    size=20, ha='center', va='top',
                    bbox=dict(boxstyle='round', fc='w'))
    axs[0].set_xlabel(r'$\theta_0$');
    axs[0].set_ylabel(r'MSE');

    axs[1].plot(thetas, l1_loss_thetas, color = 'green', lw=3);
    axs[1].scatter([thetahat_mae], [mae_constant(thetahat_mae, data)], color='green', s=100)
    axs[1].annotate(r"$\hat{\theta}_0$ = " + f"{thetahat_mae:.1f}",
                    xy=(thetahat_mae, np.average(axs[1].get_ylim())),
                    size=20, ha='center', va='top',
                    bbox=dict(boxstyle='round', fc='w'))
    axs[1].set_xlabel(r'$\theta_0$');
    axs[1].set_ylabel(r'MAE');
    if title:
        fig.suptitle(title)
    fig.tight_layout()
    return fig

fig = plot_losses(boba)
plt.figure(fig)
# plt.savefig('loss_compare.png', bbox_inches = 'tight');
plt.show()

More loss comparison: Outliers

boba_outlier = np.array([20, 21, 22, 29, 33, 1033])
fig = plot_losses(boba_outlier, theta_range=[-10, 300])
plt.figure(fig)
# plt.savefig('loss_outlier.png', bbox_inches = 'tight');
plt.show()

Uniqueness under Different Loss Functions

boba_even = np.array([20, 21, 22, 29, 33, 35])
fig = plot_losses(boba_even)
plt.figure(fig)
#plt.savefig('loss_unique.png', bbox_inches = 'tight');
plt.show()

Bonus: MAE loss curve for SLR?

We saw earlier that the MSE (average L2) loss curve for Simple Linear Regression is smooth and parabolic in 3 dimensions. What does the MAE for SLR curve look like?

It's beyond the scope of this course, but for the curious, the below cell plots the mean absolute error (average L1 loss) on the boba dataset for a simple linear regression model . I don't plot the minimum because I don't know if there's a closed form solution (you could solve it numerically with techniques you learn in lab).

Since the boba dataset didn't have input, I've arbitrarily used "temperature of that day" as input. I made up the numbers (assume degrees Farenheit).

boba_df = pd.DataFrame({"sales": [20, 21, 22, 29, 33],
                        "temps": [40, 44, 55, 70, 85]},
                       columns=["temps", "sales"])
boba_df

	temps	sales
0	40	20
1	44	21
2	55	22
3	70	29
4	85	33

# Just run this cell
import itertools
import plotly.graph_objects as go

data_linear = boba_df

def mAe_linear(theta_0, theta_1, data_linear):
    data_x, data_y = data_linear.iloc[:,0], data_linear.iloc[:,1]
    return np.mean(np.array([np.abs(y - (theta_0+theta_1*x)) for x, y in zip(data_x, data_y)]), axis=0)

theta_0_values = np.linspace(-20, 20, 80)
theta_1_values = np.linspace(-50, 50, 80)
mAe_values = [mAe_linear(theta_0, theta_1, data_linear) \
              for theta_0, theta_1 in itertools.product(theta_0_values, theta_1_values)]
mAe_values = np.reshape(mAe_values, (len(theta_0_values), len(theta_1_values)), order='F')

# Optimal point
fig_lin_mae = go.Figure(data=[go.Surface(x=theta_0_values, y=theta_1_values, z=mAe_values,name='z=MAE')])

fig_lin_mae.update_layout(
    autosize=False,
    scene = dict(
                    xaxis_title='x=theta_0',
                    yaxis_title='y=theta_1',
                    zaxis_title='z=MAE'),
                    width=500,
                    margin=dict(r=20, b=10, l=10, t=10))
fig_lin_mae.show()

Now observe the difference between the MAE loss and the MSE loss for one application.

# Just run this cell

def mSe_linear(theta_0, theta_1, data_linear):
    data_x, data_y = data_linear.iloc[:,0], data_linear.iloc[:,1]
    return np.mean(np.array([(y - (theta_0+theta_1*x))**2 for x, y in zip(data_x, data_y)]), axis=0)

theta_0_values = np.linspace(-20, 20, 80)
theta_1_values = np.linspace(-50, 50, 80)
mSe_values = [mSe_linear(theta_0, theta_1, data_linear) \
              for theta_0, theta_1 in itertools.product(theta_0_values, theta_1_values)]
mSe_values = np.reshape(mSe_values, (len(theta_0_values), len(theta_1_values)), order='F')

# Optimal point
fig_lin_mae = go.Figure(data=[go.Surface(x=theta_0_values, y=theta_1_values, z=mSe_values,name='z=MSE')])

fig_lin_mae.update_layout(
    autosize=False,
    scene = dict(
                    xaxis_title='x=theta_0',
                    yaxis_title='y=theta_1',
                    zaxis_title='z=MSE'),
                    width=500,
                    margin=dict(r=20, b=10, l=10, t=10))
fig_lin_mae.show()

🦭 Dugongs Part 2

The content in the rest of this notebook will be covered just in slides.

The code below exists just so you can reproduce the figures in the slides.

Residual Plot

dugongs = pd.read_csv("data/dugongs.csv")
dugongs.head()

	Length	Age
0	1.77	1.5
1	1.80	1.0
2	1.85	1.5
3	1.87	1.5
4	2.02	2.5

yobs = dugongs["Age"]      # The true observations y
xs = dugongs["Length"]     # Needed for linear predictions

theta_1_hat = np.corrcoef(xs, yobs)[0, 1] * np.std(yobs) / np.std(xs)
theta_0_hat = np.mean(yobs) - theta_1_hat * np.mean(xs)
yhats_linear = theta_0_hat + theta_1_hat * xs

np.corrcoef(xs, yobs)[0, 1]

0.8296474554905716

The correlation coefficient is pretty high...but there's an issue.

Let's first plot the Dugong linear fit again. It doesn't look so bad if we see it here.

# In case we're in a weird style state
sns.set_theme()
adjust_fontsize(size=16)
%matplotlib inline

sns.scatterplot(x=xs, y=yobs)
# plt.savefig('dugong_scatter.pdf', bbox_inches = 'tight');
plt.show()

sns.scatterplot(x=xs, y=yobs)
plt.plot(xs, yhats_linear, color='red', lw=1)
# plt.savefig('dugong_line.pdf', bbox_inches = 'tight');
plt.show()

Let's further inspect by plotting residuals.

residuals = yobs - yhats_linear

sns.scatterplot(x=xs, y=residuals, color='red', lw=1)
plt.axhline(0, color='black')
plt.ylabel(r"$Age - \widehat{Age}$")
# plt.savefig('dugong_residuals.pdf', bbox_inches = 'tight');
plt.show()

Log transformation of y

We could fit a line to the linear model that relates $ z = \log(y)$ to $x$:

$$ \Large \hat{z} = \theta_0 + \theta_1 x $$

dugongs["Log(Age)"] = np.log(dugongs["Age"])
zobs = dugongs["Log(Age)"]      # The LOG of true observations y
xs = dugongs["Length"]     # Needed for linear predictions

ztheta_1_hat = np.corrcoef(xs, zobs)[0, 1] * np.std(zobs) / np.std(xs)
ztheta_0_hat = np.mean(zobs) - ztheta_1_hat * np.mean(xs)
zhats_linear = ztheta_0_hat + ztheta_1_hat * xs

sns.scatterplot(x=xs, y=zobs)
# plt.savefig('dugong_zscatter.pdf', bbox_inches = 'tight');

sns.scatterplot(x=xs, y=zobs)
plt.plot(xs, zhats_linear, color='red', lw=1)
# plt.savefig('dugong_zline.pdf', bbox_inches = 'tight');
plt.show()

zresiduals = zobs - zhats_linear

sns.scatterplot(x=xs, y=zresiduals, color='red', lw=4)
plt.axhline(0, color='black')
plt.ylabel(r"$\log(Age) - \widehat{\log(Age)}$")
# plt.savefig('dugong_zresiduals.pdf', bbox_inches = 'tight');
plt.show()

Map back to the original coordinates

$$ \begin{align*} \hat{z} &= \theta_0 + \theta_1 x\\ \widehat{\log(y)}&= \theta_0 + \theta_1 x\\ e^{\widehat{\log(y)}}&= e^{\theta_0 + \theta_1 x}\\ \hat{y}&=e^{\theta_0 + \theta_1 x}\ \end{align*} $$

ypred = np.exp(zhats_linear)
sns.scatterplot(x=xs, y=yobs)
plt.plot(xs, ypred, color='red', lw=1)
# plt.savefig('dugong_zcurve.pdf', bbox_inches = 'tight');
plt.show()

Squaring transformation of x

We could fit a line to the linear model that relates $y$ to $x^2$:

$$ \Large y = \theta_0 + \theta_1 x^2 $$

Note: This is a bad way to include a quadratic term. To allow extra flexibility in the quadratic fit, we should model this with multiple regression as $y = \theta_0 + \theta_1 x + \theta_2 x^2$. So, this section is just for exposition.

dugongs["Length^2"] = dugongs["Length"] ** 2
xs = dugongs["Length^2"]   # Needed for linear predictions

quad_theta_1_hat = np.corrcoef(xs, yobs)[0, 1] * np.std(yobs) / np.std(xs)
quad_theta_0_hat = np.mean(yobs) - quad_theta_1_hat * np.mean(xs)
yhats_quad = quad_theta_0_hat + quad_theta_1_hat * xs

sns.scatterplot(x=xs, y=yobs)
plt.xlabel(r"$Length^2$")
# plt.savefig('dugong_quad_scatter.pdf', bbox_inches = 'tight');

sns.scatterplot(x=xs, y=yobs)

# Need more granularity or plotted line looks curved
xrange = np.linspace(xs.min(), xs.max(), 1000)
quad_preds_range = quad_theta_0_hat + quad_theta_1_hat * xrange
plt.plot(xrange, quad_preds_range, color='red', lw=1)

plt.xlabel(r"$Length^2$")
# plt.savefig('dugong_quad_line.pdf', bbox_inches = 'tight');
plt.show()

quad_residuals = yobs - quad_hats_linear

sns.scatterplot(x=xs, y=quad_residuals, color='red', lw=4)
plt.axhline(0, color='black')
plt.ylabel(r"$Age - \widehat{Age}$")
plt.xlabel(r"$Length^2$")
# plt.savefig('dugong_quad_residuals.pdf', bbox_inches = 'tight');
plt.show()

xs = dugongs["Length"]  
sns.scatterplot(x=xs, y=yobs)
plt.plot(xs, yhats_quad, color='red', lw=1)
# plt.savefig('dugong_quad_curve.pdf', bbox_inches = 'tight');
plt.show()