# HIDDEN
import warnings
# Ignore numpy dtype warnings. These warnings are caused by an interaction
# between numpy and Cython and can be safely ignored.
# Reference: https://stackoverflow.com/a/40846742
warnings.filterwarnings("ignore", message="numpy.dtype size changed")
warnings.filterwarnings("ignore", message="numpy.ufunc size changed")

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import seaborn as sns
%matplotlib inline
import ipywidgets as widgets
from ipywidgets import interact, interactive, fixed, interact_manual
import nbinteract as nbi

sns.set()
sns.set_context('talk')
np.set_printoptions(threshold=20, precision=2, suppress=True)
pd.options.display.max_rows = 7
pd.options.display.max_columns = 8
pd.set_option('precision', 2)
# This option stops scientific notation for pandas
# pd.set_option('display.float_format', '{:.2f}'.format)
# HIDDEN
tips = sns.load_dataset('tips')
# HIDDEN
def minimize(loss_fn, grad_loss_fn, x_vals, y_vals,
             alpha=0.0005, progress=True):
    '''
    Uses gradient descent to minimize loss_fn. Returns the minimizing value of
    theta once the loss changes less than 0.0001 between iterations.
    '''
    theta = np.array([0., 0.])
    loss = loss_fn(theta, x_vals, y_vals)
    while True:
        if progress:
            print(f'theta: {theta} | loss: {loss}')
        gradient = grad_loss_fn(theta, x_vals, y_vals)
        new_theta = theta - alpha * gradient
        new_loss = loss_fn(new_theta, x_vals, y_vals)
        
        if abs(new_loss - loss) < 0.0001:
            return new_theta
        
        theta = new_theta
        loss = new_loss

13.2. Fitting a Linear Model Using Gradient Descent

We want to fit a linear model that predicts the tip amount based on the total bill of the table:

\[ f_\boldsymbol\theta (x) = \theta_1 x + \theta_0 \]

In order to optimize \( \theta_1 \) and \( \theta_0 \), we need to first choose a loss function. We will choose the mean squared error loss function:

\[\begin{split} \begin{aligned} L(\boldsymbol\theta, \textbf{x}, \textbf{y}) &= \frac{1}{n} \sum_{i = 1}^{n}(y_i - f_\boldsymbol\theta (x_i))^2\\ \end{aligned} \end{split}\]

Note that we have modified our loss function to reflect the addition of an explanatory variable in our new model. Now, \( \textbf{x} \) is a vector containing the individual total bills, \( \textbf{y} \) is a vector containing the individual tip amounts, and \( \boldsymbol\theta \) is a vector: \( \boldsymbol\theta = [ \theta_1, \theta_0 ] \).

Using a linear model with the squared error also goes by the name of least-squares linear regression. We can use gradient descent to find the \( \boldsymbol\theta \) that minimizes the loss.

An Aside on Using Correlation

If you have seen least-squares linear regression before, you may recognize that we can compute the correlation coefficient and use it to determine \( \theta_1 \) and \( \theta_0 \). This is simpler and faster to compute than using gradient descent for this specific problem, similar to how computing the mean was simpler than using gradient descent to fit a constant model. We will use gradient descent anyway because it is a general-purpose method of loss minimization that still works when we later introduce models that do not have analytic solutions. In fact, in many real-world scenarios, we will use gradient descent even when an analytic solution exists because computing the analytic solution can take longer than gradient descent, especially on large datasets.

13.2.1. Derivative of the MSE Loss

In order to use gradient descent, we have to compute the derivative of the MSE loss with respect to \( \boldsymbol\theta \). Now that \( \boldsymbol\theta \) is a vector of length 2 instead of a scalar, \( \nabla_{\boldsymbol\theta} L(\boldsymbol\theta, \textbf{x}, \textbf{y}) \) will also be a vector of length 2.

\[\begin{split} \begin{aligned} \nabla_{\boldsymbol\theta} L(\boldsymbol\theta, \textbf{x}, \textbf{y}) &= \nabla_{\boldsymbol\theta} \left[ \frac{1}{n} \sum_{i = 1}^{n}(y_i - f_\boldsymbol\theta (x_i))^2 \right] \\ &= \frac{1}{n} \sum_{i = 1}^{n}2 (y_i - f_\boldsymbol\theta (x_i))(- \nabla_{\boldsymbol\theta} f_\boldsymbol\theta (x_i))\\ &= -\frac{2}{n} \sum_{i = 1}^{n}(y_i - f_\boldsymbol\theta (x_i))(\nabla_{\boldsymbol\theta} f_\boldsymbol\theta (x_i))\\ \end{aligned} \end{split}\]

We know:

\[ f_\boldsymbol\theta (x) = \theta_1 x + \theta_0 \]

We now need to compute \( \nabla_{\boldsymbol\theta} f_\boldsymbol\theta (x_i) \) which is a length 2 vector.

\[\begin{split} \begin{aligned} \nabla_{\boldsymbol\theta} f_\boldsymbol\theta (x_i) &= \begin{bmatrix} \frac{\partial}{\partial \theta_0} f_\boldsymbol\theta (x_i)\\ \frac{\partial}{\partial \theta_1} f_\boldsymbol\theta (x_i) \end{bmatrix} \\ &= \begin{bmatrix} \frac{\partial}{\partial \theta_0} [\theta_1 x_i + \theta_0]\\ \frac{\partial}{\partial \theta_1} [\theta_1 x_i + \theta_0] \end{bmatrix} \\ &= \begin{bmatrix} 1 \\ x_i \end{bmatrix} \\ \end{aligned} \end{split}\]

Finally, we plug back into our formula above to get

\[\begin{split} \begin{aligned} \nabla_{\boldsymbol\theta} L(\theta, \textbf{x}, \textbf{y}) &= -\frac{2}{n} \sum_{i = 1}^{n}(y_i - f_\boldsymbol\theta (x_i))(\nabla_{\boldsymbol\theta} f_\boldsymbol\theta (x_i))\\ &= -\frac{2}{n} \sum_{i = 1}^{n} (y_i - f_\boldsymbol\theta (x_i)) \begin{bmatrix} 1 \\ x_i \end{bmatrix} \\ &= -\frac{2}{n} \sum_{i = 1}^{n} \begin{bmatrix} (y_i - f_\boldsymbol\theta (x_i)) \\ (y_i - f_\boldsymbol\theta (x_i)) x_i \end{bmatrix} \\ \end{aligned} \end{split}\]

This is a length 2 vector since \( (y_i - f_\boldsymbol\theta (x_i)) \) is scalar.

13.2.2. Running Gradient Descent

Now, let’s fit a linear model on the tips dataset to predict the tip amount from the total table bill.

First, we define a Python function to compute the loss:

def simple_linear_model(thetas, x_vals):
    '''Returns predictions by a linear model on x_vals.'''
    return thetas[0] + thetas[1] * x_vals

def mse_loss(thetas, x_vals, y_vals):
    return np.mean((y_vals - simple_linear_model(thetas, x_vals)) ** 2)

Then, we define a function to compute the gradient of the loss:

def grad_mse_loss(thetas, x_vals, y_vals):
    n = len(x_vals)
    grad_0 = y_vals - simple_linear_model(thetas, x_vals)
    grad_1 = (y_vals - simple_linear_model(thetas, x_vals)) * x_vals
    return -2 / n * np.array([np.sum(grad_0), np.sum(grad_1)])
# HIDDEN
thetas = np.array([1, 1])
x_vals = np.array([3, 4])
y_vals = np.array([4, 5])
assert np.allclose(grad_mse_loss(thetas, x_vals, y_vals), [0, 0])

We’ll use the previously defined minimize function that runs gradient descent, accounting for our new explanatory variable. It has the function signature (body omitted):

minimize(loss_fn, grad_loss_fn, x_vals, y_vals)

Finally, we run gradient descent!

%%time

thetas = minimize(mse_loss, grad_mse_loss, tips['total_bill'], tips['tip'])
theta: [0. 0.] | cost: 10.896283606557377
theta: [0.   0.07] | cost: 3.8937622006094705
theta: [0.  0.1] | cost: 1.9359443267168215
theta: [0.01 0.12] | cost: 1.388538448286097
theta: [0.01 0.13] | cost: 1.235459416905535
theta: [0.01 0.14] | cost: 1.1926273731479433
theta: [0.01 0.14] | cost: 1.1806184944517062
theta: [0.01 0.14] | cost: 1.177227251696266
theta: [0.01 0.14] | cost: 1.1762453624313751
theta: [0.01 0.14] | cost: 1.1759370980989148
theta: [0.01 0.14] | cost: 1.175817178966766
CPU times: user 272 ms, sys: 67.3 ms, total: 339 ms
Wall time: 792 ms

We can see that gradient descent converges to the theta values of \(\hat\theta_0 = 0.01\) and \(\hat\theta_1 = 0.14\). Our linear model is:

\[y = 0.14x + 0.01\]

We can use our estimated thetas to make and plot our predictions alongside the original data points.

# HIDDEN
x_vals = np.array([0, 55])
sns.lmplot(x='total_bill', y='tip', data=tips, fit_reg=False)
plt.plot(x_vals, simple_linear_model(thetas, x_vals), c='goldenrod')
plt.title('Tip amount vs. Total Bill')
plt.xlabel('Total Bill')
plt.ylabel('Tip Amount');
../../_images/linear_grad_16_0.png

We can see that if a table’s bill is \(\$10\), our model will predict that the waiter gets around \(\$1.50\) in tip. Similarly, if a table’s bill is \(\$40\), our model will predict a tip of around \(\$6.00\).