# Set up packages for lecture. Don't worry about understanding this code,
# but  sure to run it if you're following along.
import numpy as np
import babypandas as bpd
import pandas as pd
from matplotlib_inline.backend_inline import set_matplotlib_formats
import matplotlib.pyplot as plt
from scipy import stats
set_matplotlib_formats("svg")
plt.style.use('ggplot')

np.set_printoptions(threshold=20, precision=2, suppress=True)
pd.set_option("display.max_rows", 7)
pd.set_option("display.max_columns", 8)
pd.set_option("display.precision", 2)

import warnings
warnings.filterwarnings('ignore')

# New minimize function (wrapper around scipy.optimize.minimize)
from inspect import signature
from scipy import optimize

def minimize(function):
    n_args = len(signature(function).parameters)
    initial = np.zeros(n_args)
    return optimize.minimize(lambda x: function(*x), initial).x

# All of the following code is for visualization.
def plot_regression_line(df, x, y, margin=.02):
    '''Computes the slope and intercept of the regression line between columns x and y in df (in original units) and plots it.'''
    m = slope(df, x, y)
    b = intercept(df, x, y)
    
    df.plot(kind='scatter', x=x, y=y, s=100, figsize=(10, 5), label='original data')
    left = df.get(x).min()*(1 - margin)
    right = df.get(x).max()*(1 + margin)
    domain = np.linspace(left, right, 10)
    plt.plot(domain, m*domain + b, color='orange', label='regression line', lw=4)
    plt.suptitle(format_equation(m, b), fontsize=18)
    plt.legend();
    
def format_equation(m, b):
    if b > 0:
        return r'$y = %.2fx + %.2f$' % (m, b)
    elif b == 0:
        return r'$y = %.2fx' % m
    else:
        return r'$y = %.2fx %.2f$' % (m, b)
    
def plot_errors(df, m, b, ax=None):
    x = df.get('x')
    y = m * x + b
    df.plot(kind='scatter', x='x', y='y', s=100, label='original data', ax=ax, figsize=(10, 5) if ax is None else None)
    
    if ax:
        plotter = ax
    else:
        plotter = plt
    
    plotter.plot(x, y, color='orange', lw=4)
    
    for k in np.arange(df.shape[0]):
        xk = df.get('x').iloc[k]
        yk = np.asarray(y)[k]
        if k == df.shape[0] - 1:
            plotter.plot([xk, xk], [yk, df.get('y').iloc[k]], linestyle=(0, (1, 1)), c='r', lw=4, label='errors')
        else:
            plotter.plot([xk, xk], [yk, df.get('y').iloc[k]], linestyle=(0, (1, 1)), c='r', lw=4)
    
    plt.title(format_equation(m, b), fontsize=18)
    plt.xlim(50, 90)
    plt.ylim(40, 100)
    plt.legend();

Lecture 25 – Regression and Least Squares

DSC 10, Fall 2023

Announcements

• Quiz 4 scores are released, and the Grade Report has been updated – it reflects your scores on everything other than Homework 6, Lab 7, the Final Project, and the Final Exam.
• Lab 7 is due Thursday 12/7 at 11:59 PM.
  • It is true that your lowest lab is dropped.
  • However, it's a bad idea to simply ignore this lab, because it's the only assignment on regression, which will be tested on the Final Exam.
• Homework 6 is due today at 11:59PM.
• The Final Project is due Tuesday 12/5 at 11:59PM.
• The Final Exam is on Saturday 12/9 from 7-10PM.
  • We'll send out more details early next week, including your room and seat assignment.
  • Start preparing by working through old Final Exams at practice.dsc10.com. Take them as if they are real exams – time yourself, and don't use resources other than the reference sheet.
  • We'll work through the solutions to the Spring 2023 Final Exam in class on Wednesday.

Agenda

• The regression line in standard units.
• The regression line in original units.
• Outliers.
• Errors in prediction.

The regression line in standard units

Example: Predicting heights 👪 📏

Recall, in the last lecture, we aimed to use a mother's height to predict her adult son's height.

galton = bpd.read_csv('data/galton.csv')
male_children = galton[galton.get('gender') == 'male']
mom_son = bpd.DataFrame().assign(mom = male_children.get('mother'), 
                                 son = male_children.get('childHeight'))
mom_son

	mom	son
0	67.0	73.2
4	66.5	73.5
5	66.5	72.5
...	...	...
925	60.0	66.0
929	66.0	64.0
932	63.0	66.5

481 rows × 2 columns

mom_son.plot(kind='scatter', x='mom', y='son', figsize=(10, 5));

Correlation

Recall, the correlation coefficient $r$ of two variables $x$ and $y$ is defined as the

• average value of the
• product of $x$ and $y$
• when both are measured in standard units.

def standard_units(col):
    return (col - col.mean()) / np.std(col)

def calculate_r(df, x, y):
    '''Returns the average value of the product of x and y, 
       when both are measured in standard units.'''
    x_su = standard_units(df.get(x))
    y_su = standard_units(df.get(y))
    return (x_su * y_su).mean()

r_mom_son = calculate_r(mom_son, 'mom', 'son')
r_mom_son

0.32300498368490554

The regression line

• The regression line is the line through $(0,0)$ with slope $r$, when both variables are measured in standard units.

• We use the regression line to make predictions!

• Example: If a mother's height is 0.5 SDs above the average mother's height, and $r = 0.32$, our prediction is that her son's height will be $0.5 \cdot 0.32 = 0.16$ SDs above the average son's height.

• Issue: To use this form of the regression line, we need to know mothers' heights in standard units, but it would be more convenient to think in terms of inches.

Concept Check ✅ – Answer at cc.dsc10.com

A course has a midterm (mean 80, standard deviation 15) and a really hard final (mean 50, standard deviation 12).

If the scatter plot comparing midterm & final scores for students looks linearly associated with correlation 0.75, then what is the predicted final exam score for a student who received a 90 on the midterm?

• A. 54
• B. 56
• C. 58
• D. 60
• E. 62

The regression line in original units

Reflection

Each time we want to predict the height of an adult son given the height of his mother, we have to:

1. Convert the mother's height from inches to standard units.

1. Multiply by the correlation coefficient to predict the son's height in standard units.

1. Convert the son's predicted height from standard units back to inches.

This is inconvenient – wouldn't it be great if we could express the regression line itself in inches?

From standard units to original units

When $x$ and $y$ are in standard units, the regression line is given by

What is the regression line when $x$ and $y$ are in their original units (e.g. inches)?

The regression line in original units

• We can work backwards from the relationship $$\text{predicted } y_{\text{(su)}} = r \cdot x_{\text{(su)}}$$ to find the line in original units.

$$\frac{\text{predicted } y - \text{mean of }y}{\text{SD of }y} = r \cdot \frac{x - \text{mean of } x}{\text{SD of }x}$$

• Note that $r, \text{mean of } x$, $\text{mean of } y$, $\text{SD of } x$, and $\text{SD of } y$ are constants – if you have a DataFrame with two columns, you can determine all 5 values.

• Re-arranging the above equation into the form $\text{predicted } y = mx + b$ yields the formulas:

$$\boxed{m = r \cdot \frac{\text{SD of } y}{\text{SD of }x}, \: \: b = \text{mean of } y - m \cdot \text{mean of } x}$$

• $m$ is the slope of the regression line and $b$ is the intercept.

Let's implement these formulas in code and try them out.

def slope(df, x, y):
    "Returns the slope of the regression line between columns x and y in df (in original units)."
    r = calculate_r(df, x, y)
    return r * np.std(df.get(y)) / np.std(df.get(x))

def intercept(df, x, y):
    "Returns the intercept of the regression line between columns x and y in df (in original units)."
    return df.get(y).mean() - slope(df, x, y) * df.get(x).mean()

Below, we compute the slope and intercept of the regression line between mothers' heights and sons' heights (in inches).

m_heights = slope(mom_son, 'mom', 'son')
m_heights

0.36506116024257595

b_heights = intercept(mom_son, 'mom', 'son')
b_heights

45.85803797199311

So, the regression line is

$$\text{predicted son's height in inches} = 0.365 \cdot \text{mother's height in inches} + 45.858$$

Making predictions

def predict_son(mom):
    return m_heights * mom + b_heights

What's the predicted height of a son whose mother is 62 inches tall?

predict_son(62)

68.49182990703282

What if the mother is 55 inches tall? 73 inches tall?

predict_son(55)

65.9364017853348

predict_son(73)

72.50750266970115

xs = np.arange(57, 72)
ys = predict_son(xs)
mom_son.plot(kind='scatter', x='mom', y='son', figsize=(10, 5), title='Regression line predictions, in original units', label='original data');
plt.plot(xs, ys, color='orange', lw=4, label='regression line')
plt.legend();

Outliers

The effect of outliers on correlation

Consider the dataset below. What is the correlation between $x$ and $y$?

outlier = bpd.read_csv('data/outlier.csv')
outlier.plot(kind='scatter', x='x', y='y', s=100, figsize=(10, 5));

calculate_r(outlier, 'x', 'y')

-0.02793982443854457

plot_regression_line(outlier, 'x', 'y')

Removing the outlier

without_outlier = outlier[outlier.get('y') > 40]

calculate_r(without_outlier, 'x', 'y')

0.9851437295364016

plot_regression_line(without_outlier, 'x', 'y')

Takeaway: Even a single outlier can have a massive impact on the correlation, and hence the regression line. Look for these before performing regression. Always visualize first!

Errors in prediction

Motivation

• We've presented the regression line in standard units as the line through the origin with slope $r$, given by $\text{predicted } y_{\text{(su)}} = r \cdot x_{\text{(su)}}$. Then, we used this equation to find a formula for the regression line in original units.

• In examples we've seen so far, the regression line seems to fit our data pretty well.
  • But how well?
  • What makes the regression line good?
  • Would another line be better?

Example: Without the outlier

outlier.plot(kind='scatter', x='x', y='y', s=100, figsize=(10, 5));

m_no_outlier = slope(without_outlier, 'x', 'y')
b_no_outlier = intercept(without_outlier, 'x', 'y')

m_no_outlier, b_no_outlier

(0.975927715724588, 3.042337135297444)

plot_errors(without_outlier, m_no_outlier, b_no_outlier)

We think our regression line is pretty good because most data points are pretty close to the regression line. The red lines are quite short.

Measuring the error in prediction

$$\text{error} = \text{actual value} - \text{prediction}$$

• A good prediction line is one where the errors tend to be small.

• To measure the rough size of the errors, for a particular set of predictions:
  1. Square the errors so that they don't cancel each other out.
  2. Take the mean of the squared errors.
  3. Take the square root to fix the units.

• This is called root mean square error (RMSE).
  • Notice the similarities to computing the SD!

Root mean squared error (RMSE) of the regression line's predictions

without_outlier

	x	y
0	50	53.53
1	55	54.21
2	60	65.65
...	...	...
6	80	79.61
7	85	88.17
8	90	91.05

9 rows × 2 columns

First, let's compute the regression line's predictions for the entire dataset.

predicted_y = m_no_outlier * without_outlier.get('x') + b_no_outlier
predicted_y

array([51.84, 56.72, 61.6 , 66.48, 71.36, 76.24, 81.12, 86.  , 90.88])

To find the RMSE, we need to start by finding the errors and squaring them.

# Errors.
without_outlier.get('y') - predicted_y

0    1.69
1   -2.51
2    4.06
     ... 
6   -1.51
7    2.18
8    0.18
Name: y, Length: 9, dtype: float64

# Squared errors.
(without_outlier.get('y') - predicted_y) ** 2

0     2.86
1     6.31
2    16.45
     ...  
6     2.27
7     4.74
8     0.03
Name: y, Length: 9, dtype: float64

Now, we need to find the mean of the squared errors, and take the square root of that. The result is the RMSE of the regression line's predictions.

# Mean squared error.
((without_outlier.get('y') - predicted_y) ** 2).mean()

4.823770221019621

# Root mean squared error.
np.sqrt(((without_outlier.get('y') - predicted_y) ** 2).mean())

2.19630831647554

The RMSE of the regression line's predictions is about 2.2. Is this big or small, relative to the predictions of other lines? 🤔

Root mean squared error (RMSE) in an arbirtrary line's predictions

• We've been using the regression line to make predictions. But we could use a different line!
  • To make a prediction for x using an arbitrary line defined by slope and intercept, compute x * slope + intercept.

• For this dataset, if we choose a different line, we will end up with different predictions, and hence a different RMSE.

def rmse(slope, intercept):
    '''Calculates the RMSE of the line with the given slope and intercept, 
    using the 'x' and 'y' columns of without_outlier.'''

    # The true values of y.
    true = without_outlier.get('y')
    
    # The predicted values of y, from plugging the x values from the 
    # given DataFrame into the line with the given slope and intercept.
    predicted = slope * without_outlier.get('x') + intercept
    
    return np.sqrt(((true - predicted) ** 2).mean())

# Check that our function works on the regression line.
rmse(m_no_outlier, b_no_outlier)

2.19630831647554

Let's compute the RMSEs of several different lines on the same dataset.

# Experiment by changing one of these!
lines = [(1.2, -15), (0.75, 11.5), (-0.4, 100)]

fig, ax = plt.subplots(1, 3, figsize=(14, 4))
for i, line in enumerate(lines):
    plt.subplot(1, 3, i + 1)
    m, b = line
    plot_errors(without_outlier, m, b, ax=ax[i])
    ax[i].set_title(format_equation(m, b) + f'\nRMSE={np.round(rmse(m, b), 2)}')

Finding the "best" prediction line by minimizing RMSE

• RMSE describes how well a line fits the data. The lower the RMSE of a line is, the better it fits the data.

• There are infinitely many slopes and intercepts, and thus infinitely many RMSEs. How do we find which combination of slope and intercept have the lowest RMSE?

• If you take DSC 40A, you'll learn how to do this using calculus. For now, we'll use a Python function that can do it automatically – minimize.

Aside: minimize

• The function minimize takes in a function as an argument, and returns the inputs to that function that produce the smallest output.

• For instance, we know that the minimizing input to the function $f(x) = (x - 5)^2 + 4$ is $x = 5$. minimize can find this, too:

def f(x):
    return (x - 5) ** 2 + 4

# Plot of f(x).
x = np.linspace(0, 10)
y = f(x)
plt.plot(x, y)
plt.title(r'$f(x) = (x - 5)^2 + 4$');