Project: Linear Regression
Reggie is a mad scientist who has been hired by the local fast food joint to build their newest ball pit in the play area. As such, he is working on researching the bounciness of different balls so as to optimize the pit. He is running an experiment to bounce different sizes of bouncy balls, and then fitting lines to the data points he records. He has heard of linear regression, but needs your help to implement a version of linear regression in Python.
Linear Regression is when you have a group of points on a graph, and you find a line that approximately resembles that group of points. A good Linear Regression algorithm minimizes the error, or the distance from each point to the line. A line with the least error is the line that fits the data the best. We call this a line of best fit.
We will use loops, lists, and arithmetic to create a function that will find a line of best fit when given a set of data.
Part 1: Calculating Error
The line we will end up with will have a formula that looks like:
y = m*x + b
m
is the slope of the line and b
is the intercept, where the line crosses the yaxis.
Fill in the function called get_y()
that takes in m
, b
, and x
. It should return what the y
value would be for that x
on that line!
def get_y(m, b, x):
return m*x+b
print(get_y(1, 0, 7) == 7)
print(get_y(5, 10, 3) == 25)
True
True
Reggie wants to try a bunch of different m
values and b
values and see which line produces the least error. To calculate error between a point and a line, he wants a function called calculate_error()
, which will take in m
, b
, and an [x, y] point called point
and return the distance between the line and the point.
To find the distance:
 Get the xvalue from the point and store it in a variable called
x_point
 Get the yvalue from the point and store it in a variable called
y_point
 Use
get_y()
to get the yvalue thatx_point
would be on the line  Find the difference between the y from
get_y
andy_point
 Return the absolute value of the distance (you can use the builtin function
abs()
to do this)
The distance represents the error between the line y = m*x + b
and the point
given.
def calculate_error(m, b, point):
x_point = point[0]
y_point = point[1]
y_line = get_y(m, b, x_point)
return abs(y_liney_point)
Let’s test this function!
#this is a line that looks like y = x, so (3, 3) should lie on it. thus, error should be 0:
print(calculate_error(1, 0, (3, 3)))
#the point (3, 4) should be 1 unit away from the line y = x:
print(calculate_error(1, 0, (3, 4)))
#the point (3, 3) should be 1 unit away from the line y = x  1:
print(calculate_error(1, 1, (3, 3)))
#the point (3, 3) should be 5 units away from the line y = x + 1:
print(calculate_error(1, 1, (3, 3)))
0
1
1
5
Great! Reggie’s datasets will be sets of points. For example, he ran an experiment comparing the width of bouncy balls to how high they bounce:
datapoints = [(1, 2), (2, 0), (3, 4), (4, 4), (5, 3)]
The first datapoint, (1, 2)
, means that his 1cm bouncy ball bounced 2 meters. The 4cm bouncy ball bounced 4 meters.
As we try to fit a line to this data, we will need a function called calculate_all_error
, which takes m
and b
that describe a line, and points
, a set of data like the example above.
calculate_all_error
should iterate through each point
in points
and calculate the error from that point to the line (using calculate_error
). It should keep a running total of the error, and then return that total after the loop.
def calculate_all_error(m, b, points):
total_error = 0
for point in points:
total_error += calculate_error(m, b, point)
return total_error
Let’s test this function!
#every point in this dataset lies upon y=x, so the total error should be zero:
datapoints = [(1, 1), (3, 3), (5, 5), (1, 1)]
print(calculate_all_error(1, 0, datapoints))
#every point in this dataset is 1 unit away from y = x + 1, so the total error should be 4:
datapoints = [(1, 1), (3, 3), (5, 5), (1, 1)]
print(calculate_all_error(1, 1, datapoints))
#every point in this dataset is 1 unit away from y = x  1, so the total error should be 4:
datapoints = [(1, 1), (3, 3), (5, 5), (1, 1)]
print(calculate_all_error(1, 1, datapoints))
#the points in this dataset are 1, 5, 9, and 3 units away from y = x + 1, respectively, so total error should be
# 1 + 5 + 9 + 3 = 18
datapoints = [(1, 1), (3, 3), (5, 5), (1, 1)]
print(calculate_all_error(1, 1, datapoints))
0
4
4
18
Great! It looks like we now have a function that can take in a line and Reggie’s data and return how much error that line produces when we try to fit it to the data.
Our next step is to find the m
and b
that minimizes this error, and thus fits the data best!
Part 2: Try a bunch of slopes and intercepts!
The way Reggie wants to find a line of best fit is by trial and error. He wants to try a bunch of different slopes (m
values) and a bunch of different intercepts (b
values) and see which one produces the smallest error value for his dataset.
Using a list comprehension, let’s create a list of possible m
values to try. Make the list possible_ms
that goes from 10 to 10 inclusive, in increments of 0.1.
Hint (to view this hint, either doubleclick this cell or highlight the following white space): you can go through the values in range(100, 100) and multiply each one by 0.1
possible_ms = [i/1010 for i in range(201)]
Now, let’s make a list of possible_bs
to check that would be the values from 20 to 20 inclusive, in steps of 0.1:
possible_bs = [i/1020 for i in range(401)]
We are going to find the smallest error. First, we will make every possible y = m*x + b
line by pairing all of the possible m
s with all of the possible b
s. Then, we will see which y = m*x + b
line produces the smallest total error with the set of data stored in datapoint
.
First, create the variables that we will be optimizing:

smallest_error
— this should start at infinity (float("inf")
) so that any error we get at first will be smaller than our value ofsmallest_error

best_m
— we can start this at0

best_b
— we can start this at0
We want to:
 Iterate through each element
m
inpossible_ms
 For every
m
value, take everyb
value inpossible_bs
 If the value returned from
calculate_all_error
on thism
value, thisb
value, anddatapoints
is less than our currentsmallest_error
,  Set
best_m
andbest_b
to be these values, and setsmallest_error
to this error.
By the end of these nested loops, the smallest_error
should hold the smallest error we have found, and best_m
and best_b
should be the values that produced that smallest error value.
Print out best_m
, best_b
and smallest_error
after the loops.
datapoints = [(1, 2), (2, 0), (3, 4), (4, 4), (5, 3)]
smallest_error = float("inf")
best_m = 0
best_b = 0
for m in possible_ms:
for b in possible_bs:
error = calculate_all_error(m, b, datapoints)
if error < smallest_error:
best_m = m
best_b = b
smallest_error = error
print(best_m, best_b, smallest_error)
0.3000000000000007 1.6999999999999993 5.0
Part 3: What does our model predict?
Now we have seen that for this set of observations on the bouncy balls, the line that fits the data best has an m
of 0.3 and a b
of 1.7:
y = 0.3x + 1.7
This line produced a total error of 5.
Using this m
and this b
, what does your line predict the bounce height of a ball with a width of 6 to be?
In other words, what is the output of get_y()
when we call it with:
 m = 0.3
 b = 1.7
 x = 6
print(get_y(0.3, 1.7, 6))
3.5
Our model predicts that the 6cm ball will bounce 3.5m.
Now, Reggie can use this model to predict the bounce of all kinds of sizes of balls he may choose to include in the ball pit!
def get_bounce_height(x):
return get_y(0.3, 1.7, x)
print(get_bounce_height(6))
3.5