norm() Explaination

In your knn_single_prediction function for task 1, you are asked to use the np.linalg.norm(). I have received quite a few questions about what this is doing, and how to properly use it in the program, so I will go over that here.

As the instructions state, you are trying to calculate the ‘distance’ away your new_example input is from all of the test data.

You may recall from a prior math or physics class where you calcualted distance between two points using a “distance formula” that likely looked something like : \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

if (x1,y1) is (0,0), you may be able to recognize this as solving for the hypotenuse of a right triangle using the Pythagorean theorem.

In calc2 and later physics classes, you will use a similar function to determine the magnitude of vectors.

In Calc3, you will learn you can use this same thing for finding the distance between two points in 3 dimensions too (and as many as you may want): \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2+ (z_2 - z_1)^2}\)

Per the original function name, this process is called normalization. ‘normalizing’ values calculated the magnitude of the values provided. You can provide 2 values, 3 values, even dozens of values if you wanted to, and the function will add up the square of all of the values, and then take the square root to provide a normalized value.

Numpy arrays make this go very quickly, as they have a normalization function already. You can replicate this using standard numpy operations too.

import numpy as np
# test points.  Feel free to test out multiple sets
p1 = np.array([1,5])
p2 = np.array([3,6])

#simmilar to what you would get from your csv file
#p1 = np.array([77.5901, 74.89, 87.53, 76.63, 16.03, 17.25, 0, 0])
#p2 = np.array([42.2627, 68.19, 101.12, 90.50, 112.23, 107.99, 25, 1])


# take the differences of the array
d1 = p2- p1

norm= np.linalg.norm(d1)

dist= np.sqrt(sum(d1**2))

print(dist)
print(norm)

2.23606797749979
2.23606797749979