• Create a new variable, \(x = 5\)
• Compute \(\pi x^2\)
• Compute \(\frac{1}{\sqrt{2\pi}}e^{-x^2}\)
  • This is the probability density function of the standard normal distribution (don’t worry if you don’t know what it is yet! 🤓)
• Don’t forget how to exponentiate in Python 😉

x = 10
x ** 5

# Not x^5

100000

• Appendix 01

• NumPy arrays are created using the np.array() function
• We can create arrays from lists
• We can also create arrays with a sequence of numbers using np.arange()
• We can create arrays with zeros or ones using np.zeros() and np.ones()
• And we’re going to learn all of this in this lecture! 🤓

• We can access elements of an array using square brackets []

• Remember that Python is zero-indexed

• Access the first and the third element of \(a\)

print(vec_a)
print(vec_a[0])
print(vec_a[2])

[1 2 3]
1
3

\(a + b = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} +\) \(\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} =\) \(\begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}\)

print(vec_a)
print(vec_b)

# When you add two arrays of the same size,
# Python adds the individual elements in each position
print(vec_a + vec_b)

[1 2 3]
[0 1 0]
[1 3 3]

\(a * b = \begin{pmatrix} a_1 * b_1 \\ a_2 * b_2 \\ a_3 * b_3 \end{pmatrix}\)

print(vec_a)
print(vec_b)

# When you multiply two arrays of the same size,
# Python multiplies the individual elements in each position
print(vec_a * vec_b)

# We can do other similar element-by-element operations
# such as subtraction, and division.
print(vec_a - vec_b)
print(vec_a / vec_b)

[1 2 3]
[0 1 0]
[0 2 0]
[1 1 3]
[inf  2. inf]

• NumPy provides several functions to compute summary statistics of an array
• For instance, we can compute the mean, median, standard deviation, variance, minimum, and maximum
• We can also compute the sum, product, and cumulative sum

print(np.mean(vec_a))
print(np.std(vec_a))
print(np.min(vec_a))
print(np.median(vec_a))
print(np.max(vec_a))

2.0
0.816496580927726
1
2.0
3

• Compute the mean of:

\(e = \begin{pmatrix} 10 \\ 8 \\ 15 \\ 0 \\ 24 \end{pmatrix}\)

• Appendix 02

• Avoid this problem by drawing from a “pregenerated” set.
• This is called a seed, and it is set using np.random.seed()
• This allows for reproducibility of results

np.random.seed(151)

random_var_b = np.random.normal(loc=0, scale=1, size=10)
print(random_var_b)

[-0.63673759  0.53155853  0.99020835 -0.6241344   1.46778078  0.40501276
  1.29817371 -2.61363271  1.35643373  1.87316055]

• We can use the plt.hist() function to compute a histogram

# Create a histogram of the random variable
randomvar_x = np.random.normal(loc=0, scale=1, size=1000)

plt.hist(x = randomvar_x)
plt.xlabel("Variable a")
plt.ylabel("Frequency")
plt.title("Histogram of random variable a")
plt.show()

• Try this again with \(size = 100, 1000, 10000\) and see how the histogram changes

• We can test whether two values are equal using the == operator (two equal signs)

• For example, 5 == 5 returns True

• We can also test whether two values are not equal using the != operator (exclamation mark followed by an equal sign, rendered as != here)

• Let’s see how this works in Python

• First, let’s load the matplotlib and numpy libraries

import matplotlib.pyplot as plt
import numpy as np

# You can compare two strings by using a double "equal sign"
# This can be useful if you're trying to evaluate whether data was entered correctly

"Is this the real life?" == "is this just fantasy?" 

False

Note: the order of the strings matter!

"ab" == "ba" 

False

• Equality of strings is most useful when you’re comparing an unknown variable to a benchmark

• Below, try switching the value of any_questions

any_questions = "no"
print(any_questions == "no")

True

any_questions = "yes" 
print(any_questions == "no")

False

• We can test whether a keyword is present in a sentence or a list using the in operator
• For example, "apple" in "I like apples" returns True
• Let’s see how this works in Python

"apple" in "I like apples"

True

keyword = "economic"
sentence = "The Federal Reserve makes forecasts about many economic outcomes"

keyword in sentence

# Try changing the keyword!

True

current_month = "September"
list_summer_months = ["June","July","August"]

print(current_month in list_summer_months)
print('June' in list_summer_months)

False
True

• Tests with numbers
  • Strictly less than (<), less than or equal (<=)
  • Equal (==)
  • Strictly more than (>), greater than or equal to (>=)
• Let’s see how this works in Python

x = 5

print(x < 5)
print(x <= 5)
print(x == 5)
print(x >= 5)
print(x > 5)

False
True
True
True
False

• We can test whether a variable is of a certain data type using the isinstance() function
• For example, isinstance(5, int) returns True
• Other data types include float, str, list, tuple, dict, set, bool

y = 10

print(isinstance(y,int))
print(isinstance(y,float))
print(isinstance(y,str))

True
False
False

• We can test whether two vectors are equal using the == operator
• For example, [1,2,3] == [1,2,3] returns True
• Please note that the equality of vectors is done element-wise

vec_a = np.array([1,2,3])
vec_b = np.array([1,2,4])

vec_a == vec_b

array([ True,  True, False])

• Define \(x= -1\). Check whether \(x^2 + 2x + 1 = 0\) is true
  • Please remember that \(x^2\) is written as x**2 in Python
  • Please note the difference between == and =
• Appendix 03

• We can test multiple conditions using the & operator
• The & operator returns True if both conditions are True
• For example, 5 > 3 & 5 < 10 returns True

# We use the "&" symbol to separate "AND" conditions
age = 31

# Is this age between 20 and 30 (including these ages)?
(age >= 20) & (age <= 30)

False

• To test whether at least one condition is True, we use the | operator
• For example, 5 > 3 | 5 > 10 returns True

# We use the "|" symbol to separate "OR" conditions.
age = 31

# Is this age higher than 20 or lower than 30?
(age >= 20) | (age <= 30)

True

# Another example
student_status = "freshman" 

# Is the student in the first two years of undergrad?
(student_status == "freshman") | (student_status == "sophomore")

True

• Now, let’s test whether you can identify the correct expression

• Create a new variable, \(x = 5\)
• Compute \(\pi x^2\)
• Compute \(\frac{1}{\sqrt{2\pi}}e^{-x^2}\)

x = 5
print(np.pi * x ** 2)
print(1 / np.sqrt(2 * np.pi) * np.exp(-x ** 2))

78.53981633974483
5.540487995575833e-12

Back to the main text

• Compute the mean of
• \(e = \begin{pmatrix} 10 \\ 8 \\ 15 \\ 0 \\ 24 \end{pmatrix}\)

vec_e = np.array([10, 8, 15, 0, 24])
print(np.mean(vec_e))

11.4

Back to the main text

• Whether \(x = -1\) is a solution to the equation \(x^2 + 2x + 1 = 0\)
• Please remember that \(x^2\) is written as x**2 in Python

x = -1
print(x ** 2 + 2 * x + 1 == 0)

True

Back to the main text

• Whether age (age = 31) is strictly less than 20, or greater than 30
• Not in the age range 25-27

age = 31

(age < 20) | (age > 30) 

True

(age < 25) | (age > 27)

True

The second answer uses | because & evaluates both statements at the same time, and one cannot be less than 25 and greater than 27 at the same time. Therefore, it must be |.

Back to the main text