Exercise A: basic python

Exercise A: basic python#

a = 0
# Run this cell using 'Shift'+'Enter

b = 0 # this is a markdown cell, change it to code (‘Esc’ then ‘y’) and execute (‘Ctrl’+’Enter’)

# Insert a cell below this one ('Esc' then 'b'), 
# write a=1 and execute with 'Shift'+'Enter' (which will run and move to next) 

# Delete this cell ('Esc' then 'd' then 'd')

### Change this code cell to markdown ('Esc' then 'm') and execute with 'Shift'+'Enter'

# Create variable called var_a with the value 2.0

# Now change the value of the variable var_a to "my_string"

# check the type of the variable var_a by using the type() function

# Create a list called LA with three elements: 1.0, "txt" and 3

# Create a variable LB and set it equal to LA

# Change the first element of LB to "b"

# Print out the first element of LA. Did it change? Why?

# Append a new element None to the list LA

# Print the list LB

# Create a new list LC, set it to the sum of LA and LB and print the result

# Create a tuple T with two elements "1" and 2.0

# Create two variables t1, t2 and set them to the first and second element of T

# Change the first element to 1.0. What happens? Why?

# Define an empty dictionary D with dict()

# add a key-value pair using D[key] = value, with your initials and 
# full name as key and value

# add a another key-value pair with initials and name of a colleague

# Print the value of the first item using square brackets []

# Define the same dictionary an alternative way using D={k1:v1, k2:v2}

# print all the keys using the keys() method on the D object, D.___()

# print all the items with with items()

v1 = 4.0
v2 = 5.0
# write the code that will print the statement "great" if v2 is greater than v1

# write the code that will do the same as above but also has an else statement "not so great"

L = ["Hello", "World"]
# write a for loop that loops through the elements of L and prints them