sets in python

my_dict = {2:3, 5:6, 8:9}

new_dict = {}
for k, v in my_dict.items():
    new_dict[v] = k

x = {"apple", "banana", "cherry"}
y = {"google", "microsoft", "apple"}

z = x.intersection(y)
    
print(z)

# Output: {"google"}

>>> A = {0, 2, 4, 6, 8};
>>> B = {1, 2, 3, 4, 5};

>>> print("Union :", A | B)  
Union : {0, 1, 2, 3, 4, 5, 6, 8}

>>> print("Intersection :", A & B)
Intersection : {2, 4}

>>> print("Difference :", A - B)
Difference : {0, 8, 6}

# elements not present both sets
>>> print("Symmetric difference :", A ^ B)   
Symmetric difference : {0, 1, 3, 5, 6, 8}

# You can't create a set like this in Python
my_set = {} # ---- This is a Dictionary/Hashmap

# To create a empty set you have to use the built in method:
my_set = set() # Correct!


set_example = {1,3,2,5,3,6}
print(set_example)

# OUTPUT
# {1,3,2,5,6} ---- Sets do not contain duplicates and are unordered

# Program to perform different set operations 
# as we do in  mathematics 
  
# sets are define 
A = {0, 2, 4, 6, 8}; 
B = {1, 2, 3, 4, 5}; 
  
# union 
print("Union :", A | B) 
  
# intersection 
print("Intersection :", A & B) 
  
# difference 
print("Difference :", A - B) 
  
# symmetric difference 
print("Symmetric difference :", A ^ B) 

The simplest way to create set is:
1. from list
code:
	s = [1,2,3]
	set = set(s)
	print(set)

2. s,add() method
code:
	set.add(1)
	set.add(2)
	set.remove(2)
	print(set)  // 1

3. Set conatins unique elements

set_example = {1, 2, 3, 4, 5, 5, 5}

print(set_example)

# OUTPUT
# {1, 2, 3, 4, 5} ----- Does not print repetitions

# Create a set
seta = {5,10,3,15,2,20}
# Or
setb = set([5, 10, 3, 15, 2, 20])

# Find the length use len()
print(len(setb))

#help set the array in python in order

thisset = {"apple", "banana", "cherry"}
print(thisset) 

set_of_base10_numbers = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
set_of_base2_numbers = {1, 0}

intersection = set_of_base10_numbers.intersection(set_of_base2_numbers)
union = set_of_base10_numbers.union(set_of_base2_numbers)

'''
intersection: {0, 1}:
	if the number is contained in both sets it becomes part of the intersection
union: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}:
	if the number exists in at lease one of the sets it becomes part of the union
'''