python sets

>>> 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) 

#get name1 and name2 as a list input
name1 =list("".join(str(x)for x in input("Enter name1").replace(" ","")))
name2 =list("".join(str(x)for x in input("Enter name2").replace(" ","")))
#check using list comprehension if x in name1 is in name2
#this will return multiple instances of the same character from name1 that matches with the name2
common = [x for x in name1 if x in name2]
#create a set out of the output so as to have only unique values of the repeated characters
unique = set(common)
#thus the above set will have common unrepeated characters from both names
#create a variable and initialize it to zero
d=0
#run a loop that checks the minimum occurrence of 
#the character from the set in name1 & name2
#Minimum because for ex: a might exist thrice in name1, but only twice in name2
#we will need to take only 2 common occurrences from the name2
#thus finding the minimum occurrence of the character from both names
for x in unique:
    d = d + min(name1.count(x),name2.count(x))
#multiplying by two, because if one character from name 1 matches with one,
#character from name2, then it makes two in total
difference = (len(name1) + len(name2)) - d*2
print(difference)

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
'''