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  • Writer's pictureGianluca Turcatel

Python Sets


Sets are used to store multiple UNIQUE items in a single variable. Sets are unordered data structure.


Initializing sets

Place items inside curly brackets separated by comma.

Syntax: set = { element1, element2, .... }

##Initialize some sets
set1 = {}        #Empty set
set2 = {1, 2, 3} 
set2 = {4, 8, 'r'} 
--------------------------------------------------------------------------
##Print the content of the set
print(set1)
: {}
print(set2)
: {1, 2, 3}
print(set3)
: {4, 8, 'r'} 
--------------------------------------------------------------------------
##Check the type:
print(type(set1))
: <class 'set'>
--------------------------------------------------------------------------
##Sets are collections of unique elements
set_num = {1, 2, 3, 2, 1}
print(set_num)
: {1, 2, 3}  #--> only unique elements are stored

Common methods & functions with sets

##To add the element to the set --> add()
set_num = {1, 2, 3}
set_num.add(20)     #Add the integer 20
print(set_num)
: {1, 2, 3, 20} 
--------------------------------------------------------------------------
##To remove a value from set --> remove()  #Option 1
set_num.remove(2) #Remove the integer 2          
print(set_num)
: {1, 3, 20}

##To remove a value from set --> discard() #Option 2
set_num.discard(3) #Remove the integer 3          
print(set_num)
: {1, 20}

##What's the difference between remove() and discard()?
#--> remove() WILL throw an error if the item we are trying to remove is       #non present in the set
#--> discard() WON'T throw an error if the item we are trying to remove      #is non present in the set
set_num = {1, 20, 8}
set_num.discard(5)  #No error is thrown
set_num.remove(5)   #TypeError is thrown
: TypeError: 5
--------------------------------------------------------------------------
##To add multiple values to a set --> update() 
set_num = {1, 20, 8, 10}
set_num.update([22, 85]) #Add 22 and 85 to the set
print(set_num)
: {1, 8, 10, 20, 85, 22}

Main operations with sets

These are Union, Intersection, Difference:

##Initialize two sets
set1 = {1, 20, 8, 10}
set2 = {88, 8, 10, 12}
--------------------------------------------------------------------------
##UNION: merge the elements of two sets into one set --> union() or |
set_final = set1.union(set2) #Syntax option 1
print(set_final)
: {1, 8, 10, 12, 20, 88}

set_final2 = set1 | set2     #Syntax option 2
print(set_final2)
: {1, 8, 10, 12, 20, 88}
--------------------------------------------------------------------------
##INTERSECTION: Find common elements from two sets --> intersection() or & 
set_intersect1 = set1.intersection(set2) #Syntax option 1
print(set_intersect1)
: {8, 10}

set_intersect2 = set1 & set2              #Syntax option 2
print(set_intersect2)
: {8, 10}
--------------------------------------------------------------------------
##DIFFERENCE: Find difference of elements from one set to the second set
## --> difference()  or -
set_difference1 = set1.difference(set2) #Syntax option 1
print(set_difference1)
: {1, 20}

set_difference2 = set1 - set2           #Syntax option 1
print(set_difference2)
: {1, 20}

# NOTE set1 - set2 != set2 - set1
set1_minus_set2 = set1 - set2
set2_minus_set1 = set2 - set1
print(set1_minus_set2 , set2_minus_set1)
print(set1_minus_set2 == set2_minus_set1)
: {1, 20}  {12, 88}
: False

Identity between sets

Two sets are identical if they contains the same UNIQUE items

set1 = {1, 8, 10}
set2 = {8, 8, 10, 1}
print(set1 == set2)
: True

Lists to sets and backward

Lists can be converted to sets and vice versa. Lists are often converted to sets to remove duplicated items.

##Initialize a list with duplicated elements
nums = [8, 8, 10, 1]
print(type(nums))
: <class 'list'>

#Convert the list to a set to remove duplicates
nums_set = set(nums)

#Convert the set back to a list
unique_num_list = list(nums_set)
print(unique_num_list)
: [8, 1, 10]
print(type(unique_num_list))
: <class 'list'>

Use case

Given a list of unique and duplicated integers, create a list containing only its unique squares.

##Initialize a list with duplicated elements
nums = [1, 3, 7, 7, 3, 8, 9, 10]

#Create a list containing the squares
square_nums = []
for num in nums:
    square_nums.append(num**2)
  
#Remove the duplicated squares
result = list(set(square_nums))
print(result)
: [64, 1, 100, 9, 81, 49]

Concluding remarks

Sets are optimized for checking whether a specific element is present or not in the set. Sets are unordered data structure, thus we can't access their item by their position/index.


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