Python Set

Python Set

Set data structure characteristics:

• Set is an object of unordered items which cannot have duplicate items.
• Set is a mutable object, this means that same instance of set can be modified to add, update and remove items.
• Set can be empty or made up of either homogenous items(same datatype) or heterogenous items(different datatype or data structures).
• Set are iterable object, this means that user can loop items in the set from 1st through last item or vice versa.
• Index cannot be used in set to retrieve record at a particular position.
• Set can be defined using curly brackets {} but empty set cannot be defined using curly brackets, it can be done with set() command. Below listed example defines a set in Python.

  
  a_set = {'D', 'B', 'M', 'S', 'T', 'U', 'T', 'O', 'R', 'I', 'A', 'L', 'S'}
  
  print(a_set)  #Set cannot have duplicates
  {'I', 'B', 'T', 'A', 'D', 'U', 'L', 'S', 'R', 'M', 'O'}
  

• Reading and printing items in a set using for loop.

  
  a_set = {'D', 'B', 'M', 'S', 'T', 'U', 'T', 'O', 'R', 'I', 'A', 'L', 'S'}
  print(a_set)
  
  Output:   {'I', 'B', 'T', 'A', 'D', 'U', 'L', 'S', 'R', 'M', 'O'}
  
  
  for item in a_set:
  	print(item)
  
  Output:  
  I
  B
  T
  A
  D
  U
  L
  S
  R
  M
  O
  

• Keyword 'del' can be used to del set variable as shown in below example.

  
  a_set = {'D', 'B', 'M', 'S', 'T', 'U', 'T', 'O', 'R', 'I', 'A', 'L', 'S'}
  
  del a_set
  a_set
  Traceback (most recent call last):
    File "", line 1, in 
  NameError: name 'a_set' is not defined
  

• Available built-in Python functions helps to perform variety of operations on set objects, click on the required function to get more detail and sample examples.
  • Length
  • Add
  • Update
  • Remove
  • Discard
  • Pop
  • Copy
  • Clear
  • issubset
  • issuperset
  • union
  • difference
  • difference_update
  • intersection
  • intersection_update
  • isdisjoint
  • symmetric_difference
  • symmetric_difference_update
• Some of miscellaneous topics explained with examples
  • Remove Vs Pop Vs Discard
  • Set Comprehensions
  • Create a copy of Set (3 approaches)
  • Create a reference to a existing set
  • Add Vs Update

➠ Remove Vs Pop Vs Discard

• pop() function will throw error if set is empty.
• remove() function will throw error if input item is not present in the set.
• discard() function will not throw any error if item is not present in the set.
  
  
• pop() function does not requires any input to remove item.
• remove() function requires value of item as input which need to be removed.
• discard() function requires value of item as input which need to be removed.
  
  
• pop() function will randomly remove item from the set.
• remove() function will remove input value(item) from the set.
• discard() function will remove input value(item) from the set.
  
  
• pop() function will return the item's value which was removed from the set.
• remove() function will not return anything, it will just remove the item from the set.
• discard() function will not return anything, it will just remove the item from the set.
  
  Pop() Example:

  
  set_1 = {1, 2, 3, 6, 'A'}
  print(set_1.pop())      # pop() function returning 1
  
  Output:  3
  

  
  Remove() Example:

  
  set_1 = {1, 2, 3, 6, 'A'}
  print(set_1.remove(2))    # remove() function returning None
  
  Output:  None
  

  
  Discard() Example:

  
  set_1 = {1, 2, 3, 6, 'A'}
  print(set_1.discard(2))    # discard() function returning None
  print(set_1.discard(2))    # discard() function returning None
  
  Output:  None
  

➠ Set Comprehensions: Python provide a concise way to create set without using map(), filter() or lambda functions. Set comprehension statement consists of an expression followed by a for loop clause, then zero or more for loops or if clauses.

• Example 1: Creating a simple set from 0-4 using Set comprehension.

  
  set_1 = {i for i in range(0,5)}
  print(set_1)
  
  Output: {0, 1, 2, 3, 4}
  

• Example 2: Creating a set from a range with a condition using comprehension.

  
  set_1 = {i for i in range(0,10) if i%3==0}
  print(set_1)
  
  Output: {0, 9, 3, 6}
  

• Example 3: Creating a set with nested loop using comprehension.

  
  set_1 = {i*j for i in range(1,3) for j in range(2,5)}
  print(set_1)
  
  Output: {2, 3, 4, 6, 8}
  

• Example 4: Creating a set with nested loop and a condition using comprehension.

  
  set_1 = {i*j for i in range(1,3) for j in range(2,5) if i*j//4!=0}
  print(set_1)
  
  Output: {8, 4, 6}
  

➠ Create a Copy of Set: Set object can not be copied using assignment operator in Python. Assignment operator creates a reference to an existing set object. Copy of a set is sometimes needed so one set(copy) can be changed without changing the content of other set copy.

• Approach 1 : Using built-in 'set' class to create a copy of existing set.

  
  set_1 = {1, 2, 3, 6, 'A'}
  print(set_1)
  Output 1: {1, 2, 3, 6, 'A'}
  
  
  set_2 = set(set_1)  # Creating a copy here
  print(set_2)
  Output 2: {1, 2, 3, 6, 'A'}
  
  
  set_2.pop()   # Removing item from set_2
  Output 3: 1
  
  
  print(set_2)   # 'A' value is not present in set_2
  Output 4: {2, 3, 6, 'A'}
  
  
  print(set_1)   # '1' value is still present in set_1
  Output 5: {1, 2, 3, 6, 'A'}
  

  
  
• Approach 2 : Using built-in copy() function to creating a copy of input set.

  
  Syntax: set_object.copy()
  
  set_1 = {1, 2, 3, 6, 'A'}
  set_2 = set_1.copy()
  print(set_2)
  Output 2: {1, 2, 3, 6, 'A'}
  
  
  set_2.pop()   # Removing item from set_2
  Output 3:  1
  
  
  print(set_2)   # '1' value is not present in set_2
  Output 4: {2, 3, 6, 'A'}
  
  
  print(set_1)   # '1' value is still present in set_1
  Output 5: {1, 2, 3, 6, 'A'}
  

  
  
• Approach 3 : Using built-in 'copy' module to creating a copy of existing set.

  
  set_1 = {1, 2, 3, 6, 'A'}
  print(set_1)
  Output 1: {1, 2, 3, 6, 'A'}
  
  
  import copy
  set_4=copy.copy(set_1)  # Creating a copy here
  print(set_4)
  Output 2: {1, 2, 3, 6, 'A'}
  
  
  set_4.pop()   # Removing item from set_4
  Output 3:  1
  
  
  print(set_4)   # '1' value is not present in set_4
  Output 4: {2, 3, 6, 'A'}
  
  
  print(set_1)   # '1' value is still present in set_1
  Output 5: {1, 2, 3, 6, 'A'}
  

➠ Create a reference to a existing set : Assignment operator can be used create a reference to an existing set object. If set is referenced, then any change in either of set variable will result in changes in the content of other set(because it is just a reference) as well.

• 
  set_1 = {1, 2, 3, 6, 'A'}
  print(set_1)
  Output 1: {1, 2, 3, 6, 'A'}
  
  
  set_5 = set_1  # Creating a reference here
  print(set_5)
  Output 2: {1, 2, 3, 6, 'A'}
  
  
  set_5.add('P')   # Adding 'P' item to set_5
  
  
  print(set_5)   # 'P' value is present in set_5
  Output 4: {'P', 1, 2, 3, 6, 'A'}
  
  
  print(set_1)   # 'P' value is also present in set_1
  Output 5: {'P', 1, 2, 3, 6, 'A'}
  
  
  set_1.pop()   # Removing item from set_1
  Output 6:  P
  
  
  print(set_1)   # 'P' value not present in set_1
  Output 7: {1, 2, 3, 6, 'A'}
  
  
  print(set_5)   # '6' value also not present in set_5
  Output 8: {1, 2, 3, 6, 'A'}
  

➠ Update Vs Add

• update() function requires iterable object(such as set,list) as input otherwise it will throw error that 'object is not iterable'.
• add() function cannot have iterable object as input otherwise it will throw error 'TypeError: unhashable type'.
  
  
• update() function will iterate through all the elements in the passed input and add them as individual item in the existing set.
• add() function will add the passed input as single element in the existing set.
  
  Update() Example:

  
  set_1 = {1, 2, 3, 6, 'A'}
  set_2 = {4, 5, 7}
  
  set_1.update(set_2)   # Using update() function here
  print(set_1)           # set_2 elements got added in set_1
  
  Output:  {1, 2, 3, 4, 5, 6, 7, 'A'}
  
  
  set_1.update(5)      # Using update() function here
  
  Output: TypeError: 'int' object is not iterable
  
  Output: {1, 2, 3, 5, 6, 'A'}
  

  
  Add() Example:

  
  set_1 = {1, 2, 3, 6, 'A'}
  set_2 = {4, 5, 7}
  
  set_1.add(set_2)   # Using add() function here
  
  Output:  TypeError: unhashable type: 'set'
  
  
  set_1.add(5)      # Using add() function here
  print(set_1)           # Element 5 got added in set_1
  
  Output: {1, 2, 3, 5, 6, 'A'}