Problem Statement – Here are the major tasks that are needed to be performed.

1. Create a dictionary and display its keys alphabetically.
2. Display both the keys and values sorted in alphabetical order by the key.
3. Same as part (ii), but sorted in alphabetical order by the value.

Approach –
Load the Dictionary and perform the following operations:

1. First, sort the keys alphabetically using key_value.iterkeys() function.
2. Second, sort the keys alphabetically using sorted (key_value) function & print the value corresponding to it.
3. Third, sort the values alphabetically using key_value.iteritems(), key = lambda (k, v) : (v, k))

Let’s try performing the above-mentioned tasks:

1. Displaying the Keys Alphabetically:Examples:

   Input:
   key_value[2] = '64'       
   key_value[1] = '69'
   key_value[4] = '23'
   key_value[5] = '65'
   key_value[6] = '34'
   key_value[3] = '76'
   
   Output:
   1 2 3 4 5 6 
   

   Program:

   # Function calling def dictionairy():   # Declare hash function        key_value ={}       # Initializing value   key_value[2] = 56  key_value[1] = 2  key_value[5] = 12  key_value[4] = 24  key_value[6] = 18  key_value[3] = 323    print ("Task 1:-\n")  print ("Keys are")     # iterkeys() returns an iterator over the   # dictionary’s keys.  for i in sorted (key_value.keys()) :       print(i, end = " ")   def main():     # function calling     dictionairy()                    # Main function calling if __name__=="__main__":           main()

   Output:

   Task 1:-
   
   Keys are
   1 2 3 4 5 6
   

2. Sorting the Keys and Values in Alphabetical Order using the Key.Examples:

   Input:
   key_value[2] = '64'       
   key_value[1] = '69'
   key_value[4] = '23'
   key_value[5] = '65'
   key_value[6] = '34'
   key_value[3] = '76'
   
   Output:
   (1, 69) (2, 64) (3, 76) (4, 23) (5, 65) (6, 34)
   

   Program:

   # function calling def dictionairy():     # Declaring the hash function        key_value ={}        # Initialize value   key_value[2] = 56  key_value[1] = 2  key_value[5] = 12  key_value[4] = 24  key_value[6] = 18  key_value[3] = 323     print ("Task 2:-\nKeys and Values sorted in",              "alphabetical order by the key  ")      # sorted(key_value) returns an iterator over the   # Dictionary’s value sorted in keys.    for i in sorted (key_value) :     print ((i, key_value[i]), end =" ")   def main():     # function calling     dictionairy()                    # main function calling if __name__=="__main__":          main()

   Output:

   Task 2:-
   Keys and Values sorted in alphabetical order by the key  
   (1, 2) (2, 56) (3, 323) (4, 24) (5, 12) (6, 18)
   

3. Sorting the Keys and Values in alphabetical using the valueExamples:

   Input:
   key_value[2] = '64'       
   key_value[1] = '69'
   key_value[4] = '23'
   key_value[5] = '65'
   key_value[6] = '34'
   key_value[3] = '76'
   
   Output:
   (4, 23), (6, 34), (2, 64), (5, 65), (1, 69), (3, 76)
   

   Program:

   # Function calling def dictionairy():     # Declaring hash function        key_value ={}        # Initializing the value   key_value[2] = 56  key_value[1] = 2  key_value[5] = 12  key_value[4] = 24  key_value[6] = 18  key_value[3] = 323       print ("Task 3:-\nKeys and Values sorted",     "in alphabetical order by the value")     # Note that it will sort in lexicographical order  # For mathematical way, change it to float  print(sorted(key_value.items(), key =              lambda kv:(kv[1], kv[0])))        def main():     # function calling     dictionairy()                   # main function calling if __name__=="__main__":           main()
   Output:

   Task 3:-
   Keys and Values sorted in alphabetical order by the value
   [(1, 2), (5, 12), (6, 18), (4, 24), (2, 56), (3, 323)]
   

Learning Outcome:

• How to handle a dictionary.
• Dictionary has O(1) search time complexity whereas List has O(n) time complexity. So it is recommended to use the dictionary where ever possible.