Prerequisite:

• Dictionary
• List
• Merging Two Dictionaries
• Dictionary Methods

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:

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   # 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()

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   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:

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   # 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()

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   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 value

   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:
   (4, 23), (6, 34), (2, 64), (5, 65), (1, 69), (3, 76)
   

   Program:

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   # 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()

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   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)]
   

Sort the dictionary by key
Note:it will sort in lexicogrphical order
Taking key’s type as string
Program:

Here, Output is Soretd Dictionary using key in Lexicolographical order
Lexicographical order:https://en.wikipedia.org/wiki/Lexicographical_order

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.
• The best application of dictionary can be seen in Twitter Sentiment Analysis where analysis Twitter sentiments are analysed, using Lexicon approach.

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