Python – Frequency of unequal items in Dictionary

Sometimes, while working with Python, we can come across a problem in which we need to check for the unequal items count among two dictionaries. This has an application in cases of web development and other domains as well. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using dictionary comprehension
This particular task can be performed in one line using dictionary comprehension which offers a way of compacting lengthy brute logic and just checks for unequal items and increments count.

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 code to demonstrate working of
# Dissimilar items frequency in Dictionary
# Using dictionary comprehension
  
# initializing dictionaries
test_dict1 = {'gfg' : 1, 'is' : 2, 'best' : 3}
test_dict2 = {'gfg' : 1, 'is' : 2, 'good' : 3}
  
# printing original dictionaries
print("The original dictionary 1 is : " + str(test_dict1))
print("The original dictionary 2 is : " + str(test_dict2))
  
# Dissimilar items frequency in Dictionary
# Using dictionary comprehension
res = {key: test_dict1[key] for key in test_dict1 if key not in test_dict2}
  
# printing result
print("The number of uncommon items are : " + str(len(res)))

chevron_right






                        filter_none









Output :          


The original dictionary 1 is : {'best': 3, 'is': 2, 'gfg': 1}
The original dictionary 2 is : {'good': 3, 'is': 2, 'gfg': 1}
The number of uncommon items are : 1





 


Method #2 : Using set() + XOR operator + items()

The combination of above methods can be used to perform this particular task. In this, the set function removes duplicates and XOR operator computes the matched items. Result can be computed by subtracting original dict. length by new dict. length.











                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 code to demonstrate working of 


# Dissimilar items frequency in Dictionary 


# Using set() + XOR operator + items() 


  


# initializing dictionaries 


test_dict1 = {'gfg' : 1, 'is' : 2, 'best' : 3} 


test_dict2 = {'gfg' : 1, 'is' : 2, 'good' : 3} 


  


# printing original dictionaries 


print("The original dictionary 1 is : " + str(test_dict1)) 


print("The original dictionary 2 is : " + str(test_dict2)) 


  


# Dissimilar items frequency in Dictionary 


# Using set() + XOR operator + items() 


res = set(test_dict1.items()) ^ set(test_dict2.items()) 


  


# printing result 


print("The number of uncommon items are : " + str(len(test_dict1) - len(res))) 



















                        chevron_right







                        filter_none









Output :          


The original dictionary 1 is : {'best': 3, 'is': 2, 'gfg': 1}
The original dictionary 2 is : {'good': 3, 'is': 2, 'gfg': 1}
The number of uncommon items are : 1








    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
manjeet_04
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.