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Python – Remove Duplicate Dictionaries characterized by Key

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  • Last Updated : 01 Aug, 2020

Given a list of dictionaries, remove all the dictionaries which are duplicate with respect to K key.

Input : test_list = [{“Gfg” : 6, “is” : 9, “best” : 10}, {“Gfg” : 8, “is” : 11, “best” : 10}, {“Gfg” : 2, “is” : 16, “best” : 10}], K = “best”
Output : [{“Gfg” : 6, “is” : 9, “best” : 10}]
Explanation : All keys have 10 value, only 1st record is retained.

Input : test_list = [{“Gfg” : 6, “is” : 9, “best” : 10}, {“Gfg” : 8, “is” : 11, “best” : 12}, {“Gfg” : 2, “is” : 16, “best” : 15}], K = “best”
Output : [{“Gfg” : 6, “is” : 9, “best” : 10}, {“Gfg” : 8, “is” : 11, “best” : 12}, {“Gfg” : 2, “is” : 16, “best” : 15}]
Explanation : All values of “best” are unique, hence no removal of dictionaries.

Method : Using loop 

This is brute way in which this task can be performed. In this, we iterate for each dictionary and memoize the Key, if similar key’s same value occur, then that dictionary is avoided in resultant list of dictionaries.

Python3




# Python3 code to demonstrate working of 
# Remove Duplicate Dictionaries characterized by Key
# Using loop
  
# initializing lists
test_list = [{"Gfg" : 6, "is" : 9, "best" : 10}, 
             {"Gfg" : 8, "is" : 11, "best" : 19},
             {"Gfg" : 2, "is" : 16, "best" : 10},
             {"Gfg" : 12, "is" : 1, "best" : 8},
             {"Gfg" : 22, "is" : 6, "best" : 8}]
  
# printing original list
print("The original list : " + str(test_list))
  
# initializing Key 
K = "best"
  
memo = set()
res = []
for sub in test_list:
      
    # testing for already present value
    if sub[K] not in memo:
        res.append(sub)
          
        # adding in memo if new value
        memo.add(sub[K])
      
# printing result 
print("The filtered list : " + str(res))

Output

The original list : [{‘Gfg’: 6, ‘is’: 9, ‘best’: 10}, {‘Gfg’: 8, ‘is’: 11, ‘best’: 19}, {‘Gfg’: 2, ‘is’: 16, ‘best’: 10}, {‘Gfg’: 12, ‘is’: 1, ‘best’: 8}, {‘Gfg’: 22, ‘is’: 6, ‘best’: 8}]
The filtered list : [{‘Gfg’: 6, ‘is’: 9, ‘best’: 10}, {‘Gfg’: 8, ‘is’: 11, ‘best’: 19}, {‘Gfg’: 12, ‘is’: 1, ‘best’: 8}]


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