Python | Set Difference in list of dictionaries

Last Updated : 27 Apr, 2023

The difference of two lists have been discussed many times, but sometimes we have a large number of data and we need to find the difference i.e the elements in dict2 not in 1 to reduce the redundancies. Let’s discuss certain ways in which this can be done.

Method #1 : Using list comprehension The naive method to iterate both the list and extract the difference can be shortened to the method in which we shorten the code and increase the readability using list comprehension.

Python3

# Python3 code to demonstrate 
# set difference in dictionary list 
# using list comprehension
 
# initializing list 
test_list1 = [{"HpY" : 22}, {"BirthdaY" : 2}, ]
test_list2 = [{"HpY" : 22}, {"BirthdaY" : 2}, {"Shambhavi" : 2019}]
 
# printing original lists
print ("The original list 1 is : " + str(test_list1))
print ("The original list 2 is : " +  str(test_list2))
 
# using list comprehension
# set difference in dictionary list 
res = [i for i in test_list1 if i not in test_list2] \
      + [j for j in test_list2 if j not in test_list1]
 
# printing result 
print ("The set difference of list is : " +  str(res))

Output :

The original list 1 is : [{‘HpY’: 22}, {‘BirthdaY’: 2}] The original list 2 is : [{‘HpY’: 22}, {‘BirthdaY’: 2}, {‘Shambhavi’: 2019}] The set difference of list is : [{‘Shambhavi’: 2019}]

Time Complexity: O(n*n), where n is the length of the list test_list
Auxiliary Space: O(n*n) additional space of size n is created where n is the number of elements in the res list

Method #2 : Using itertools.filterfalse() This is a different way in which this particular task can be performed using the in built python function. The filterfalse method filters the not present element of one list with respect to other.

Python3

# Python3 code to demonstrate 
# set difference in dictionary list 
# using itertools.filterfalse()
import itertools
 
# initializing list 
test_list1 = [{"HpY" : 22}, {"BirthdaY" : 2}, ]
test_list2 = [{"HpY" : 22}, {"BirthdaY" : 2}, {"Shambhavi" : 2019}]
 
# printing original lists
print ("The original list 1 is : " + str(test_list1))
print ("The original list 2 is : " +  str(test_list2))
 
# using itertools.filterfalse()
# set difference in dictionary list 
res = list(itertools.filterfalse(lambda i: i in test_list1, test_list2)) \
    + list(itertools.filterfalse(lambda j: j in test_list2, test_list1))
 
# printing result 
print ("The set difference of list is : " +  str(res))

Output :

The original list 1 is : [{‘HpY’: 22}, {‘BirthdaY’: 2}] The original list 2 is : [{‘HpY’: 22}, {‘BirthdaY’: 2}, {‘Shambhavi’: 2019}] The set difference of list is : [{‘Shambhavi’: 2019}]

Time Complexity: O(n*n), where n is the length of the list test_list
Auxiliary Space: O(n*n) additional space of size n is created where n is the number of elements in the res list