# Python | Set Difference in list of dictionaries

• Last Updated : 26 Feb, 2019

The difference of two lists have been discussed many a 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 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}]

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 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}]

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