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Python – Remove dictionary from a list of dictionaries if a particular value is not present
• Last Updated : 01 Oct, 2020

Given a list of dictionaries, remove all dictionaries which don’t have K as a value.

Examples:

Input : test_list = [{“Gfg” : 4, “is” : 8, “best” : 9}, {“Gfg” : 3, “is”: 7, “best” : 5}], K = 7
Output : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}]
Explanation : Resultant dictionary doesn’t contain 7 as any element.

Input : test_list = [{“Gfg” : 4, “is” : 7, “best” : 9}, {“Gfg” : 3, “is”: 7, “best” : 5}], K = 7
Output : []
Explanation : All contain 7 as element in List.

Method #1 : Using loop + values()

This is one of the ways in which this task can be performed. In this, we perform task of iterating for all the dictionaries using loop and check for value presence using values().

## Python3

 `# Python3 code to demonstrate working of  ` `# Remove dictionary if K value not present ` `# Using loop + values() ` ` `  `# initializing lists ` `test_list ``=` `[{``"Gfg"` `: ``4``, ``"is"` `: ``8``, ``"best"` `: ``9``}, ` `             ``{``"Gfg"` `: ``5``, ``"is"``: ``8``, ``"best"` `: ``1``}, ` `             ``{``"Gfg"` `: ``3``, ``"is"``: ``7``, ``"best"` `: ``6``},  ` `             ``{``"Gfg"` `: ``3``, ``"is"``: ``7``, ``"best"` `: ``5``}] ` ` `  `# printing original list ` `print``(``"The original list : "` `+` `str``(test_list)) ` ` `  `# initializing K  ` `K ``=` `7` ` `  `res ``=` `[] ` ` `  `# using loop to check for K element  ` `for` `sub ``in` `test_list: ` `    ``if` `K ``not` `in` `list``(sub.values()): ` `        ``res.append(sub) ` ` `  `# printing result  ` `print``(``"Filtered dictionaries : "` `+` `str``(res))`

Output:

The original list : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 6}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 5}]
Filtered dictionaries : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}]

Method #2 : Using list comprehension

This is yet another way in which this task can be performed. In this, we extract all the values using one-liner using list comprehension. The values are extracted using values().

## Python3

 `# Python3 code to demonstrate working of  ` `# Remove dictionary if K value not present ` `# Using list comprehension ` ` `  `# initializing lists ` `test_list ``=` `[{``"Gfg"` `: ``4``, ``"is"` `: ``8``, ``"best"` `: ``9``}, ` `             ``{``"Gfg"` `: ``5``, ``"is"``: ``8``, ``"best"` `: ``1``}, ` `             ``{``"Gfg"` `: ``3``, ``"is"``: ``7``, ``"best"` `: ``6``},  ` `             ``{``"Gfg"` `: ``3``, ``"is"``: ``7``, ``"best"` `: ``5``}] ` ` `  `# printing original list ` `print``(``"The original list : "` `+` `str``(test_list)) ` ` `  `# initializing K  ` `K ``=` `7` ` `  `res ``=` `[] ` ` `  `# using one-liner to extract dicts with NO K value ` `# using not in operator to check presence ` `res ``=` `[sub ``for` `sub ``in` `test_list ``if` `K ``not` `in` `list``(sub.values())] ` ` `  `# printing result  ` `print``(``"Filtered dictionaries : "` `+` `str``(res))`

Output:

The original list : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 6}, {‘Gfg’: 3, ‘is’: 7, ‘best’: 5}]
Filtered dictionaries : [{‘Gfg’: 4, ‘is’: 8, ‘best’: 9}, {‘Gfg’: 5, ‘is’: 8, ‘best’: 1}]

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