# Python – Remove Keys with K value

• Last Updated : 29 Aug, 2020

Given a dictionary, remove all keys with value equals K.

Input : test_dict = {‘Gfg’ : 8, ‘is’ : 7, ‘best’ : 8, ‘for’ : 6, ‘geeks’ : 11}, K = 8
Output : {‘is’ : 7, ‘for’ : 6, ‘geeks’ : 11}
Explanation : “Gfg” and “Best”, valued 8, are removed.

Input : test_dict = {‘Gfg’ : 8, ‘is’ : 8, ‘best’ : 8, ‘for’ : 8, ‘geeks’ : 8}, K = 8
Output : {}
Explanation : All keys, valued 8, are removed.

Method #1 : Using dictionary comprehension

This is one of the ways in which this task can be performed. In this, we check for only elements that are not equal to K and retain it, inside dictionary comprehension as one-liner.

## Python3

 `# Python3 code to demonstrate working of ``# Remove Keys with K value``# Using dictionary comprehension`` ` `# initializing dictionary``test_dict ``=` `{``'Gfg'` `: ``6``, ``'is'` `: ``7``, ``'best'` `: ``9``, ``'for'` `: ``6``, ``'geeks'` `: ``11``} `` ` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))`` ` `# initializing K ``K ``=` `6`` ` `# using dictionary comprehension ``# to compare not equal to K and retain ``res ``=` `{key: val ``for` `key, val ``in` `test_dict.items() ``if` `val !``=` `K}``         ` `# printing result ``print``(``"The filtered dictionary : "` `+` `str``(res)) `
Output
```The original dictionary is : {'Gfg': 6, 'is': 7, 'best': 9, 'for': 6, 'geeks': 11}
The filtered dictionary : {'is': 7, 'best': 9, 'geeks': 11}
```

Method #2 : Using dict() + filter() + lambda

The combination of above functions can be used to solve this problem. In this, we filter all the non-K elements and retain. Finally the result is converted to dictionary using dict().

## Python3

 `# Python3 code to demonstrate working of ``# Remove Keys with K value``# Using dict() + filter() + lambda`` ` `# initializing dictionary``test_dict ``=` `{``'Gfg'` `: ``6``, ``'is'` `: ``7``, ``'best'` `: ``9``, ``'for'` `: ``6``, ``'geeks'` `: ``11``} `` ` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))`` ` `# initializing K ``K ``=` `6`` ` `# employing lambda for computation ``# filter() to perform filter according to lambda``res ``=` `dict``(``filter``(``lambda` `key: key[``1``] !``=` `K, test_dict.items()))``         ` `# printing result ``print``(``"The filtered dictionary : "` `+` `str``(res)) `
Output
```The original dictionary is : {'Gfg': 6, 'is': 7, 'best': 9, 'for': 6, 'geeks': 11}
The filtered dictionary : {'is': 7, 'best': 9, 'geeks': 11}
```

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