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# Python | Count keys with particular value in dictionary

• Difficulty Level : Hard
• Last Updated : 26 Jul, 2019

Sometimes, while working with Python dictionaries, we can come across a problem in which we have a particular value, and we need to find frequency if it’s occurrence. Let’s discuss certain ways in which this problem can be solved.

Method #1 : Using loop
This problem can be solved using naive method of loop. In this we just iterate through each key in dictionary and when a match is found, the counter is increased.

 `# Python3 code to demonstrate working of``# Count keys with particular value in dictionary``# Using loop`` ` `# Initialize dictionary``test_dict ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``2``, ``'best'` `: ``3``, ``'for'` `: ``2``, ``'CS'` `: ``2``}`` ` `# printing original dictionary``print``(``"The original dictionary : "` `+`  `str``(test_dict))`` ` `# Initialize value ``K ``=` `2`` ` `# Using loop``# Selective key values in dictionary``res ``=` `0``for` `key ``in` `test_dict:``    ``if` `test_dict[key] ``=``=` `K:``        ``res ``=` `res ``+` `1``     ` `# printing result ``print``(``"Frequency of K is : "` `+` `str``(res))`
Output :

The original dictionary : {‘gfg’: 1, ‘CS’: 2, ‘best’: 3, ‘for’: 2, ‘is’: 2}
Frequency of K is : 3

Method #2 : Using `sum() + values()`
This can also be solved using the combination of `sum() and value()`. In this, sum is used to perform the summation of values filtered and values of dictionary are extracted using `values()`

 `# Python3 code to demonstrate working of``# Count keys with particular value in dictionary``# Using sum() + values()`` ` `# Initialize dictionary``test_dict ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``2``, ``'best'` `: ``3``, ``'for'` `: ``2``, ``'CS'` `: ``2``}`` ` `# printing original dictionary``print``(``"The original dictionary : "` `+`  `str``(test_dict))`` ` `# Initialize value ``K ``=` `2`` ` `# Using sum() + values()``# Selective key values in dictionary``res ``=` `sum``(x ``=``=` `K ``for` `x ``in` `test_dict.values())``     ` `# printing result ``print``(``"Frequency of K is : "` `+` `str``(res))`
Output :

The original dictionary : {‘gfg’: 1, ‘CS’: 2, ‘best’: 3, ‘for’: 2, ‘is’: 2}
Frequency of K is : 3

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