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# Python | Count keys with particular value in dictionary

• Difficulty Level : Hard
• Last Updated : 01 Feb, 2023

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

 `# 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, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3```

Time Complexity: O(n)
Auxiliary Space : O(1)

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

 `# 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, 'is': 2, 'best': 3, 'for': 2, 'CS': 2}
Frequency of K is : 3```

Time Complexity: O(n)
Auxiliary Space : O(1)

Method #3 : Using count() and values().These methods can be used together to find number of keys with particular value.

## Python3

 `# Python3 code to demonstrate working of` `# Count keys with particular value in dictionary` `# Using count() + 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 count() + values()` `list1``=``list``(test_dict.values())`   `# Selective key values in dictionary` `res ``=` `list1.count(K)` `   `  `# printing result ` `print``(``"Frequency of K is : "` `+` `str``(res))`

Output

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

#### Complexity Analysis:

In all the solutions time and space complexity are the same:

Time Complexity: O(n)
Auxiliary Space : O(n)

Method #4 : Using lambda functions

## Python3

 `# Python3 code to demonstrate working of` `# Count keys with particular value in dictionary`   `# 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`   `keys ``=` `list``(test_dict.keys())` `res ``=` `len``(``list``(``filter``(``lambda` `x: test_dict[x] ``=``=` `K, keys)))`   `# printing result` `print``(``"Frequency of K is : "` `+` `str``(res))`

Output

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

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #5 : Using collections.Counter()

## Python3

 `# Python3 code to demonstrate working of` `# Count keys with particular value in dictionary` `# Using collections.Counter()` ` `  `# Importing collections for Counter` `import` `collections` ` `  `# 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 collections.Counter()` `res ``=` `collections.Counter(test_dict.values())[K]` `     `  `# printing result` `print``(``"Frequency of K is : "` `+` `str``(res))` `#This code is contributed by Edula Vinay Kumar Reddy`

Output

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

Time Complexity: O(n)
Auxiliary Space: O(1)

Method #6: Using operator.countOf() method:

## Python3

 `# Python3 code to demonstrate working of` `# Count keys with particular value in dictionary` `import` `operator as op` `# 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 count() + values()` `list1``=``list``(test_dict.values())`   `# Selective key values in dictionary` `res ``=` `op.countOf(list1,K)` `   `  `# printing result ` `print``(``"Frequency of K is : "` `+` `str``(res))`

Output

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

Time Complexity: O(N)
Auxiliary Space: O(N)

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