# Python | Get the number of keys with given value N in dictionary

• Last Updated : 19 Jan, 2023

In this article, we will see how to get the number of keys with some given value N in a given dictionary. There are multiple methods to do this task. Let’s see them with help of examples. Simple method –

## Python3

 `# Python3 code to Get the number of keys``# with given value N 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``N ``=` `2` `# Using loop``# Selective key values in dictionary``res ``=` `0``for` `key ``in` `test_dict:``    ``if` `test_dict[key] ``=``=` `N:``        ``res ``=` `res ``+` `1``    ` `# printing result``print``(``"Frequency of N is : "` `+` `str``(res))`

Output

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

Method #2: Using sum() + values()

## Python3

 `# Python3 code to Get the number of keys``# with given value N 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``N ``=` `2` `# Using sum() + values()``# Selective key values in dictionary``res ``=` `sum``(x ``=``=` `N ``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
```

Method #3: Using Counter() function

## Python3

 `# Python3 code to Get the number of keys``# with given value N in dictionary``from` `collections ``import` `Counter``# 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``N ``=` `2` `freq ``=` `Counter(test_dict.values())``res ``=` `freq[N]` `# printing result``print``(``"Frequency of N is : "` `+` `str``(res))`

Output

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

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

Method #4: Using list comprehension

This code uses a list comprehension to iterate through the keys in the dictionary, and for each key it checks if the value associated with that key is equal to the given value N. If it is, it adds that key to a new list. The length of this list is then calculated, which represents the number of keys in the dictionary that have a value of N.

## Python3

 `# Method #4: Using list comprehension` `# 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``N ``=` `2` `# Using list comprehension``# Selective key values in dictionary``res ``=` `len``([key ``for` `key ``in` `test_dict ``if` `test_dict[key] ``=``=` `N])` `# printing result``print``(``"Frequency of N 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 N is : 3
```

This method has a time complexity of O(n) as it iterates through all the keys of the dictionary. The auxiliary space of this method is O(n) as it creates a new list to store the keys that have the given value N.

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