# Python | N largest values in dictionary

• Last Updated : 22 Feb, 2023

Many times while working with Python dictionary, we can have a particular problem to find the N maxima of values in numerous keys. This problem is quite common while working with web development domain. Let’s discuss several ways in which this task can be performed.

Method #1 : itemgetter() + items() + sorted() The combination of above method is used to perform this particular task. In this, we just reverse sort the dictionary values expressed using itemgetter() and accessed using items().

## Python3

 `# Python3 code to demonstrate working of``# N largest values in dictionary``# Using sorted() + itemgetter() + items()``from` `operator ``import` `itemgetter` `# Initialize dictionary``test_dict ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``4``, ``'best'` `: ``6``, ``'for'` `: ``7``, ``'geeks'` `: ``3` `}` `# Initialize N``N ``=` `3` `# printing original dictionary``print``("The original dictionary ``is` `: " ``+` `str``(test_dict))` `# N largest values in dictionary``# Using sorted() + itemgetter() + items()``res ``=` `dict``(``sorted``(test_dict.items(), key ``=` `itemgetter(``1``), reverse ``=` `True``)[:N])` `# printing result``print``("The top N value pairs are  " ``+` `str``(res))`

Output :

```The original dictionary is : {'best': 6, 'gfg': 1, 'geeks': 3, 'for': 7, 'is': 4}
The top N value pairs are  {'for': 7, 'is': 4, 'best': 6}```

Time complexity: O(n logn)

Auxiliary space: O(n)

Method #2 : Using nlargest() This task can be performed using the nlargest function. This is inbuilt function in heapq library which internally performs this task and can be used to do it externally. Has the drawback of printing just keys not values.

## Python3

 `# Python3 code to demonstrate working of``# N largest values in dictionary``# Using nlargest``from` `heapq ``import` `nlargest` `# Initialize dictionary``test_dict ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``4``, ``'best'` `: ``6``, ``'for'` `: ``7``, ``'geeks'` `: ``3` `}` `# Initialize N``N ``=` `3` `# printing original dictionary``print``("The original dictionary ``is` `: " ``+` `str``(test_dict))` `# N largest values in dictionary``# Using nlargest``res ``=` `nlargest(N, test_dict, key ``=` `test_dict.get)` `# printing result``print``("The top N value pairs are  " ``+` `str``(res))`

Output :

```The original dictionary is : {'gfg': 1, 'best': 6, 'geeks': 3, 'for': 7, 'is': 4}
The top N value pairs are  ['for', 'best', 'is']```

Using lambda function with the sorted() function:

This approach uses a lambda function as the key argument in the sorted() function to extract the values from the dictionary and sort them in descending order.
Example:

## Python3

 `# Initialize dictionary``test_dict ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``4``, ``'best'` `: ``6``, ``'for'` `: ``7``, ``'geeks'` `: ``3` `}``  ` `# Initialize N``N ``=` `3``  ` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))` `# Sort the dictionary by value using a lambda function to extract the values``# and then reverse the sort to get the largest values first``res ``=` `dict``(``sorted``(test_dict.items(), key ``=` `lambda` `x: x[``1``], reverse ``=` `True``)[:N])``  ` `# printing result``print``(``"The top N value pairs are  "` `+` `str``(res))``#This code is contributed by Edula Vinay Kumar Reddy`

Output

```The original dictionary is : {'gfg': 1, 'is': 4, 'best': 6, 'for': 7, 'geeks': 3}
The top N value pairs are  {'for': 7, 'best': 6, 'is': 4}```

This approach’s time complexity is O(n log n) and space complexity is O(n) as it stores all the key-value pairs in a list before slicing the first N key-value pairs.

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