# Python | Top N pairs by Kth element from list

• Last Updated : 03 Nov, 2019

Sometimes, while working with data, we can have a problem in which we need to get the maximum of elements filtered by the Kth element of record. This has a very important utility in web development domain. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using `filter() + lambda + set()` + list comprehension

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The combination of above functions can be used to perform this particular function. In this, we first filter the top N elements from Kth index and then apply these values to the list and return the result.

 `# Python3 code to demonstrate working of``# Top N pairs by Kth element from list``# Using filter() + lambda + set() + list comprehension`` ` `# initialize list ``test_list ``=` `[(``'gfg'``, ``4``, ``'good'``), (``'gfg'``, ``2``, ``'better'``), ``             ``(``'gfg'``, ``1``, ``'best'``), (``'gfg'``, ``3``, ``'geeks'``)]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initialize N ``N ``=` `3` ` ` `# initialize K ``K ``=` `1`` ` `# Top N pairs by Kth element from list``# Using filter() + lambda + set() + list comprehension``temp ``=` `set``(``list``({sub[K] ``for` `sub ``in` `test_list})[``-``N:])``res ``=` `list``(``filter``(``lambda` `sub: sub[K] ``in` `temp, test_list))`` ` `# printing result``print``(``"Top N elements of Kth index are : "` `+` `str``(res))`
Output :

The original list is : [(‘gfg’, 4, ‘good’), (‘gfg’, 2, ‘better’), (‘gfg’, 1, ‘best’), (‘gfg’, 3, ‘geeks’)]
Top N elements of Kth index are : [(‘gfg’, 4, ‘good’), (‘gfg’, 2, ‘better’), (‘gfg’, 3, ‘geeks’)]

Method #2 : Using `groupby() + sorted()` + loop

This task can also be performed using above functionalities. In this, we first group the top N elements together and then limit by N while constructing the result list.

 `# Python3 code to demonstrate working of``# Top N pairs by Kth element from list``# Using groupby() + sorted() + loop``import` `itertools`` ` `# initialize list ``test_list ``=` `[(``'gfg'``, ``4``, ``'good'``), (``'gfg'``, ``2``, ``'better'``),``             ``(``'gfg'``, ``1``, ``'best'``), (``'gfg'``, ``3``, ``'geeks'``)]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initialize N ``N ``=` `3` ` ` `# initialize K ``K ``=` `1`` ` `# Top N pairs by Kth element from list``# Using groupby() + sorted() + loop``res ``=` `[]``temp ``=` `itertools.groupby(``sorted``(test_list, key ``=` `lambda` `sub : sub[K], ``                          ``reverse ``=` `True``), key ``=` `lambda` `sub : sub[K])``for` `i ``in` `range``(N):``    ``res.extend(``list``(``next``(temp)[K]))`` ` `# printing result``print``(``"Top N elements of Kth index are : "` `+` `str``(res))`
Output :

The original list is : [(‘gfg’, 4, ‘good’), (‘gfg’, 2, ‘better’), (‘gfg’, 1, ‘best’), (‘gfg’, 3, ‘geeks’)]
Top N elements of Kth index are : [(‘gfg’, 4, ‘good’), (‘gfg’, 2, ‘better’), (‘gfg’, 3, ‘geeks’)]

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