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Python | Minimum K records of Nth index in tuple list

• Last Updated : 17 Dec, 2019

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

Method #1 : Using filter() + lambda + set() + list comprehension
The combination of above functions can be used to perform this particular function. In this, we first filter the min K elements from Nth index and then apply this values to the list and return the result.

 # Python3 code to demonstrate working of# Minimum K records of Nth index in tuple 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 listprint("The original list is : " + str(test_list))  # initialize N N = 1  # initialize K K = 2  # Minimum K records of Nth index in tuple list# Using filter() + lambda + set() + list comprehensiontemp = set(list({sub[N] for sub in test_list})[ :K])res = list(filter(lambda sub: sub[N] in temp, test_list))  # printing resultprint("Min K elements of Nth index are : " + str(res))
Output :

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

Method #2 : Using groupby() + sorted() + loop
This task can also be performed using above functionalities. In this, we first group the min K elements together and then limit by K while constructing the result list.

 # Python3 code to demonstrate working of# Minimum K records of Nth index in tuple list# Using groupby() + sorted() + loopimport itertools  # initialize list test_list = [('gfg', 4, 'good'), ('gfg', 2, 'better'),              ('gfg', 1, 'best'), ('gfg', 3, 'geeks')]  # printing original listprint("The original list is : " + str(test_list))  # initialize N N = 1  # initialize K K = 2  # Minimum K records of Nth index in tuple list# Using groupby() + sorted() + loopres = []temp = itertools.groupby(sorted(test_list, key = lambda sub : sub[N]),                                             key = lambda sub : sub[N])  for i in range(K):    res.extend(list(next(temp)[N]))  # printing resultprint("Min K elements of Nth index are : " + str(res))
Output :

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

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