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Python – Unique Kth positioned tuples

  • Last Updated : 02 Jun, 2020

Sometimes, while working with Python records, we can have a problem in which we need to extract only the unique tuples, based on some particular index of tuples. This kind of problem can have applications in domains such as web development. Let’s discuss certain ways in which this task can be performed.

Input :
test_list = [(5, 6, 5), (4, 2, 7), (1, 2, 3), (9, 6, 5)]
K = 3
Output : [(1, 2, 3), (5, 6, 5), (4, 2, 7)]
Input :
test_list = [(5, ), (1, ), (1, ), (9, )]
K = 1
Output : [(1, ), (5, ), (9, )]

Method #1 : Using map() + next() + lambda
The combination of above functions can be used to solve this problem. In this, we extend the logic of getting unique elements extracted using lambda function and next(), using map().




# Python3 code to demonstrate working of 
# Unique Kth index tuples
# Using map() + next() + lambda
  
# initializing list
test_list = [(5, 6, 8), (4, 2, 7), (1, 2, 3), (9, 6, 5)]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K 
K = 2
  
# Unique Kth index tuples
# Using map() + next() + lambda
res = [*map(lambda ele: next(tup for tup in test_list if tup[K - 1] == ele),
      {tup[K - 1] for tup in test_list})]
  
# printing result 
print("The extracted elements : " + str(res)) 
Output :
The original list is : [(5, 6, 8), (4, 2, 7), (1, 2, 3), (9, 6, 5)]
The extracted elements : [(4, 2, 7), (5, 6, 8)]

 



Method #2 : Using next() + groupby() + lambda
The combination of above functions can also be used to solve this problem. In this, we perform task of map() in above using groupby(), in a more compact way.




# Python3 code to demonstrate working of 
# Unique Kth index tuples
# Using next() + groupby() + lambda
from itertools import groupby
  
# initializing list
test_list = [(5, 6, 8), (4, 2, 7), (1, 2, 3), (9, 6, 5)]
  
# printing original list
print("The original list is : " + str(test_list))
  
# initializing K 
K = 2
  
# Unique Kth index tuples
# Using next() + groupby() + lambda
temp = lambda ele : ele[K - 1]
res = [next(val) for _, val in groupby(sorted(test_list, key = temp), key = temp)]
  
# printing result 
print("The extracted elements : " + str(res)) 
Output :
The original list is : [(5, 6, 8), (4, 2, 7), (1, 2, 3), (9, 6, 5)]
The extracted elements : [(4, 2, 7), (5, 6, 8)]

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