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# Python | K-th element value indices

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Method # 1: Using a Loop

This is a brute force method by which this problem can be solved. In this case, we keep a counter for the indices and add to the list if we find a specific entry at the Kth position in the tuple.

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` # Python3 code to demonstrate how it works # Indexes of the value of the Kth element # Using a loop   # initialize the list test_list = [( 3 , 1 , 5 ), ( 1 , 3 , 6 ), ( 2 , 5 , 7 ), ( 5 , 2 , 8 ), ( 6 , 3 , 0 )]   # print original list print ( "The original list is:" + str (test_list))   # initialize ele ele = 3   # initialize K K = 1     # Indexes of the value of the Kth element # Using a loop # using y for K = 1 res = [] count = 0 for x, y, z in test_list: if y = = ele: res.append (count) count = count + 1   # print result print ( "The indices of element at Kth position: " + str (res)) `

` ` Output:

The original list is: [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)]
The indices of element at Kth position: [1, 4]

Method # 2: Using ` en umerate () ` + list comprehension

A combination of the above functions can be used to solve this problem. In this we enumerate the indexes using ` enumerate () `, the rest are performed as in the previous method, but compactly.

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` # Python3 code for demonstrations # Indexes of the value of the Kth element # Using enumerate () + list comprehension   # initialize the list test_list = [( 3 , 1 , 5 ), ( 1 , 3 , 6 ), ( 2 , 5 , 7 ),  ( 5 , 2 , 8 ), ( 6 , 3 , 0 )]    # print original list print ( "The original list is:" + str (test_list))   # initialize ele ele = 3   # initialize K K = 1     # Kth element value indices # Using enumerate () + list comprehension res = [a for a, b in enumerate (test_list) if b [K] = = ele]    # print result print ( "The indices of element at Kth position:" + str (res)) `

` ` Output:

The original list is: [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)]
The indices of element at Kth position: [1, 4]

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