# Python – Remove Rows for similar Kth column element

• Last Updated : 27 Oct, 2021

Given a Matrix, remove row if similar element has occurred in row above in Kth column.

Input : test_list = [[3, 4, 5], [2, 3, 5], [10, 4, 3], [7, 8, 9], [9, 3, 6]], K = 2
Output : [[3, 4, 5], [10, 4, 3], [7, 8, 9], [9, 3, 6]]
Explanation : In [2, 3, 5], we already has list [3, 4, 5] having 5 at K, i.e 2nd pos.
Input : test_list = [[3, 4, 5], [2, 3, 3], [10, 4, 3], [7, 8, 9], [9, 3, 6]], K = 2
Output : [[3, 4, 5], [2, 3, 3], [7, 8, 9], [9, 3, 6]]
Explanation : In [10, 4, 3], we already has list [2, 3, 3] having 3 at K, i.e 2nd pos.

Method : Using loop

In this, we maintain a memoization container which keeps track of elements in Kth column, if row’s Kth column element is present already, that row is omitted from result.

## Python3

 `# Python3 code to demonstrate working of``# Remove Rows for similar Kth column element``# Using loop` `# initializing list``test_list ``=` `[[``3``, ``4``, ``5``], [``2``, ``3``, ``5``], [``10``, ``4``, ``3``], [``7``, ``8``, ``9``], [``9``, ``3``, ``6``]]` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))` `# initializing K``K ``=` `1` `res ``=` `[]``memo ``=` `[]``for` `sub ``in` `test_list:``    ` `    ``# in operator used to check if present or not``    ``if` `not` `sub[K] ``in` `memo:``        ``res.append(sub)``        ``memo.append(sub[K])` `# printing result``print``(``"The filtered Matrix : "` `+` `str``(res))`

Output

```The original list is : [[3, 4, 5], [2, 3, 5], [10, 4, 3], [7, 8, 9], [9, 3, 6]]
The filtered Matrix : [[3, 4, 5], [2, 3, 5], [7, 8, 9]]```
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