How to find the last common value in all the contents of a list and get the position number of the contents in the list?

Question

Python

There is such a list (the number of contents in it may vary):

rows = [('oops', 0, 'new1', 'AgACAgIAAxkBAAISxl78VN0h', 4, 4821, 2654700348373, '111111'),
        ('aaapp', 7000, 'test22222', 'AgACAgIAAxkBAAIVEl78g75SvLP7MlZTBkncHQ', 4, 5429, 7965418332586, '111111')]

How do I find the last common elements of a list's content and get the position number of that content? For example: there are two tuples in the rows list, these tuples have the last common elements '111111', I have to get the sequence number of the tuples in the rows list where there are common elements. That is, if I run this list through the script, I should get 0 and 1.

Example 2:

rows = [('oops', 0, 'new1', 'AgACAgIAAxkBAAISxl78VN0h', 4, 4821, 2654700348373, '111111'),
        ('aaapp', 7000, 'test22222', 'AgACAgIAAxkBAAIVEl78g75SvLP7MlZTBkncHQ', 4, 5429, 7965418332586, '111111'),
        ('oops', 0, 'new1', 'AgACAgIAAxkBAAISxl78VN0h', 4, 4821, 2654700348373, '22222'),
        ('oops', 0, 'new1', 'AgACAgIAAxkBAAISxl78VN0h', 4, 4821, 2654700348373, '22222'),]

If I run this list through the script, I should get 0 and 1 divided by 2 and 3. That is, the script should split the common elements into pairs. How can this be done?

Answer 1

from itertools import groupby
print(
    [
        (k, list(l))
        for k, l in
        groupby(
            enumerate(rows),
            key=lambda row: row[1][-1]
        )
    ]
)