Remove lists within a nested list with duplicates based on last two elements

Question

I have a nested list like this:

my_list = [['10', 'A', 'A'], ['5', 'B', 'A'], ['2', 'B', 'B'], ['10','A','B']]

I would like to check if there is a duplicate letter in the last two positions and if there is not, print those lists.

final_list = [['5','B','A'],['10','A','B']

Finally, I would like to print the numerical value from each of these lists:

only_numbers = ['5','10']

However, I'm getting stuck at the identifying duplicates within the lists. I have found this answer Removing Duplicates from Nested List Based on First 2 Elements, but I when I tried to apply this on my example code above (and actual code), I got some lists with duplicates and some without duplicates.

seen = set() 
seen_add = seen.add
final_list = [x for x in my_list if tuple(x[-2:]) not in seen and not seen_add(tuple(x[-2:]))]

But I get:

final_list = [['10', 'A', 'A'], ['5', 'B', 'A'], ['2', 'B', 'B'], ['10', 'A', 'B']]

What am I missing?

EDIT: Fixed my personal coding attempt below so that its functions (thank you to Karl Knechtel for your explanation)

only_numbers = []
for x in my_list:  
     if not x[1] == x[2]:
         add_number.append(x[0])
print(only_numbers)

"I have found this answer Removing Duplicates from Nested List Based on First 2 Elements" When you say "duplicate", you mean something completely different from the other person. You want to check two different elements within each sublist; the other code is about comparing of elements between different sublists. In your example data, you have one sublist each with a different pair of letters - AA, AB, BA, and BB - none of those is equal to any other, so no duplicates are detected. — Karl Knechtel
– Karl Knechtel, Commented Sep 22, 2020 at 3:54
In general, the way you do list comprehensions that care about each element in isolation is: first you come up with a rule that tells you what to do with the element, then you write the comprehension so it applies the rule to every element. In your case, the rule is to take that element (which is also a list) and compare the last two values in that list. So, write the code that does the comparison first. — Karl Knechtel
– Karl Knechtel, Commented Sep 22, 2020 at 3:56
"And if someone has the time, please explain why my logic for this code below I wrote is wrong" The way in which it is wrong is that you get three copies of each number. The reason for this is that the i loop is unwanted: you only want to .append the selected x[0] value once, but you append it every time through the inner i loop. It loops three times because your sublists have 3 elements in them. — Karl Knechtel
– Karl Knechtel, Commented Sep 22, 2020 at 3:59
Perhaps you automatically wrote a nested loop because you have nested lists. The issue is that you do not treat the inner lists as something to loop over; you have a specific way of looking at specific indexes within the inner list. — Karl Knechtel
– Karl Knechtel, Commented Sep 22, 2020 at 4:00

Rakesh · Accepted Answer · 2020-09-22 03:49:07Z

2

This is one approach using filter and a list comprehension

Ex:

my_list = [['10', 'A', 'A'], ['5', 'B', 'A'], ['2', 'B', 'B'], ['10','A','B']]
result = [i for i, *_ in filter(lambda x: x[-1] != x[-2], my_list)]
# OR 
# [i for i, j, k in my_list if j != k]
print(result)

Output:

['5', '10']

answered Sep 22, 2020 at 3:49

Rakesh

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