How to delete the dulplicate list in list of list if I want to check by spetific index?

So I try to delete the duplicate list in list of list by checking just from only one element in that list. I only know to delete the duplicate from normal list, can you guys help me? Thanks!

So what I want to do is append just ONLY one list per name which I have sort it and the most value of each person will be the first one of another value in the same name.

input : liz = [['Lina', 60], ['Gaga', 0], ['Dene', 100], ['Dene', 80], ['Dene', 0], ['Helen', 0], ['Will', 100]]
expected output : lix = [['Lina', 60], ['Gaga', 0], ['Dene', 100], ['Helen', 0], ['Will', 100]]

What I have try to reach it and it just append all of list into list because it isn't equal to recent of all because its value of index[1]:

lix = []
for ele in liz:
        if ele not in lix:
            lix.append(ele)

To do so, you can simply have a record of all the names that are already in the list and compare them.

lix = {} # use dict instead of list
for pair in liz:
    if pair[0] in lix:
        if lix[pair[0]] < pair[1]: # if it is not the largest, add the largest
            lix[pair[0]] = pair[1]
    else:
        lix[pair[0]] = pair[1]

To keep the first number encountered for each distinct string, you can feed your list in reverse to a dictionary constructors. Then convert the items back into a list of lists. Because the dictionary constructor will only hold one item per key it updates the existing item when duplicates are supplied. So, the reversed order ensures that the first occurrences are loaded last in the dictionary and end up being the value at the end.

liz = [['Lina', 60], ['Gaga', 0], ['Dene', 100], ['Dene', 80], 
       ['Dene', 0], ['Helen', 0], ['Will', 100]]

lix = [*{L[0]:L for L in liz[::-1]}.values()][::-1]

print(lix)

[['Lina', 60], ['Gaga', 0], ['Dene', 100], ['Helen', 0], ['Will', 100]]

Another way to do this is to use a set of seen values that you check to filter and augment as you go through the items:

seen = set()
lix = [L for L in liz if not (L[0] in seen or seen.add(L[0]))]

the same can be done without a set but would run much slower because of the sequential search process used to check every item:

lix = list()
lix.extend(L for L in liz if all(L[0] != used for used,*_ in lix) )

If the sort order of your list already has the duplicate names in consecutive groups (and numbers in decreasing order for duplicate names), you can use the groupby function from itertools directly:

from itertools import groupby
lix = [L for _,(L,*_) in groupby(liz,lambda i:i[0])]

If it is not sorted, you can use a composite key to sort it:

liz.sort(key=lambda x:(x[0],-x[1]))  

# negating x[1] produces a decreasing order of numbers for identical names

Note that liz.sort(reverse=True) would also work but would give you the names in reversed alphabetical order

If you don't want to sort your list at all but want the largest numbers for each name, you can build a dictionary of selected sub-lists with a simple for-loop and use its values as the final output:

selected = dict()
for L in liz:
    if L[0] not in selected or L[1] > selected[L[0]][1]:
        selected[L[0]] = L
lix = list(selected.values())