CODEWITHSUNDEEP
pythonlistduplicates
Ella
Ella

Reputation: 361

Remove duplicates without changing the order in two dimension list

How could I remove duplicates two dimension list without changing the order?

list_j = [[100,2,3,3], [4,98,99,98], [5,99,98,4], [5,99,98,5], [100,99,98,100,100,6]]

list_pre = [list(set(i)) for i in list_j]
print(list_pre)

[[2, 3, 100], [98, 99, 4], [98, 99, 4, 5], [98, 99, 5], [98, 99, 100, 6]]

As you can see it changed the order. What I want is [[100,2,3,],...]

Desired output [[100,2,3,], [4,98,99], [5,99,98,4], [5,99,98], [100,99,98,6]]

Upvotes: 0

Views: 98

Answers (3)

Higs
Higs

Reputation: 404

Try this:

def remove_duplicates(my_list):
    result_list = []
    for _list in my_list:
        result_list.append(sorted(set(_list), key=lambda ind: _list.index(ind)))
    return result_list

Upvotes: 0

Jarvis
Jarvis

Reputation: 8564

Use OrderedDict to maintain the order of insertion of keys:

from collections import OrderedDict
list_j = [[100,2,3,3], [4,98,99,98], [5,99,98,4], [5,99,98,5], [100,99,98,100,100,6]]
output = [list(OrderedDict.fromkeys(x)) for x in list_j]

gives

[[100, 2, 3], [4, 98, 99], [5, 99, 98, 4], [5, 99, 98], [100, 99, 98, 6]]

If you using Python 3.7 or higher, you can use normal dictionaries as well since they also maintain the order of insertion:

output = [list(dict.fromkeys(x)) for x in list_j]

Upvotes: 5

mpountou
mpountou

Reputation: 838

Same question here:

def f7(seq):
    seen = set()
    seen_add = seen.add
    return [x for x in seq if not (x in seen or seen_add(x))]

list_j = [[100,2,3,3], [4,98,99,98], [5,99,98,4], [5,99,98,5], [100,99,98,100,100,6]]

list_pre = [f7(i) for i in list_j]

print(list_pre)

This will give the desired output

Upvotes: 0

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