CODEWITHSUNDEEP
python
user2417731
user2417731

Reputation: 179

Keeping same order when using set

My task is to use a set to convert a list of duplicates to a list of unique numbers. However, I want to retain the positions.

Simple enough I thought; so I made a dictionary that first stores the original list's positions.

def get_positions(a): 
    positions = {}

    for ele in a: 
        if not ele in positions:
            positions[ele] = a.index(ele) 

    return positions

So lets say I have a list a = [1, 2, 4, 4, 5]

Positions will give me a dictionary of {0:1, 1:2, 2:4, 3:4, 4:5}.

This however was unsuccessful because I repeated numbers will not get their positions stored.

Is there a way of achieving this?

Thanks.

UPDATE:

It seems I wasn't clear. I need to use a set. So, I get a list a=[1,2,4,4,5] and I must convert it to a set to erase the duplicates. Then, I need to get a list with the elements in the same order. (It's an assignment problem)

Upvotes: 0

Views: 456

Answers (5)

rlms
rlms

Reputation: 11050

I think you are going about this the wrong way. You are trying to remove duplicates from a list, but you are having a problem which you are trying to solve by having the positions of things in the list without duplicates being removed. Instead, I think it would be better to do something more like this:

def remove_duplicates(seq):
    new_list = []
    for i in seq:
        if i not in new_list:
            new_list.append(i)
    return new_list

One fairly readable way to do this, using sets (with the corresponding O(1) membership testing (but with higher memory usage) is this:

def remove_duplicates(seq):
    seen = set()
    new_list = []
    for i in seq:
        if i not in seen:
            new_list.append(i)
            seen.add(i)
    return new_list

This answer to the same question also uses sets, and is quite possibly faster (but using a bit hacky in using and not set.add).

Upvotes: 2

Corley Brigman
Corley Brigman

Reputation: 12371

you want an OrderedSet. but this sounds like a homework problem, and i don't know if they would accept that.

Upvotes: 0

arshajii
arshajii

Reputation: 129477

You can use an OrderedDict:

>>> from collections import OrderedDict
>>> 
>>> a = [1, 2, 4, 4, 5]
>>> 
>>> list(OrderedDict.fromkeys(a))
[1, 2, 4, 5]

You can also use a plain set for this. One common way is:

>>> a = [1, 2, 4, 4, 5]
>>> 
>>> seen = set()
>>> [x for x in a if x not in seen and not seen.add(x)]
[1, 2, 4, 5]

The trick here is that not seen.add(x) will always be True because add() always returns None. In practice, I would always use the OrderedDict approach.

See also: How do you remove duplicates from a list in Python whilst preserving order?

Upvotes: 6

Joel
Joel

Reputation: 5674

Store it as a set of tuples with (position, element)

Upvotes: 0

ModulusJoe
ModulusJoe

Reputation: 1446

This can be done with a loop and an if statement:

>>> oldlist = [1,2,3,3,4,5,4,5,6,2,3,5,7,8,3,3,3,9]
>>> newlist = []
>>> for x in oldlist:
...  if not x in newlist:
...   newlist.append(x)
...
>>> newlist
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>>

Upvotes: 1

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