Converting a list to a set changes element order

In Python 3.6, set() now should keep the order, but there is another solution for Python 2 and 3:

>>> x = [1, 2, 20, 6, 210]
>>> sorted(set(x), key=x.index)
[1, 2, 20, 6, 210]

Remove duplicates and preserve order by below function

def unique(sequence):
    seen = set()
    return [x for x in sequence if not (x in seen or seen.add(x))]

How to remove duplicates from a list while preserving order in Python

Answering your first question, a set is a data structure optimized for set operations. Like a mathematical set, it does not enforce or maintain any particular order of the elements. The abstract concept of a set does not enforce order, so the implementation is not required to. When you create a set from a list, Python has the liberty to change the order of the elements for the needs of the internal implementation it uses for a set, which is able to perform set operations efficiently.

In mathematics, there are sets and ordered sets (osets).

• set: an unordered container of unique elements (Implemented)
• oset: an ordered container of unique elements (NotImplemented)

In Python, only sets are directly implemented. We can emulate osets with regular dict keys (3.7+).

Given

a = [1, 2, 20, 6, 210, 2, 1]
b = {2, 6}

Code

oset = dict.fromkeys(a).keys()
# dict_keys([1, 2, 20, 6, 210])

Demo

Replicates are removed, insertion-order is preserved.

list(oset)
# [1, 2, 20, 6, 210]

Set-like operations on dict keys.

oset - b
# {1, 20, 210}

oset | b
# {1, 2, 5, 6, 20, 210}

oset & b
# {2, 6}

oset ^ b
# {1, 5, 20, 210}

Details

Note: an unordered structure does not preclude ordered elements. Rather, maintained order is not guaranteed. Example:

assert {1, 2, 3} == {2, 3, 1}                    # sets (order is ignored)

assert [1, 2, 3] != [2, 3, 1]                    # lists (order is guaranteed)

One may be pleased to discover that a list and multiset (mset) are two more fascinating, mathematical data structures:

• list: an ordered container of elements that permits replicates (Implemented)
• mset: an unordered container of elements that permits replicates (NotImplemented)*

Summary

Container | Ordered | Unique | Implemented
----------|---------|--------|------------
set       |    n    |    y   |     y
oset      |    y    |    y   |     n
list      |    y    |    n   |     y
mset      |    n    |    n   |     n*  

^{*A multiset can be indirectly emulated with collections.Counter(), a dict-like mapping of multiplicities (counts).}

You can remove the duplicated values and keep the list order of insertion with one line of code, Python 3.8.2

mylist = ['b', 'b', 'a', 'd', 'd', 'c']


results = list({value:"" for value in mylist})

print(results)

>>> ['b', 'a', 'd', 'c']

results = list(dict.fromkeys(mylist))

print(results)

>>> ['b', 'a', 'd', 'c']

As denoted in other answers, sets are data structures (and mathematical concepts) that do not preserve the element order -

However, by using a combination of sets and dictionaries, it is possible that you can achieve wathever you want - try using these snippets:

# save the element order in a dict:
x_dict = dict(x,y for y, x in enumerate(my_list) )
x_set = set(my_list)
#perform desired set operations
...
#retrieve ordered list from the set:
new_list = [None] * len(new_set)
for element in new_set:
   new_list[x_dict[element]] = element

Building on Sven's answer, I found using collections.OrderedDict like so helped me accomplish what you want plus allow me to add more items to the dict:

import collections

x=[1,2,20,6,210]
z=collections.OrderedDict.fromkeys(x)
z
OrderedDict([(1, None), (2, None), (20, None), (6, None), (210, None)])

If you want to add items but still treat it like a set you can just do:

z['nextitem']=None

And you can perform an operation like z.keys() on the dict and get the set:

list(z.keys())
[1, 2, 20, 6, 210]

Late to answer but you can use Pandas, pd.Series to convert list while preserving the order:

import pandas as pd
x = pd.Series([1, 2, 20, 6, 210, 2, 1])
print(pd.unique(x))

Output: array([ 1, 2, 20, 6, 210])

Works for a list of strings

x = pd.Series(['c', 'k', 'q', 'n', 'p','c', 'n'])
print(pd.unique(x))

Output ['c' 'k' 'q' 'n' 'p']

One more simpler way can be two create a empty list ,let's say "unique_list" for adding the unique elements from the original list, for example:

unique_list=[]

for i in original_list:
    if i not in unique_list:
        unique_list.append(i)
    else:
        pass

This will give you all the unique elements as well as maintain the order.

I like the following solution for its conciseness. It uses PEP 448 – Additional Unpacking Generalizations which is valid in Python 3.5+, and relies on dict insertion order being preserved (CPython 3.6+ or Python 3.7+).

def unique(sequence):
    return [*dict.fromkeys(sequence)]

Also, for larger sequences, it seems to be faster than checking set membership:

from timeit import timeit


def unique(sequence):
    return [*dict.fromkeys(sequence)]


def set_unique(sequence):
    seen = set()
    return [x for x in sequence if not (x in seen or seen.add(x))]


a_large_list = [*range(100)] + [*range(10)]

assert set_unique(a_large_list) == unique(a_large_list)
print(
    timeit('set_unique(a_large_list)', globals=globals(), number=100_000)
    / timeit('unique(a_large_list)', globals=globals(), number=100_000)
)  # 2.379634026726176 times faster!

An implementation of the highest score concept above that brings it back to a list:

def SetOfListInOrder(incominglist):
    from collections import OrderedDict
    outtemp = OrderedDict()
    for item in incominglist:
        outtemp[item] = None
    return(list(outtemp))

Tested (briefly) on Python 3.6 and Python 2.7.

In case you have a small number of elements in your two initial lists on which you want to do set difference operation, instead of using collections.OrderedDict which complicates the implementation and makes it less readable, you can use:

# initial lists on which you want to do set difference
>>> nums = [1,2,2,3,3,4,4,5]
>>> evens = [2,4,4,6]
>>> evens_set = set(evens)
>>> result = []
>>> for n in nums:
...   if not n in evens_set and not n in result:
...     result.append(n)
... 
>>> result
[1, 3, 5]

Its time complexity is not that good but it is neat and easy to read.

It's interesting that people always use 'real world problem' to make joke on the definition in theoretical science.

If set has order, you first need to figure out the following problems. If your list has duplicate elements, what should the order be when you turn it into a set? What is the order if we union two sets? What is the order if we intersect two sets with different order on the same elements?

Plus, set is much faster in searching for a particular key which is very good in sets operation (and that's why you need a set, but not list).

If you really care about the index, just keep it as a list. If you still want to do set operation on the elements in many lists, the simplest way is creating a dictionary for each list with the same keys in the set along with a value of list containing all the index of the key in the original list.

def indx_dic(l):
    dic = {}
    for i in range(len(l)):
        if l[i] in dic:
            dic.get(l[i]).append(i)
        else:
            dic[l[i]] = [i]
    return(dic)

a = [1,2,3,4,5,1,3,2]
set_a  = set(a)
dic_a = indx_dic(a)

print(dic_a)
# {1: [0, 5], 2: [1, 7], 3: [2, 6], 4: [3], 5: [4]}
print(set_a)
# {1, 2, 3, 4, 5}

We can use collections.Counter for this:

# tested on python 3.7
>>> from collections import Counter
>>> lst = ["1", "2", "20", "6", "210"]

>>> for i in Counter(lst):
>>>     print(i, end=" ")
1 2 20 6 210 

>>> for i in set(lst):
>>>     print(i, end=" ")
20 6 2 1 210

You can remove the duplicated values and keep the list order of insertion, if you want

lst = [1,2,1,3]
new_lst = []

for num in lst :
    if num not in new_lst :
        new_lst.append(num)

# new_lst = [1,2,3]

don't use 'sets' for removing duplicate if 'order' is something you want,

use sets for searching i.e.
x in list
takes O(n) time
where
x in set
takes O(1) time *most cases