Recently I noticed that when I am converting a list
to set
the order of elements is changed and is sorted by character.
Consider this examp
Here's an easy way to do it:
x=[1,2,20,6,210]
print sorted(set(x))
In mathematics, there are sets and ordered sets (osets).
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:
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).
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
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.