Python has an ordered dictionary. What about an ordered set?
For many purposes simply calling sorted will suffice. For example
>>> s = set([0, 1, 2, 99, 4, 40, 3, 20, 24, 100, 60])
>>> sorted(s)
[0, 1, 2, 3, 4, 20, 24, 40, 60, 99, 100]
If you are going to use this repeatedly, there will be overhead incurred by calling the sorted function so you might want to save the resulting list, as long as you're done changing the set. If you need to maintain unique elements and sorted, I agree with the suggestion of using OrderedDict from collections with an arbitrary value such as None.
As others have said, OrderedDict
is a superset of an ordered set in terms of functionality, but if you need a set for interacting with an API and don't need it to be mutable, OrderedDict.keys()
is actually an implementation abc.collections.Set
:
import random
from collections import OrderedDict, abc
a = list(range(0, 100))
random.shuffle(a)
# True
a == list(OrderedDict((i, 0) for i in a).keys())
# True
isinstance(OrderedDict().keys(), abc.Set)
The caveats are immutability and having to build up the set like a dict, but it's simple and only uses built-ins.
If you're using the ordered set to maintain a sorted order, consider using a sorted set implementation from PyPI. The sortedcontainers module provides a SortedSet for just this purpose. Some benefits: pure-Python, fast-as-C implementations, 100% unit test coverage, hours of stress testing.
Installing from PyPI is easy with pip:
pip install sortedcontainers
Note that if you can't pip install
, simply pull down the sortedlist.py and sortedset.py files from the open-source repository.
Once installed you can simply:
from sortedcontainers import SortedSet
help(SortedSet)
The sortedcontainers module also maintains a performance comparison with several alternative implementations.
For the comment that asked about Python's bag data type, there's alternatively a SortedList data type which can be used to efficiently implement a bag.
The ParallelRegression package provides a setList( ) ordered set class that is more method-complete than the options based on the ActiveState recipe. It supports all methods available for lists and most if not all methods available for sets.
So i also had a small list where i clearly had the possibility of introducing non-unique values.
I searched for the existence of a unique list of some sort, but then realized that testing the existence of the element before adding it works just fine.
if(not new_element in my_list):
my_list.append(new_element)
I don't know if there are caveats to this simple approach, but it solves my problem.
As other answers mention, as for python 3.7+, the dict is ordered by definition. Instead of subclassing OrderedDict
we can subclass abc.collections.MutableSet
or typing.MutableSet
using the dict's keys to store our values.
class OrderedSet(typing.MutableSet[T]):
"""A set that preserves insertion order by internally using a dict."""
def __init__(self, iterable: t.Iterator[T]):
self._d = dict.fromkeys(iterable)
def add(self, x: T) -> None:
self._d[x] = None
def discard(self, x: T) -> None:
self._d.pop(x)
def __contains__(self, x: object) -> bool:
return self._d.__contains__(x)
def __len__(self) -> int:
return self._d.__len__()
def __iter__(self) -> t.Iterator[T]:
return self._d.__iter__()
Then just:
x = OrderedSet([1, 2, -1, "bar"])
x.add(0)
assert list(x) == [1, 2, -1, "bar", 0]
I put this code in a small library, so anyone can just pip install
it.
I can do you one better than an OrderedSet: boltons has a pure-Python, 2/3-compatible IndexedSet
type that is not only an ordered set, but also supports indexing (as with lists).
Simply pip install boltons
(or copy setutils.py
into your codebase), import the IndexedSet
and:
>>> from boltons.setutils import IndexedSet
>>> x = IndexedSet(list(range(4)) + list(range(8)))
>>> x
IndexedSet([0, 1, 2, 3, 4, 5, 6, 7])
>>> x - set(range(2))
IndexedSet([2, 3, 4, 5, 6, 7])
>>> x[-1]
7
>>> fcr = IndexedSet('freecreditreport.com')
>>> ''.join(fcr[:fcr.index('.')])
'frecditpo'
Everything is unique and retained in order. Full disclosure: I wrote the IndexedSet
, but that also means you can bug me if there are any issues. :)
While others have pointed out that there is no built-in implementation of an insertion-order preserving set in Python (yet), I am feeling that this question is missing an answer which states what there is to be found on PyPI.
There are the packages:
Some of these implementations are based on the recipe posted by Raymond Hettinger to ActiveState which is also mentioned in other answers here.
my_set[5]
)remove(item)
Both implementations have O(1) for add(item)
and __contains__(item)
(item in my_set
).
There's no OrderedSet
in official library.
I make an exhaustive cheatsheet of all the data structure for your reference.
DataStructure = {
'Collections': {
'Map': [
('dict', 'OrderDict', 'defaultdict'),
('chainmap', 'types.MappingProxyType')
],
'Set': [('set', 'frozenset'), {'multiset': 'collection.Counter'}]
},
'Sequence': {
'Basic': ['list', 'tuple', 'iterator']
},
'Algorithm': {
'Priority': ['heapq', 'queue.PriorityQueue'],
'Queue': ['queue.Queue', 'multiprocessing.Queue'],
'Stack': ['collection.deque', 'queue.LifeQueue']
},
'text_sequence': ['str', 'byte', 'bytearray']
}
A little late to the game, but I've written a class setlist
as part of collections-extended
that fully implements both Sequence
and Set
>>> from collections_extended import setlist
>>> sl = setlist('abracadabra')
>>> sl
setlist(('a', 'b', 'r', 'c', 'd'))
>>> sl[3]
'c'
>>> sl[-1]
'd'
>>> 'r' in sl # testing for inclusion is fast
True
>>> sl.index('d') # so is finding the index of an element
4
>>> sl.insert(1, 'd') # inserting an element already in raises a ValueError
ValueError
>>> sl.index('d')
4
GitHub: https://github.com/mlenzen/collections-extended
Documentation: http://collections-extended.lenzm.net/en/latest/
The keys of a dictionary are unique. Thus, if one disregards the values in an ordered dictionary (e.g. by assigning them None
), then one has essentially an ordered set.
As of Python 3.1 and 2.7 there is collections.OrderedDict
. The following is an example implementation of an OrderedSet. (Note that only few methods need to be defined or overridden: collections.OrderedDict
and collections.MutableSet
do the heavy lifting.)
import collections
class OrderedSet(collections.OrderedDict, collections.MutableSet):
def update(self, *args, **kwargs):
if kwargs:
raise TypeError("update() takes no keyword arguments")
for s in args:
for e in s:
self.add(e)
def add(self, elem):
self[elem] = None
def discard(self, elem):
self.pop(elem, None)
def __le__(self, other):
return all(e in other for e in self)
def __lt__(self, other):
return self <= other and self != other
def __ge__(self, other):
return all(e in self for e in other)
def __gt__(self, other):
return self >= other and self != other
def __repr__(self):
return 'OrderedSet([%s])' % (', '.join(map(repr, self.keys())))
def __str__(self):
return '{%s}' % (', '.join(map(repr, self.keys())))
difference = __sub__
difference_update = __isub__
intersection = __and__
intersection_update = __iand__
issubset = __le__
issuperset = __ge__
symmetric_difference = __xor__
symmetric_difference_update = __ixor__
union = __or__
In case you're already using pandas in your code, its Index
object behaves pretty like an ordered set, as shown in this article.
Examples from the article:
indA = pd.Index([1, 3, 5, 7, 9])
indB = pd.Index([2, 3, 5, 7, 11])
indA & indB # intersection
indA | indB # union
indA - indB # difference
indA ^ indB # symmetric difference
The answer is no, but you can use collections.OrderedDict
from the Python standard library with just keys (and values as None
) for the same purpose.
Update: As of Python 3.7 (and CPython 3.6), standard dict
is guaranteed to preserve order and is more performant than OrderedDict
. (For backward compatibility and especially readability, however, you may wish to continue using OrderedDict
.)
Here's an example of how to use dict
as an ordered set to filter out duplicate items while preserving order, thereby emulating an ordered set. Use the dict
class method fromkeys()
to create a dict, then simply ask for the keys()
back.
>>> keywords = ['foo', 'bar', 'bar', 'foo', 'baz', 'foo']
>>> list(dict.fromkeys(keywords))
['foo', 'bar', 'baz']
Source: Stackoverflow.com