I have a list of instances of a certain class. This list contains `duplicates', in the sense that duplicates share the exact same attributes. I want to remove the duplicates from this list.
I can check whether two instances share the same attributes by using
class MyClass:
def __eq__(self, other) :
return self.__dict__ == other.__dict__
I could of course iterate through the whole list of instances and compare them element by element to remove duplicates, but I was wondering if there is a more pythonic way to do this, preferably using the in operator + list comprehension.
sets (no order)
A set cannot contain duplicate elements. list(set(content)) will deduplicate a list. This is not too inefficient and is probably one of the better ways to do it :P You will need to define a __hash__ function for your class though, which must be the same for equal elements and different for unequal elements for this to work. Note that the hash value must obey the aforementioned rule but otherwise it may change between runs without causing issues.
index function (stable order)
You could do lambda l: [l[index] for index in range(len(l)) if index == l.index(l[index])]. This only keeps elements that are the first in the list.
in operator (stable order)
def uniquify(content):
result = []
for element in content:
if element not in result:
result.append(element)
return result
This will keep appending elements to the output list unless they are already in the output list.
A little more on the set approach. You can safely implement a hash by delegating to a tuple's hash - just hash a tuple of all the attributes you want to look at. You will also need to define an __eq__ that behaves properly.
class MyClass:
def __init__(self, a, b, c):
self.a = a
self.b = b
self.c = c
def __eq__(self, other):
return (self.a, self.b, self.c) == (other.a, other.b, other.c)
def __hash__(self):
return hash((self.a, self.b, self.c))
def __repr__(self):
return "MyClass({!r}, {!r}, {!r})".format(self.a, self.b, self.c)
As you're doing so much tuple construction, you could just make your class iterable:
def __iter__(self):
return iter((self.a, self.b, self.c))
This enables you to call tuple on self instead of laboriously doing .a, .b, .c etc.
You can then do something like this:
def unordered_elim(l):
return list(set(l))
If you want to preserve ordering, you can use an OrderedDict instead:
from collections import OrderedDict
def ordered_elim(l):
return list(OrderedDict.fromkeys(l).keys())
This should be faster than using in or index, while still preserving ordering. You can test it something like this:
data = [MyClass("this", "is a", "duplicate"),
MyClass("first", "unique", "datum"),
MyClass("this", "is a", "duplicate"),
MyClass("second", "unique", "datum")]
print(unordered_elim(data))
print(ordered_elim(data))
With this output:
[MyClass('first', 'unique', 'datum'), MyClass('second', 'unique', 'datum'), MyClass('this', 'is a', 'duplicate')]
[MyClass('this', 'is a', 'duplicate'), MyClass('first', 'unique', 'datum'), MyClass('second', 'unique', 'datum')]
NB if any of your attributes aren't hashable, this won't work, and you'll either need to work around it (change a list to a tuple) or use a slow, n ^ 2 approach like in.
I have nested dictionaries that may contain other dictionaries or lists. I need to be able to compare a list (or set, really) of these dictionaries to show that they are equal.
The order of the list is not uniform. Typically, I would turn the list into a set, but it is not possible since there are values that are also dictionaries.
a = {'color': 'red'}
b = {'shape': 'triangle'}
c = {'children': [{'color': 'red'}, {'age': 8},]}
test_a = [a, b, c]
test_b = [b, c, a]
print(test_a == test_b) # False
print(set(test_a) == set(test_b)) # TypeError: unhashable type: 'dict'
Is there a good way to approach this to show that test_a has the same contents as test_b?
You can use a simple loop to check if each of one list is in the other:
def areEqual(a, b):
if len(a) != len(b):
return False
for d in a:
if d not in b:
return False
return True
I suggest writing a function that turns any Python object into something orderable, with its contents, if it has any, in sorted order. If we call it canonicalize, we can compare nested objects with:
canonicalize(test_a) == canonicalize(test_b)
Here's my attempt at writing a canonicalize function:
def canonicalize(x):
if isinstance(x, dict):
x = sorted((canonicalize(k), canonicalize(v)) for k, v in x.items())
elif isinstance(x, collections.abc.Iterable) and not isinstance(x, str):
x = sorted(map(canonicalize, x))
else:
try:
bool(x < x) # test for unorderable types like complex
except TypeError:
x = repr(x) # replace with something orderable
return x
This should work for most Python objects. It won't work for lists of heterogeneous items, containers that contain themselves (which will cause the function to hit the recursion limit), nor float('nan') (which has bizarre comparison behavior, and so may mess up the sorting of any container it's in).
It's possible that this code will do the wrong thing for non-iterable, unorderable objects, if they don't have a repr function that describes all the data that makes up their value (e.g. what is tested by ==). I picked repr as it will work on any kind of object and might get it right (it works for complex, for example). It should also work as desired for classes that have a repr that looks like a constructor call. For classes that have inherited object.__repr__ and so have repr output like <Foo object at 0xXXXXXXXX> it at least won't crash, though the objects will be compared by identity rather than value. I don't think there's any truly universal solution, and you can add some special cases for classes you expect to find in your data if they don't work with repr.
If the elements in both lists are shallow, the idea of sorting them, and then comparing with equality can work. The problem with #Alex's solution is that he is only using "id" - but if instead of id, one uses a function that will sort dictionaries properly, things shuld just work:
def sortkey(element):
if isinstance(element, dict):
element = sorted(element.items())
return repr(element)
sorted(test_a, key=sortkey) == sorted(test_b, key=sotrkey)
(I use an repr to wrap the key because it will cast all elements to string before comparison, which will avoid typerror if different elements are of unorderable types - which would almost certainly happen if you are using Python 3.x)
Just to be clear, if your dictionaries and lists have nested dictionaries themselves, you should use the answer by #m_callens. If your inner lists are also unorderd, you can fix this to work, jsut sorting them inside the key function as well.
In this case they are the same dicts so you can compare ids (docs). Note that if you introduced a new dict whose values were identical it would still be treated differently. I.e. d = {'color': 'red'} would be treated as not equal to a.
sorted(map(id, test_a)) == sorted(map(id, test_b))
As #jsbueno points out, you can do this with the kwarg key.
sorted(test_a, key=id) == sorted(test_b, key=id)
An elegant and relatively fast solution:
class QuasiUnorderedList(list):
def __eq__(self, other):
"""This method isn't as ineffiecient as you think! It runs in O(1 + 2 + 3 + ... + n) time,
possibly better than recursively freezing/checking all the elements."""
for item in self:
for otheritem in other:
if otheritem == item:
break
else:
# no break was reached, item not found.
return False
return True
This runs in O(1 + 2 + 3 + ... + n) flat. While slow for dictionaries of low depth, this is faster for dictionaries of high depth.
Here's a considerably longer snippet which is faster for dictionaries where depth is low and length is high.
class FrozenDict(collections.Mapping, collections.Hashable): # collections.Hashable = portability
"""Adapated from http://stackoverflow.com/a/2704866/1459669"""
def __init__(self, *args, **kwargs):
self._d = dict(*args, **kwargs)
self._hash = None
def __iter__(self):
return iter(self._d)
def __len__(self):
return len(self._d)
def __getitem__(self, key):
return self._d[key]
def __hash__(self):
# It would have been simpler and maybe more obvious to
# use hash(tuple(sorted(self._d.iteritems()))) from this discussion
# so far, but this solution is O(n). I don't know what kind of
# n we are going to run into, but sometimes it's hard to resist the
# urge to optimize when it will gain improved algorithmic performance.
# Now thread safe by CrazyPython
if self._hash is None:
_hash = 0
for pair in self.iteritems():
_hash ^= hash(pair)
self._hash = _hash
return _hash
def freeze(obj):
if type(obj) in (str, int, ...): # other immutable atoms you store in your data structure
return obj
elif issubclass(type(obj), list): # ugly but needed
return set(freeze(item) for item in obj)
elif issubclass(type(obj), dict): # for defaultdict, etc.
return FrozenDict({key: freeze(value) for key, value in obj.items()})
else:
raise NotImplementedError("freeze() doesn't know how to freeze " + type(obj).__name__ + " objects!")
class FreezableList(list, collections.Hashable):
_stored_freeze = None
_hashed_self = None
def __eq__(self, other):
if self._stored_freeze and (self._hashed_self == self):
frozen = self._stored_freeze
else:
frozen = freeze(self)
if frozen is not self._stored_freeze:
self._stored_hash = frozen
return frozen == freeze(other)
def __hash__(self):
if self._stored_freeze and (self._hashed_self == self):
frozen = self._stored_freeze
else:
frozen = freeze(self)
if frozen is not self._stored_freeze:
self._stored_hash = frozen
return hash(frozen)
class UncachedFreezableList(list, collections.Hashable):
def __eq__(self, other):
"""No caching version of __eq__. May be faster.
Don't forget to get rid of the declarations at the top of the class!
Considerably more elegant."""
return freeze(self) == freeze(other)
def __hash__(self):
"""No caching version of __hash__. See the notes in the docstring of __eq__2"""
return hash(freeze(self))
Test all three (QuasiUnorderedList, FreezableList, and UncachedFreezableList) and see which one is faster in your situation. I'll betcha it's faster than the other solutions.
I am in need of a Python (2.7) object that functions like a set (fast insertion, deletion, and membership checking) but has the ability to return a random value. Previous questions asked on stackoverflow have answers that are things like:
import random
random.sample(mySet, 1)
But this is quite slow for large sets (it runs in O(n) time).
Other solutions aren't random enough (they depend on the internal representation of python sets, which produces some results which are very non-random):
for e in mySet:
break
# e is now an element from mySet
I coded my own rudimentary class which has constant time lookup, deletion, and random values.
class randomSet:
def __init__(self):
self.dict = {}
self.list = []
def add(self, item):
if item not in self.dict:
self.dict[item] = len(self.list)
self.list.append(item)
def addIterable(self, item):
for a in item:
self.add(a)
def delete(self, item):
if item in self.dict:
index = self.dict[item]
if index == len(self.list)-1:
del self.dict[self.list[index]]
del self.list[index]
else:
self.list[index] = self.list.pop()
self.dict[self.list[index]] = index
del self.dict[item]
def getRandom(self):
if self.list:
return self.list[random.randomint(0,len(self.list)-1)]
def popRandom(self):
if self.list:
index = random.randint(0,len(self.list)-1)
if index == len(self.list)-1:
del self.dict[self.list[index]]
return self.list.pop()
returnValue = self.list[index]
self.list[index] = self.list.pop()
self.dict[self.list[index]] = index
del self.dict[returnValue]
return returnValue
Are there any better implementations for this, or any big improvements to be made to this code?
I think the best way to do this would be to use the MutableSet abstract base class in collections. Inherit from MutableSet, and then define add, discard, __len__, __iter__, and __contains__; also rewrite __init__ to optionally accept a sequence, just like the set constructor does. MutableSet provides built-in definitions of all other set methods based on those methods. That way you get the full set interface cheaply. (And if you do this, addIterable is defined for you, under the name extend.)
discard in the standard set interface appears to be what you have called delete here. So rename delete to discard. Also, instead of having a separate popRandom method, you could just define popRandom like so:
def popRandom(self):
item = self.getRandom()
self.discard(item)
return item
That way you don't have to maintain two separate item removal methods.
Finally, in your item removal method (delete now, discard according to the standard set interface), you don't need an if statement. Instead of testing whether index == len(self.list) - 1, simply swap the final item in the list with the item at the index of the list to be popped, and make the necessary change to the reverse-indexing dictionary. Then pop the last item from the list and remove it from the dictionary. This works whether index == len(self.list) - 1 or not:
def discard(self, item):
if item in self.dict:
index = self.dict[item]
self.list[index], self.list[-1] = self.list[-1], self.list[index]
self.dict[self.list[index]] = index
del self.list[-1] # or in one line:
del self.dict[item] # del self.dict[self.list.pop()]
One approach you could take is to derive a new class from set which salts itself with random objects of a type derived from int.
You can then use pop to select a random element, and if it is not of the salt type, reinsert and return it, but if it is of the salt type, insert a new, randomly-generated salt object (and pop to select a new object).
This will tend to alter the order in which objects are selected. On average, the number of attempts will depend on the proportion of salting elements, i.e. amortised O(k) performance.
Can't we implement a new class inheriting from set with some (hackish) modifications that enable us to retrieve a random element from the list with O(1) lookup time? Btw, on Python 2.x you should inherit from object, i.e. use class randomSet(object). Also PEP8 is something to consider for you :-)
Edit:
For getting some ideas of what hackish solutions might be capable of, this thread is worth reading:
http://python.6.n6.nabble.com/Get-item-from-set-td1530758.html
Here's a solution from scratch, which adds and pops in constant time. I also included some extra set functions for demonstrative purposes.
from random import randint
class RandomSet(object):
"""
Implements a set in which elements can be
added and drawn uniformly and randomly in
constant time.
"""
def __init__(self, seq=None):
self.dict = {}
self.list = []
if seq is not None:
for x in seq:
self.add(x)
def add(self, x):
if x not in self.dict:
self.dict[x] = len(self.list)
self.list.append(x)
def pop(self, x=None):
if x is None:
i = randint(0,len(self.list)-1)
x = self.list[i]
else:
i = self.dict[x]
self.list[i] = self.list[-1]
self.dict[self.list[-1]] = i
self.list.pop()
self.dict.pop(x)
return x
def __contains__(self, x):
return x in self.dict
def __iter__(self):
return iter(self.list)
def __repr__(self):
return "{" + ", ".join(str(x) for x in self.list) + "}"
def __len__(self):
return len(self.list)
Yes, I'd implement an "ordered set" in much the same way you did - and use a list as an internal data structure.
However, I'd inherit straight from "set" and just keep track of the added items in an
internal list (as you did) - and leave the methods I don't use alone.
Maybe add a "sync" method to update the internal list whenever the set is updated
by set-specific operations, like the *_update methods.
That if using an "ordered dict" does not cover your use cases. (I just found that trying to cast ordered_dict keys to a regular set is not optmized, so if you need set operations on your data that is not an option)
If you don't mind only supporting comparable elements, then you could use blist.sortedset.
As an exercise, and mostly for my own amusement, I'm implementing a backtracking packrat parser. The inspiration for this is i'd like to have a better idea about how hygenic macros would work in an algol-like language (as apposed to the syntax free lisp dialects you normally find them in). Because of this, different passes through the input might see different grammars, so cached parse results are invalid, unless I also store the current version of the grammar along with the cached parse results. (EDIT: a consequence of this use of key-value collections is that they should be immutable, but I don't intend to expose the interface to allow them to be changed, so either mutable or immutable collections are fine)
The problem is that python dicts cannot appear as keys to other dicts. Even using a tuple (as I'd be doing anyways) doesn't help.
>>> cache = {}
>>> rule = {"foo":"bar"}
>>> cache[(rule, "baz")] = "quux"
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'dict'
>>>
I guess it has to be tuples all the way down. Now the python standard library provides approximately what i'd need, collections.namedtuple has a very different syntax, but can be used as a key. continuing from above session:
>>> from collections import namedtuple
>>> Rule = namedtuple("Rule",rule.keys())
>>> cache[(Rule(**rule), "baz")] = "quux"
>>> cache
{(Rule(foo='bar'), 'baz'): 'quux'}
Ok. But I have to make a class for each possible combination of keys in the rule I would want to use, which isn't so bad, because each parse rule knows exactly what parameters it uses, so that class can be defined at the same time as the function that parses the rule.
Edit: An additional problem with namedtuples is that they are strictly positional. Two tuples that look like they should be different can in fact be the same:
>>> you = namedtuple("foo",["bar","baz"])
>>> me = namedtuple("foo",["bar","quux"])
>>> you(bar=1,baz=2) == me(bar=1,quux=2)
True
>>> bob = namedtuple("foo",["baz","bar"])
>>> you(bar=1,baz=2) == bob(bar=1,baz=2)
False
tl'dr: How do I get dicts that can be used as keys to other dicts?
Having hacked a bit on the answers, here's the more complete solution I'm using. Note that this does a bit extra work to make the resulting dicts vaguely immutable for practical purposes. Of course it's still quite easy to hack around it by calling dict.__setitem__(instance, key, value) but we're all adults here.
class hashdict(dict):
"""
hashable dict implementation, suitable for use as a key into
other dicts.
>>> h1 = hashdict({"apples": 1, "bananas":2})
>>> h2 = hashdict({"bananas": 3, "mangoes": 5})
>>> h1+h2
hashdict(apples=1, bananas=3, mangoes=5)
>>> d1 = {}
>>> d1[h1] = "salad"
>>> d1[h1]
'salad'
>>> d1[h2]
Traceback (most recent call last):
...
KeyError: hashdict(bananas=3, mangoes=5)
based on answers from
http://stackoverflow.com/questions/1151658/python-hashable-dicts
"""
def __key(self):
return tuple(sorted(self.items()))
def __repr__(self):
return "{0}({1})".format(self.__class__.__name__,
", ".join("{0}={1}".format(
str(i[0]),repr(i[1])) for i in self.__key()))
def __hash__(self):
return hash(self.__key())
def __setitem__(self, key, value):
raise TypeError("{0} does not support item assignment"
.format(self.__class__.__name__))
def __delitem__(self, key):
raise TypeError("{0} does not support item assignment"
.format(self.__class__.__name__))
def clear(self):
raise TypeError("{0} does not support item assignment"
.format(self.__class__.__name__))
def pop(self, *args, **kwargs):
raise TypeError("{0} does not support item assignment"
.format(self.__class__.__name__))
def popitem(self, *args, **kwargs):
raise TypeError("{0} does not support item assignment"
.format(self.__class__.__name__))
def setdefault(self, *args, **kwargs):
raise TypeError("{0} does not support item assignment"
.format(self.__class__.__name__))
def update(self, *args, **kwargs):
raise TypeError("{0} does not support item assignment"
.format(self.__class__.__name__))
# update is not ok because it mutates the object
# __add__ is ok because it creates a new object
# while the new object is under construction, it's ok to mutate it
def __add__(self, right):
result = hashdict(self)
dict.update(result, right)
return result
if __name__ == "__main__":
import doctest
doctest.testmod()
Here is the easy way to make a hashable dictionary. Just remember not to mutate them after embedding in another dictionary for obvious reasons.
class hashabledict(dict):
def __hash__(self):
return hash(tuple(sorted(self.items())))
Hashables should be immutable -- not enforcing this but TRUSTING you not to mutate a dict after its first use as a key, the following approach would work:
class hashabledict(dict):
def __key(self):
return tuple((k,self[k]) for k in sorted(self))
def __hash__(self):
return hash(self.__key())
def __eq__(self, other):
return self.__key() == other.__key()
If you DO need to mutate your dicts and STILL want to use them as keys, complexity explodes hundredfolds -- not to say it can't be done, but I'll wait until a VERY specific indication before I get into THAT incredible morass!-)
All that is needed to make dictionaries usable for your purpose is to add a __hash__ method:
class Hashabledict(dict):
def __hash__(self):
return hash(frozenset(self))
Note, the frozenset conversion will work for all dictionaries (i.e. it doesn't require the keys to be sortable). Likewise, there is no restriction on the dictionary values.
If there are many dictionaries with identical keys but with distinct values, it is necessary to have the hash take the values into account. The fastest way to do that is:
class Hashabledict(dict):
def __hash__(self):
return hash((frozenset(self), frozenset(self.itervalues())))
This is quicker than frozenset(self.iteritems()) for two reasons. First, the frozenset(self) step reuses the hash values stored in the dictionary, saving unnecessary calls to hash(key). Second, using itervalues will access the values directly and avoid the many memory allocator calls using by items to form new many key/value tuples in memory every time you do a lookup.
The given answers are okay, but they could be improved by using frozenset(...) instead of tuple(sorted(...)) to generate the hash:
>>> import timeit
>>> timeit.timeit('hash(tuple(sorted(d.iteritems())))', "d = dict(a=3, b='4', c=2345, asdfsdkjfew=0.23424, x='sadfsadfadfsaf')")
4.7758948802947998
>>> timeit.timeit('hash(frozenset(d.iteritems()))', "d = dict(a=3, b='4', c=2345, asdfsdkjfew=0.23424, x='sadfsadfadfsaf')")
1.8153600692749023
The performance advantage depends on the content of the dictionary, but in most cases I've tested, hashing with frozenset is at least 2 times faster (mainly because it does not need to sort).
A reasonably clean, straightforward implementation is
import collections
class FrozenDict(collections.Mapping):
"""Don't forget the docstrings!!"""
def __init__(self, *args, **kwargs):
self._d = dict(*args, **kwargs)
def __iter__(self):
return iter(self._d)
def __len__(self):
return len(self._d)
def __getitem__(self, key):
return self._d[key]
def __hash__(self):
return hash(tuple(sorted(self._d.iteritems())))
I keep coming back to this topic... Here's another variation. I'm uneasy with subclassing dict to add a __hash__ method; There's virtually no escape from the problem that dict's are mutable, and trusting that they won't change seems like a weak idea. So I've instead looked at building a mapping based on a builtin type that is itself immutable. although tuple is an obvious choice, accessing values in it implies a sort and a bisect; not a problem, but it doesn't seem to be leveraging much of the power of the type it's built on.
What if you jam key, value pairs into a frozenset? What would that require, how would it work?
Part 1, you need a way of encoding the 'item's in such a way that a frozenset will treat them mainly by their keys; I'll make a little subclass for that.
import collections
class pair(collections.namedtuple('pair_base', 'key value')):
def __hash__(self):
return hash((self.key, None))
def __eq__(self, other):
if type(self) != type(other):
return NotImplemented
return self.key == other.key
def __repr__(self):
return repr((self.key, self.value))
That alone puts you in spitting distance of an immutable mapping:
>>> frozenset(pair(k, v) for k, v in enumerate('abcd'))
frozenset([(0, 'a'), (2, 'c'), (1, 'b'), (3, 'd')])
>>> pairs = frozenset(pair(k, v) for k, v in enumerate('abcd'))
>>> pair(2, None) in pairs
True
>>> pair(5, None) in pairs
False
>>> goal = frozenset((pair(2, None),))
>>> pairs & goal
frozenset([(2, None)])
D'oh! Unfortunately, when you use the set operators and the elements are equal but not the same object; which one ends up in the return value is undefined, we'll have to go through some more gyrations.
>>> pairs - (pairs - goal)
frozenset([(2, 'c')])
>>> iter(pairs - (pairs - goal)).next().value
'c'
However, looking values up in this way is cumbersome, and worse, creates lots of intermediate sets; that won't do! We'll create a 'fake' key-value pair to get around it:
class Thief(object):
def __init__(self, key):
self.key = key
def __hash__(self):
return hash(pair(self.key, None))
def __eq__(self, other):
self.value = other.value
return pair(self.key, None) == other
Which results in the less problematic:
>>> thief = Thief(2)
>>> thief in pairs
True
>>> thief.value
'c'
That's all the deep magic; the rest is wrapping it all up into something that has an interface like a dict. Since we're subclassing from frozenset, which has a very different interface, there's quite a lot of methods; we get a little help from collections.Mapping, but most of the work is overriding the frozenset methods for versions that work like dicts, instead:
class FrozenDict(frozenset, collections.Mapping):
def __new__(cls, seq=()):
return frozenset.__new__(cls, (pair(k, v) for k, v in seq))
def __getitem__(self, key):
thief = Thief(key)
if frozenset.__contains__(self, thief):
return thief.value
raise KeyError(key)
def __eq__(self, other):
if not isinstance(other, FrozenDict):
return dict(self.iteritems()) == other
if len(self) != len(other):
return False
for key, value in self.iteritems():
try:
if value != other[key]:
return False
except KeyError:
return False
return True
def __hash__(self):
return hash(frozenset(self.iteritems()))
def get(self, key, default=None):
thief = Thief(key)
if frozenset.__contains__(self, thief):
return thief.value
return default
def __iter__(self):
for item in frozenset.__iter__(self):
yield item.key
def iteritems(self):
for item in frozenset.__iter__(self):
yield (item.key, item.value)
def iterkeys(self):
for item in frozenset.__iter__(self):
yield item.key
def itervalues(self):
for item in frozenset.__iter__(self):
yield item.value
def __contains__(self, key):
return frozenset.__contains__(self, pair(key, None))
has_key = __contains__
def __repr__(self):
return type(self).__name__ + (', '.join(repr(item) for item in self.iteritems())).join('()')
#classmethod
def fromkeys(cls, keys, value=None):
return cls((key, value) for key in keys)
which, ultimately, does answer my own question:
>>> myDict = {}
>>> myDict[FrozenDict(enumerate('ab'))] = 5
>>> FrozenDict(enumerate('ab')) in myDict
True
>>> FrozenDict(enumerate('bc')) in myDict
False
>>> FrozenDict(enumerate('ab', 3)) in myDict
False
>>> myDict[FrozenDict(enumerate('ab'))]
5
The accepted answer by #Unknown, as well as the answer by #AlexMartelli work perfectly fine, but only under the following constraints:
The dictionary's values must be hashable. For example, hash(hashabledict({'a':[1,2]})) will raise TypeError.
Keys must support comparison operation. For example, hash(hashabledict({'a':'a', 1:1})) will raise TypeError.
The comparison operator on keys imposes total ordering. For example, if the two keys in a dictionary are frozenset((1,2,3)) and frozenset((4,5,6)), they compare unequal in both directions. Therefore, sorting the items of a dictionary with such keys can result in an arbitrary order, and therefore will violate the rule that equal objects must have the same hash value.
The much faster answer by #ObenSonne lifts the constraints 2 and 3, but is still bound by constraint 1 (values must be hashable).
The faster yet answer by #RaymondHettinger lifts all 3 constraints because it does not include .values() in the hash calculation. However, its performance is good only if:
Most of the (non-equal) dictionaries that need to be hashed have do not identical .keys().
If this condition isn't satisfied, the hash function will still be valid, but may cause too many collisions. For example, in the extreme case where all the dictionaries are generated from a website template (field names as keys, user input as values), the keys will always be the same, and the hash function will return the same value for all the inputs. As a result, a hashtable that relies on such a hash function will become as slow as a list when retrieving an item (O(N) instead of O(1)).
I think the following solution will work reasonably well even if all 4 constraints I listed above are violated. It has an additional advantage that it can hash not only dictionaries, but any containers, even if they have nested mutable containers.
I'd much appreciate any feedback on this, since I only tested this lightly so far.
# python 3.4
import collections
import operator
import sys
import itertools
import reprlib
# a wrapper to make an object hashable, while preserving equality
class AutoHash:
# for each known container type, we can optionally provide a tuple
# specifying: type, transform, aggregator
# even immutable types need to be included, since their items
# may make them unhashable
# transformation may be used to enforce the desired iteration
# the result of a transformation must be an iterable
# default: no change; for dictionaries, we use .items() to see values
# usually transformation choice only affects efficiency, not correctness
# aggregator is the function that combines all items into one object
# default: frozenset; for ordered containers, we can use tuple
# aggregator choice affects both efficiency and correctness
# e.g., using a tuple aggregator for a set is incorrect,
# since identical sets may end up with different hash values
# frozenset is safe since at worst it just causes more collisions
# unfortunately, no collections.ABC class is available that helps
# distinguish ordered from unordered containers
# so we need to just list them out manually as needed
type_info = collections.namedtuple(
'type_info',
'type transformation aggregator')
ident = lambda x: x
# order matters; first match is used to handle a datatype
known_types = (
# dict also handles defaultdict
type_info(dict, lambda d: d.items(), frozenset),
# no need to include set and frozenset, since they are fine with defaults
type_info(collections.OrderedDict, ident, tuple),
type_info(list, ident, tuple),
type_info(tuple, ident, tuple),
type_info(collections.deque, ident, tuple),
type_info(collections.Iterable, ident, frozenset) # other iterables
)
# hash_func can be set to replace the built-in hash function
# cache can be turned on; if it is, cycles will be detected,
# otherwise cycles in a data structure will cause failure
def __init__(self, data, hash_func=hash, cache=False, verbose=False):
self._data=data
self.hash_func=hash_func
self.verbose=verbose
self.cache=cache
# cache objects' hashes for performance and to deal with cycles
if self.cache:
self.seen={}
def hash_ex(self, o):
# note: isinstance(o, Hashable) won't check inner types
try:
if self.verbose:
print(type(o),
reprlib.repr(o),
self.hash_func(o),
file=sys.stderr)
return self.hash_func(o)
except TypeError:
pass
# we let built-in hash decide if the hash value is worth caching
# so we don't cache the built-in hash results
if self.cache and id(o) in self.seen:
return self.seen[id(o)][0] # found in cache
# check if o can be handled by decomposing it into components
for typ, transformation, aggregator in AutoHash.known_types:
if isinstance(o, typ):
# another option is:
# result = reduce(operator.xor, map(_hash_ex, handler(o)))
# but collisions are more likely with xor than with frozenset
# e.g. hash_ex([1,2,3,4])==0 with xor
try:
# try to frozenset the actual components, it's faster
h = self.hash_func(aggregator(transformation(o)))
except TypeError:
# components not hashable with built-in;
# apply our extended hash function to them
h = self.hash_func(aggregator(map(self.hash_ex, transformation(o))))
if self.cache:
# storing the object too, otherwise memory location will be reused
self.seen[id(o)] = (h, o)
if self.verbose:
print(type(o), reprlib.repr(o), h, file=sys.stderr)
return h
raise TypeError('Object {} of type {} not hashable'.format(repr(o), type(o)))
def __hash__(self):
return self.hash_ex(self._data)
def __eq__(self, other):
# short circuit to save time
if self is other:
return True
# 1) type(self) a proper subclass of type(other) => self.__eq__ will be called first
# 2) any other situation => lhs.__eq__ will be called first
# case 1. one side is a subclass of the other, and AutoHash.__eq__ is not overridden in either
# => the subclass instance's __eq__ is called first, and we should compare self._data and other._data
# case 2. neither side is a subclass of the other; self is lhs
# => we can't compare to another type; we should let the other side decide what to do, return NotImplemented
# case 3. neither side is a subclass of the other; self is rhs
# => we can't compare to another type, and the other side already tried and failed;
# we should return False, but NotImplemented will have the same effect
# any other case: we won't reach the __eq__ code in this class, no need to worry about it
if isinstance(self, type(other)): # identifies case 1
return self._data == other._data
else: # identifies cases 2 and 3
return NotImplemented
d1 = {'a':[1,2], 2:{3:4}}
print(hash(AutoHash(d1, cache=True, verbose=True)))
d = AutoHash(dict(a=1, b=2, c=3, d=[4,5,6,7], e='a string of chars'),cache=True, verbose=True)
print(hash(d))
You might also want to add these two methods to get the v2 pickling protocol work with hashdict instances. Otherwise cPickle will try to use hashdict.____setitem____ resulting in a TypeError. Interestingly, with the other two versions of the protocol your code works just fine.
def __setstate__(self, objstate):
for k,v in objstate.items():
dict.__setitem__(self,k,v)
def __reduce__(self):
return (hashdict, (), dict(self),)
serialize the dict as string with json package:
d = {'a': 1, 'b': 2}
s = json.dumps(d)
restore the dict when you need:
d2 = json.loads(s)
If you don't put numbers in the dictionary and you never lose the variables containing your dictionaries, you can do this:
cache[id(rule)] = "whatever"
since id() is unique for every dictionary
EDIT:
Oh sorry, yeah in that case what the other guys said would be better. I think you could also serialize your dictionaries as a string, like
cache[ 'foo:bar' ] = 'baz'
If you need to recover your dictionaries from the keys though, then you'd have to do something uglier like
cache[ 'foo:bar' ] = ( {'foo':'bar'}, 'baz' )
I guess the advantage of this is that you wouldn't have to write as much code.