What would a "frozen dict" be?


  • A frozen set is a frozenset.
  • A frozen list could be a tuple.
  • What would a frozen dict be? An immutable, hashable dict.

I guess it could be something like collections.namedtuple, but that is more like a frozen-keys dict (a half-frozen dict). Isn’t it?

A “frozendict” should be a frozen dictionary, it should have keys, values, get, etc., and support in, for, etc.

update :
* there it is : https://www.python.org/dev/peps/pep-0603

Asked By: dugres



Assuming the keys and values of the dictionary are themselves immutable (e.g. strings) then:

>>> d
{'forever': 'atones', 'minks': 'cards', 'overhands': 'warranted', 
 'hardhearted': 'tartly', 'gradations': 'snorkeled'}
>>> t = tuple((k, d[k]) for k in sorted(d.keys()))
>>> hash(t)
Answered By: msw

Python doesn’t have a builtin frozendict type. It turns out this wouldn’t be useful too often (though it would still probably be useful more often than frozenset is).

The most common reason to want such a type is when memoizing function calls for functions with unknown arguments. The most common solution to store a hashable equivalent of a dict (where the values are hashable) is something like tuple(sorted(kwargs.items())).

This depends on the sorting not being a bit insane. Python cannot positively promise sorting will result in something reasonable here. (But it can’t promise much else, so don’t sweat it too much.)

You could easily enough make some sort of wrapper that works much like a dict. It might look something like

import collections

class FrozenDict(collections.Mapping):
    """Don't forget the docstrings!!"""
    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.
        if self._hash is None:
            hash_ = 0
            for pair in self.items():
                hash_ ^= hash(pair)
            self._hash = hash_
        return self._hash

It should work great:

>>> x = FrozenDict(a=1, b=2)
>>> y = FrozenDict(a=1, b=2)
>>> x is y
>>> x == y
>>> x == {'a': 1, 'b': 2}
>>> d = {x: 'foo'}
>>> d[y]
Answered By: Mike Graham

I think of frozendict everytime I write a function like this:

def do_something(blah, optional_dict_parm=None):
    if optional_dict_parm is None:
        optional_dict_parm = {}
Answered By: Mark Visser

Here is the code I’ve been using. I subclassed frozenset. The advantages of this are the following.

  1. This is a truly immutable object. No relying on the good behavior of future users and developers.
  2. It’s easy to convert back and forth between a regular dictionary and a frozen dictionary. FrozenDict(orig_dict) –> frozen dictionary. dict(frozen_dict) –> regular dict.

Update Jan 21 2015: The original piece of code I posted in 2014 used a for-loop to find a key that matched. That was incredibly slow. Now I’ve put together an implementation which takes advantage of frozenset’s hashing features. Key-value pairs are stored in special containers where the __hash__ and __eq__ functions are based on the key only. This code has also been formally unit-tested, unlike what I posted here in August 2014.

MIT-style license.

if 3 / 2 == 1:
    version = 2
elif 3 / 2 == 1.5:
    version = 3

def col(i):
    ''' For binding named attributes to spots inside subclasses of tuple.'''
    g = tuple.__getitem__
    def _col(self):
        return g(self,i)
    return _col

class Item(tuple):
    ''' Designed for storing key-value pairs inside
        a FrozenDict, which itself is a subclass of frozenset.
        The __hash__ is overloaded to return the hash of only the key.
        __eq__ is overloaded so that normally it only checks whether the Item's
        key is equal to the other object, HOWEVER, if the other object itself
        is an instance of Item, it checks BOTH the key and value for equality.

        WARNING: Do not use this class for any purpose other than to contain
        key value pairs inside FrozenDict!!!!

        The __eq__ operator is overloaded in such a way that it violates a
        fundamental property of mathematics. That property, which says that
        a == b and b == c implies a == c, does not hold for this object.
        Here's a demonstration:
            [in]  >>> x = Item(('a',4))
            [in]  >>> y = Item(('a',5))
            [in]  >>> hash('a')
            [out] >>> 194817700
            [in]  >>> hash(x)
            [out] >>> 194817700
            [in]  >>> hash(y)
            [out] >>> 194817700
            [in]  >>> 'a' == x
            [out] >>> True
            [in]  >>> 'a' == y
            [out] >>> True
            [in]  >>> x == y
            [out] >>> False

    __slots__ = ()
    key, value = col(0), col(1)
    def __hash__(self):
        return hash(self.key)
    def __eq__(self, other):
        if isinstance(other, Item):
            return tuple.__eq__(self, other)
        return self.key == other
    def __ne__(self, other):
        return not self.__eq__(other)
    def __str__(self):
        return '%r: %r' % self
    def __repr__(self):
        return 'Item((%r, %r))' % self

class FrozenDict(frozenset):
    ''' Behaves in most ways like a regular dictionary, except that it's immutable.
        It differs from other implementations because it doesn't subclass "dict".
        Instead it subclasses "frozenset" which guarantees immutability.
        FrozenDict instances are created with the same arguments used to initialize
        regular dictionaries, and has all the same methods.
            [in]  >>> f = FrozenDict(x=3,y=4,z=5)
            [in]  >>> f['x']
            [out] >>> 3
            [in]  >>> f['a'] = 0
            [out] >>> TypeError: 'FrozenDict' object does not support item assignment

        FrozenDict can accept un-hashable values, but FrozenDict is only hashable if its values are hashable.
            [in]  >>> f = FrozenDict(x=3,y=4,z=5)
            [in]  >>> hash(f)
            [out] >>> 646626455
            [in]  >>> g = FrozenDict(x=3,y=4,z=[])
            [in]  >>> hash(g)
            [out] >>> TypeError: unhashable type: 'list'

        FrozenDict interacts with dictionary objects as though it were a dict itself.
            [in]  >>> original = dict(x=3,y=4,z=5)
            [in]  >>> frozen = FrozenDict(x=3,y=4,z=5)
            [in]  >>> original == frozen
            [out] >>> True

        FrozenDict supports bi-directional conversions with regular dictionaries.
            [in]  >>> original = {'x': 3, 'y': 4, 'z': 5}
            [in]  >>> FrozenDict(original)
            [out] >>> FrozenDict({'x': 3, 'y': 4, 'z': 5})
            [in]  >>> dict(FrozenDict(original))
            [out] >>> {'x': 3, 'y': 4, 'z': 5}   '''

    __slots__ = ()
    def __new__(cls, orig={}, **kw):
        if kw:
            d = dict(orig, **kw)
            items = map(Item, d.items())
                items = map(Item, orig.items())
            except AttributeError:
                items = map(Item, orig)
        return frozenset.__new__(cls, items)

    def __repr__(self):
        cls = self.__class__.__name__
        items = frozenset.__iter__(self)
        _repr = ', '.join(map(str,items))
        return '%s({%s})' % (cls, _repr)

    def __getitem__(self, key):
        if key not in self:
            raise KeyError(key)
        diff = self.difference
        item = diff(diff({key}))
        key, value = set(item).pop()
        return value

    def get(self, key, default=None):
        if key not in self:
            return default
        return self[key]

    def __iter__(self):
        items = frozenset.__iter__(self)
        return map(lambda i: i.key, items)

    def keys(self):
        items = frozenset.__iter__(self)
        return map(lambda i: i.key, items)

    def values(self):
        items = frozenset.__iter__(self)
        return map(lambda i: i.value, items)

    def items(self):
        items = frozenset.__iter__(self)
        return map(tuple, items)

    def copy(self):
        cls = self.__class__
        items = frozenset.copy(self)
        dupl = frozenset.__new__(cls, items)
        return dupl

    def fromkeys(cls, keys, value):
        d = dict.fromkeys(keys,value)
        return cls(d)

    def __hash__(self):
        kv = tuple.__hash__
        items = frozenset.__iter__(self)
        return hash(frozenset(map(kv, items)))

    def __eq__(self, other):
        if not isinstance(other, FrozenDict):
                other = FrozenDict(other)
            except Exception:
                return False
        return frozenset.__eq__(self, other)

    def __ne__(self, other):
        return not self.__eq__(other)

if version == 2:
    #Here are the Python2 modifications
    class Python2(FrozenDict):
        def __iter__(self):
            items = frozenset.__iter__(self)
            for i in items:
                yield i.key

        def iterkeys(self):
            items = frozenset.__iter__(self)
            for i in items:
                yield i.key

        def itervalues(self):
            items = frozenset.__iter__(self)
            for i in items:
                yield i.value

        def iteritems(self):
            items = frozenset.__iter__(self)
            for i in items:
                yield (i.key, i.value)

        def has_key(self, key):
            return key in self

        def viewkeys(self):
            return dict(self).viewkeys()

        def viewvalues(self):
            return dict(self).viewvalues()

        def viewitems(self):
            return dict(self).viewitems()

    #If this is Python2, rebuild the class
    #from scratch rather than use a subclass
    py3 = FrozenDict.__dict__
    py3 = {k: py3[k] for k in py3}
    py2 = {}
    dct = Python2.__dict__
    py2.update({k: dct[k] for k in dct})

    FrozenDict = type('FrozenDict', (frozenset,), py2)
Answered By: Steve Zelaznik

Yes, this is my second answer, but it is a completely different approach. The first implementation was in pure python. This one is in Cython. If you know how to use and compile Cython modules, this is just as fast as a regular dictionary. Roughly .04 to .06 micro-sec to retrieve a single value.

This is the file “frozen_dict.pyx”

import cython
from collections import Mapping

cdef class dict_wrapper:
    cdef object d
    cdef int h

    def __init__(self, *args, **kw):
        self.d = dict(*args, **kw)
        self.h = -1

    def __len__(self):
        return len(self.d)

    def __iter__(self):
        return iter(self.d)

    def __getitem__(self, key):
        return self.d[key]

    def __hash__(self):
        if self.h == -1:
            self.h = hash(frozenset(self.d.iteritems()))
        return self.h

class FrozenDict(dict_wrapper, Mapping):
    def __repr__(self):
        c = type(self).__name__
        r = ', '.join('%r: %r' % (k,self[k]) for k in self)
        return '%s({%s})' % (c, r)

__all__ = ['FrozenDict']

Here’s the file “setup.py”

from distutils.core import setup
from Cython.Build import cythonize

    ext_modules = cythonize('frozen_dict.pyx')

If you have Cython installed, save the two files above into the same directory. Move to that directory in the command line.

python setup.py build_ext --inplace
python setup.py install

And you should be done.

Answered By: Steve Zelaznik

Curiously, although we have the seldom useful frozenset, there’s still no frozen mapping. The idea was rejected in PEP 416 — Add a frozendict builtin type. This idea may be revisited in a later Python release, see PEP 603 — Adding a frozenmap type to collections.

So the Python 2 solution to this:

def foo(config={'a': 1}):

Still seems to be the usual:

def foo(config=None):
    if config is None:
        config = {'a': 1}  # default config

In Python 3 you have the option of this:

from types import MappingProxyType

default_config = {'a': 1}
DEFAULTS = MappingProxyType(default_config)

def foo(config=DEFAULTS):

Now the default config can be updated dynamically, but remain immutable where you want it to be immutable by passing around the proxy instead.

So changes in the default_config will update DEFAULTS as expected, but you can’t write to the mapping proxy object itself.

Admittedly it’s not really the same thing as an "immutable, hashable dict", but it might be a decent substitute for some use cases of a frozendict.

Answered By: wim

The main disadvantage of namedtuple is that it needs to be specified before it is used, so it’s less convenient for single-use cases.

However, there is a practical workaround that can be used to handle many such cases. Let’s say that you want to have an immutable equivalent of the following dict:

    'something': 123,
    'something_else': 456

This can be emulated like this:

from collections import namedtuple

MY_CONSTANT = namedtuple('MyConstant', 'something something_else')(123, 456)

It’s even possible to write an auxiliary function to automate this:

def freeze_dict(data):
    from collections import namedtuple
    keys = sorted(data.keys())
    frozen_type = namedtuple(''.join(keys), keys)
    return frozen_type(**data)

a = {'foo':'bar', 'x':'y'}
fa = freeze_dict(data)
assert a['foo'] == fa.foo

Of course this works only for flat dicts, but it shouldn’t be too difficult to implement a recursive version.

Answered By: Berislav Lopac

You may use frozendict from utilspie package as:

>>> from utilspie.collectionsutils import frozendict

>>> my_dict = frozendict({1: 3, 4: 5})
>>> my_dict  # object of `frozendict` type
frozendict({1: 3, 4: 5})

# Hashable
>>> {my_dict: 4}
{frozendict({1: 3, 4: 5}): 4}

# Immutable
>>> my_dict[1] = 5
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/mquadri/workspace/utilspie/utilspie/collectionsutils/collections_utils.py", line 44, in __setitem__
    self.__setitem__.__name__, type(self).__name__))
AttributeError: You can not call '__setitem__()' for 'frozendict' object

As per the document:

frozendict(dict_obj): Accepts obj of dict type and returns a hashable and immutable dict

Answered By: Moinuddin Quadri

Subclassing dict

i see this pattern in the wild (github) and wanted to mention it:

class FrozenDict(dict):
    def __init__(self, *args, **kwargs):
        self._hash = None
        super(FrozenDict, self).__init__(*args, **kwargs)

    def __hash__(self):
        if self._hash is None:
            self._hash = hash(tuple(sorted(self.items())))  # iteritems() on py2
        return self._hash

    def _immutable(self, *args, **kws):
        raise TypeError('cannot change object - object is immutable')

    # makes (deep)copy alot more efficient
    def __copy__(self):
        return self

    def __deepcopy__(self, memo=None):
        if memo is not None:
            memo[id(self)] = self
        return self

    __setitem__ = _immutable
    __delitem__ = _immutable
    pop = _immutable
    popitem = _immutable
    clear = _immutable
    update = _immutable
    setdefault = _immutable

example usage:

d1 = FrozenDict({'a': 1, 'b': 2})
d2 = FrozenDict({'a': 1, 'b': 2})
assert isinstance(d1, dict)
assert len(set([d1, d2])) == 1  # hashable


  • support for get(), keys(), items() (iteritems() on py2) and all the goodies from dict out of the box without explicitly implementing them
  • uses internally dict which means performance (dict is written in c in CPython)
  • elegant simple and no black magic
  • isinstance(my_frozen_dict, dict) returns True – although python encourages duck-typing many packages uses isinstance(), this can save many tweaks and customizations


  • any subclass can override this or access it internally (you cant really 100% protect something in python, you should trust your users and provide good documentation).
  • if you care for speed, you might want to make __hash__ a bit faster.
Answered By: ShmulikA

In the absence of native language support, you can either do it yourself or use an existing solution. Fortunately Python makes it dead simple to extend off of their base implementations.

class frozen_dict(dict):
    def __setitem__(self, key, value):
        raise Exception('Frozen dictionaries cannot be mutated')

frozen_dict = frozen_dict({'foo': 'FOO' })
print(frozen['foo']) # FOO
frozen['foo'] = 'NEWFOO' # Exception: Frozen dictionaries cannot be mutated

# OR

from types import MappingProxyType

frozen_dict = MappingProxyType({'foo': 'FOO'})
print(frozen_dict['foo']) # FOO
frozen_dict['foo'] = 'NEWFOO' # TypeError: 'mappingproxy' object does not support item assignment
Answered By: efreezy

Install frozendict

pip install frozendict

Use it!

from frozendict import frozendict

def smth(param = frozendict({})):
Answered By: Andrey Korchak

There is no fronzedict, but you can use MappingProxyType that was added to the standard library with Python 3.3:

>>> from types import MappingProxyType
>>> foo = MappingProxyType({'a': 1})
>>> foo
mappingproxy({'a': 1})
>>> foo['a'] = 2
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'mappingproxy' object does not support item assignment
>>> foo
mappingproxy({'a': 1})
Answered By: Julio Marins

I needed to access fixed keys for something at one point for something that was a sort of globally-constanty kind of thing and I settled on something like this:

class MyFrozenDict:
    def __getitem__(self, key):
        if key == 'mykey1':
            return 0
        if key == 'mykey2':
            return "another value"
        raise KeyError(key)

Use it like

a = MyFrozenDict()

WARNING: I don’t recommend this for most use cases as it makes some pretty severe tradeoffs.

Answered By: Adverbly

freeze implements frozen collections (dict, list and set) that are hashable, type-hinted and will recursively freeze the data you give them (when possible) for you.

pip install frz


from freeze import FDict

a_mutable_dict = {
    "list": [1, 2],
    "set": {3, 4},

a_frozen_dict = FDict(a_mutable_dict)

# FDict: {'list': FList: (1, 2), 'set': FSet: {3, 4}}
Answered By: hagaiw