什么是“冻结命令”?

问题:什么是“冻结命令”?

  • 冻结集是冻结集。
  • 冻结列表可能是一个元组。
  • 冻结的字典是什么?一个不变的,可哈希的字典。

我猜可能是collections.namedtuple,但是更像是冰冻的字典(半冻结​​的字典)。是不是

A“frozendict”应该是一个冰冻的字典,它应该有keysvaluesget,等,并支持infor等等。

更新:
*它是:https : //www.python.org/dev/peps/pep-0603

  • 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


回答 0

Python没有内置的Frozendict类型。事实证明,这并不是太有用了(尽管它可能仍然比以前有用frozenset)。

想要这种类型的最常见原因是在记忆函数调用具有未知参数的函数时。存储dict的可哈希等效项(值是可哈希的)的最常见解决方案是tuple(sorted(kwargs.iteritems()))

这取决于排序是否有点疯狂。Python无法肯定地承诺排序将在这里产生合理的结果。(但是,它不能承诺其他任何事情,因此请不要流汗过多。)


您可以轻松地制作某种类似于dict的包装器。它可能看起来像

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

它应该很棒:

>>> x = FrozenDict(a=1, b=2)
>>> y = FrozenDict(a=1, b=2)
>>> x is y
False
>>> x == y
True
>>> x == {'a': 1, 'b': 2}
True
>>> d = {x: 'foo'}
>>> d[y]
'foo'

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.iteritems())).

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
False
>>> x == y
True
>>> x == {'a': 1, 'b': 2}
True
>>> d = {x: 'foo'}
>>> d[y]
'foo'

回答 1

奇怪的是,尽管我们很少frozenset在python中有用,但仍然没有冻结的映射。这个想法在PEP 416中被拒绝-添加一个Frozendict内置类型。可以在Python 3.9中重新考虑这个想法,请参阅PEP 603-向collections添加一个Frozenmap类型

因此,python 2解决方案:

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

似乎还是有些la脚:

def foo(config=None):
    if config is None:
        config = default_config = {'a': 1}
    ...

在python3您的选择这个

from types import MappingProxyType

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

def foo(config=DEFAULTS):
    ...

现在,默认配置可以动态更新,但是可以通过传递代理来保持默认配置不变。

因此,中的更改将按预期default_config更新DEFAULTS,但是您无法写入映射代理对象本身。

诚然,这与“不可变,可哈希的字典”不是完全一样的东西,但是考虑到我们可能希望使用“冻结字典”的相同用例,它是一个不错的替代品。

Curiously, although we have the seldom useful frozenset in python, there’s still no frozen mapping. The idea was rejected in PEP 416 — Add a frozendict builtin type. The idea may be revisited in Python 3.9, 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 somewhat lame:

def foo(config=None):
    if config is None:
        config = default_config = {'a': 1}
    ...

In python3 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 quite the same thing as an “immutable, hashable dict” – but it’s a decent substitute given the same kind of use cases for which we might want a frozendict.


回答 2

假设字典的键和值本身是不可变的(例如字符串),则:

>>> d
{'forever': 'atones', 'minks': 'cards', 'overhands': 'warranted', 
 'hardhearted': 'tartly', 'gradations': 'snorkeled'}
>>> t = tuple((k, d[k]) for k in sorted(d.keys()))
>>> hash(t)
1524953596

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)
1524953596

回答 3

没有fronzedict,但是您可以使用MappingProxyTypePython 3.3中添加到标准库中的:

>>> 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})

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})

回答 4

这是我一直在使用的代码。我把Frozenset归为一类。其优点如下。

  1. 这是一个真正的不变的对象。不依赖未来用户和开发人员的良好行为。
  2. 在常规字典和冻结字典之间来回转换很容易。FrozenDict(orig_dict)->冻结的字典。dict(frozen_dict)->常规字典

2015年1月21日更新:我在2014年发布的原始代码使用了for循环来查找匹配的键。那太慢了。现在,我整理了一个利用Frozenset的哈希功能的实现。键值对存储在特殊的容器中,其中__hash__和和__eq__函数仅基于键。与我在2014年8月发布的代码不同,该代码也已经过正式的单元测试。

MIT样式的许可证。

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__
    @property
    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())
        else:
            try:
                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

    @classmethod
    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):
            try:
                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 = {}
    py2.update(py3)
    dct = Python2.__dict__
    py2.update({k: dct[k] for k in dct})

    FrozenDict = type('FrozenDict', (frozenset,), py2)

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__
    @property
    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())
        else:
            try:
                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

    @classmethod
    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):
            try:
                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 = {}
    py2.update(py3)
    dct = Python2.__dict__
    py2.update({k: dct[k] for k in dct})

    FrozenDict = type('FrozenDict', (frozenset,), py2)

回答 5

每当我编写这样的函数时,我都会想起Frozendict:

def do_something(blah, optional_dict_parm=None):
    if optional_dict_parm is None:
        optional_dict_parm = {}

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 = {}

回答 6

您可以将frozendictfrom utilspie包用作:

>>> 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

根据文件

Frozendict(dict_obj):接受dict类型的obj并返回一个可哈希且不可变的 dict

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


回答 7

安装freezedict

pip install frozendict

用它!

from frozendict import frozendict

def smth(param = frozendict({})):
    pass

Install frozendict

pip install frozendict

Use it!

from frozendict import frozendict

def smth(param = frozendict({})):
    pass

回答 8

是的,这是我的第二个答案,但这是一种完全不同的方法。第一个实现是在纯python中实现的。这是在Cython中。如果您知道如何使用和编译Cython模块,这与常规词典一样快。大约.04到.06毫秒,以检索单个值。

这是文件“ 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']

这是文件“ setup.py”

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

setup(
    ext_modules = cythonize('frozen_dict.pyx')
)

如果您安装了Cython,请将上面的两个文件保存到同一目录中。在命令行中移至该目录。

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

并且应该完成。

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

setup(
    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.


回答 9

其主要缺点namedtuple是在使用前需要先指定它,因此对于单次使用的情况不太方便。

但是,有一种实际的解决方法可用于处理许多此类情况。假设您想拥有以下字典的不变的等同物:

MY_CONSTANT = {
    'something': 123,
    'something_else': 456
}

可以这样模拟:

from collections import namedtuple

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

甚至有可能编写一个辅助函数来自动执行此操作:

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

当然,这仅适用于简单的命令,但实现递归版本并不难。

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:

MY_CONSTANT = {
    '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.


回答 10

子类化 dict

我在野外(github)看到了这种模式,想提一下:

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')

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

用法示例:

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

优点

  • 支持get()keys()items()iteritems()上PY2)和所有从东西dict开箱没有明确执行这些
  • 在内部使用dict这意味着性能(dict用CPython用c编写)
  • 优雅简约,无黑魔法
  • isinstance(my_frozen_dict, dict)返回True-尽管python鼓励使用鸭式键入许多软件包isinstance(),但这可以节省许多调整和自定义

缺点

  • 任何子类都可以覆盖它或在内部访问它(您不能真正100%保护python中的某些内容,您应该信任您的用户并提供良好的文档)。
  • 如果您关心速度,则可能需要__hash__提高速度。

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')

    __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})
d1.keys() 
assert isinstance(d1, dict)
assert len(set([d1, d2])) == 1  # hashable

Pros

  • 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

Cons

  • 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.

回答 11

另一个选择是包中的MultiDictProxymultidict

Another option is the MultiDictProxy class from the multidict package.


回答 12

我需要在某一时刻访问某种东西的固定键,这是一种全球稳定的东西,因此我选择了以下方式:

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

像这样使用

a = MyFrozenDict()
print(a['mykey1'])

警告:对于大多数用例,我不建议这样做,因为这会带来一些非常严重的折衷。

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()
print(a['mykey1'])

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


回答 13

在没有本地语言支持的情况下,您可以自己做,也可以使用现有的解决方案。幸运的是,Python使扩展基本实现变得非常简单。

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

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