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keys
,values
,get
,等,并支持in
,for
等等。
更新:
*它是: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'
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。
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
回答 3
没有fronzedict
,但是您可以使用MappingProxyType
Python 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})
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。
- 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。
- 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)
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 = {}
回答 6
您可以将frozendict
from 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
回答 7
pip install frozendict
用它!
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。
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。
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。
回答 11
另一个选择是包中的MultiDictProxy
类multidict
。
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。
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