冻结集就是冻结集。 冻结列表可以是元组。 冷冻字典会是什么?一个不可变的、可哈希的字典。
我猜它可能是collections.namedtuple之类的东西,但那更像是一个冻结的字典(一个半冻结的字典)。不是吗?
一个“frozendict”应该是一个冻结的字典,它应该有键,值,get等,并支持in, for等。
更新: 在这里:https://www.python.org/dev/peps/pep-0603
冻结集就是冻结集。 冻结列表可以是元组。 冷冻字典会是什么?一个不可变的、可哈希的字典。
我猜它可能是collections.namedtuple之类的东西,但那更像是一个冻结的字典(一个半冻结的字典)。不是吗?
一个“frozendict”应该是一个冻结的字典,它应该有键,值,get等,并支持in, for等。
更新: 在这里:https://www.python.org/dev/peps/pep-0603
当前回答
每次写这样的函数时,我都会想到frozendict:
def do_something(blah, optional_dict_parm=None):
if optional_dict_parm is None:
optional_dict_parm = {}
其他回答
这是我一直在用的代码。我子类化了frozenset。这样做的优点如下。
这是一个真正的不可变对象。不依赖未来用户和开发人员的良好行为。 在常规字典和固定字典之间来回转换很容易。frozdict (orig_dict)—>冻结字典。字典(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for循环来查找匹配的键。这实在是太慢了。现在我已经组合了一个实现,它利用了frozenset的散列特性。键-值对存储在特殊的容器中,其中__hash__和__eq__函数仅基于键。这段代码也经过了正式的单元测试,这与我2014年8月在这里发布的代码不同。
MIT-style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)
我需要在某一时刻访问某个固定键用于某种全局常量类型的东西我确定了这样的东西:
class MyFrozenDict:
def __getitem__(self, key):
if key == 'mykey1':
return 0
if key == 'mykey2':
return "another value"
raise KeyError(key)
像这样使用它
a = MyFrozenDict()
print(a['mykey1'])
警告:对于大多数用例,我不建议这样做,因为它需要进行一些相当严重的权衡。
namedtuple的主要缺点是需要在使用之前指定它,因此对于单一用例来说不太方便。
然而,有一种实用的变通方法可以用来处理许多此类情况。让我们假设你想有一个不可变的等价物如下dict:
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)
甚至还可以编写一个辅助函数来实现自动化:
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,这只适用于平面字典,但实现递归版本应该不会太难。
每次写这样的函数时,我都会想到frozendict:
def do_something(blah, optional_dict_parm=None):
if optional_dict_parm is None:
optional_dict_parm = {}
你可以使用utispie包装的冷冻液:
>>> 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