问题:藏字典吗?

出于缓存目的,我需要根据字典中存在的GET参数生成缓存键。

目前,我正在使用sha1(repr(sorted(my_dict.items())))sha1()一种内部使用hashlib的便捷方法),但我很好奇是否有更好的方法。

For caching purposes I need to generate a cache key from GET arguments which are present in a dict.

Currently I’m using sha1(repr(sorted(my_dict.items()))) (sha1() is a convenience method that uses hashlib internally) but I’m curious if there’s a better way.


回答 0

如果您的字典未嵌套,则可以使用字典的项目进行冻结设置,然后使用hash()

hash(frozenset(my_dict.items()))

与生成JSON字符串或字典表示相比,此方法的计算强度要​​低得多。

更新:请参阅下面的注释,为什么这种方法可能无法产生稳定的结果。

If your dictionary is not nested, you could make a frozenset with the dict’s items and use hash():

hash(frozenset(my_dict.items()))

This is much less computationally intensive than generating the JSON string or representation of the dictionary.

UPDATE: Please see the comments below, why this approach might not produce a stable result.


回答 1

sorted(d.items())仅仅使用它不足以使我们保持稳定。其中的某些值d也可以是字典,并且它们的键仍将以任意顺序出现。只要所有键都是字符串,我更喜欢使用:

json.dumps(d, sort_keys=True)

就是说,如果散列需要在不同的机器或Python版本之间保持稳定,我不确定这是防弹的。您可能想要添加separatorsensure_ascii参数,以保护自己免受在那里对默认值的任何更改。我将不胜感激。

Using sorted(d.items()) isn’t enough to get us a stable repr. Some of the values in d could be dictionaries too, and their keys will still come out in an arbitrary order. As long as all the keys are strings, I prefer to use:

json.dumps(d, sort_keys=True)

That said, if the hashes need to be stable across different machines or Python versions, I’m not certain that this is bulletproof. You might want to add the separators and ensure_ascii arguments to protect yourself from any changes to the defaults there. I’d appreciate comments.


回答 2

编辑:如果您所有的键都是字符串,那么在继续阅读此答案之前,请参阅Jack O’Connor的简单得多(且更快)的解决方案(该方法也适用于哈希嵌套词典)。

尽管已经接受了答案,但是问题的标题是“哈希Python字典”,并且关于该标题的答案不完整。(关于问题的内容,答案是完整的。)

嵌套词典

如果人们在Stack Overflow上搜索如何对字典进行哈希处理,则可能会偶然发现这个标题恰当的问题,并且如果人们试图对乘法嵌套字典进行哈希处理,则可能会感到不满意。上面的答案在这种情况下不起作用,您必须实现某种递归机制才能检索哈希。

这是一种这样的机制:

import copy

def make_hash(o):

  """
  Makes a hash from a dictionary, list, tuple or set to any level, that contains
  only other hashable types (including any lists, tuples, sets, and
  dictionaries).
  """

  if isinstance(o, (set, tuple, list)):

    return tuple([make_hash(e) for e in o])    

  elif not isinstance(o, dict):

    return hash(o)

  new_o = copy.deepcopy(o)
  for k, v in new_o.items():
    new_o[k] = make_hash(v)

  return hash(tuple(frozenset(sorted(new_o.items()))))

奖励:散列对象和类

hash()当您对类或实例进行哈希处理时,该函数非常有用。但是,这是我发现的与对象有关的哈希问题:

class Foo(object): pass
foo = Foo()
print (hash(foo)) # 1209812346789
foo.a = 1
print (hash(foo)) # 1209812346789

即使我更改了foo,哈希也一样。这是因为foo的标识未更改,因此哈希是相同的。如果希望foo根据其当前定义进行不同的哈希处理,则解决方案是对实际更改的内容进行哈希处理。在这种情况下,__dict__属性:

class Foo(object): pass
foo = Foo()
print (make_hash(foo.__dict__)) # 1209812346789
foo.a = 1
print (make_hash(foo.__dict__)) # -78956430974785

las,当您尝试对类本身执行相同的操作时:

print (make_hash(Foo.__dict__)) # TypeError: unhashable type: 'dict_proxy'

class __dict__属性不是普通的字典:

print (type(Foo.__dict__)) # type <'dict_proxy'>

这是与之前类似的机制,可以适当地处理类:

import copy

DictProxyType = type(object.__dict__)

def make_hash(o):

  """
  Makes a hash from a dictionary, list, tuple or set to any level, that 
  contains only other hashable types (including any lists, tuples, sets, and
  dictionaries). In the case where other kinds of objects (like classes) need 
  to be hashed, pass in a collection of object attributes that are pertinent. 
  For example, a class can be hashed in this fashion:

    make_hash([cls.__dict__, cls.__name__])

  A function can be hashed like so:

    make_hash([fn.__dict__, fn.__code__])
  """

  if type(o) == DictProxyType:
    o2 = {}
    for k, v in o.items():
      if not k.startswith("__"):
        o2[k] = v
    o = o2  

  if isinstance(o, (set, tuple, list)):

    return tuple([make_hash(e) for e in o])    

  elif not isinstance(o, dict):

    return hash(o)

  new_o = copy.deepcopy(o)
  for k, v in new_o.items():
    new_o[k] = make_hash(v)

  return hash(tuple(frozenset(sorted(new_o.items()))))

您可以使用此方法返回包含多个元素的哈希元组:

# -7666086133114527897
print (make_hash(func.__code__))

# (-7666086133114527897, 3527539)
print (make_hash([func.__code__, func.__dict__]))

# (-7666086133114527897, 3527539, -509551383349783210)
print (make_hash([func.__code__, func.__dict__, func.__name__]))

注意:以上所有代码均假定使用Python3.x。尽管我认为make_hash()可以在2.7.2中使用,但并未在早期版本中进行测试。至于使这些示例可行,我确实知道

func.__code__ 

应该替换为

func.func_code

EDIT: If all your keys are strings, then before continuing to read this answer, please see Jack O’Connor’s significantly simpler (and faster) solution (which also works for hashing nested dictionaries).

Although an answer has been accepted, the title of the question is “Hashing a python dictionary”, and the answer is incomplete as regards that title. (As regards the body of the question, the answer is complete.)

Nested Dictionaries

If one searches Stack Overflow for how to hash a dictionary, one might stumble upon this aptly titled question, and leave unsatisfied if one is attempting to hash multiply nested dictionaries. The answer above won’t work in this case, and you’ll have to implement some sort of recursive mechanism to retrieve the hash.

Here is one such mechanism:

import copy

def make_hash(o):

  """
  Makes a hash from a dictionary, list, tuple or set to any level, that contains
  only other hashable types (including any lists, tuples, sets, and
  dictionaries).
  """

  if isinstance(o, (set, tuple, list)):

    return tuple([make_hash(e) for e in o])    

  elif not isinstance(o, dict):

    return hash(o)

  new_o = copy.deepcopy(o)
  for k, v in new_o.items():
    new_o[k] = make_hash(v)

  return hash(tuple(frozenset(sorted(new_o.items()))))

Bonus: Hashing Objects and Classes

The hash() function works great when you hash classes or instances. However, here is one issue I found with hash, as regards objects:

class Foo(object): pass
foo = Foo()
print (hash(foo)) # 1209812346789
foo.a = 1
print (hash(foo)) # 1209812346789

The hash is the same, even after I’ve altered foo. This is because the identity of foo hasn’t changed, so the hash is the same. If you want foo to hash differently depending on its current definition, the solution is to hash off whatever is actually changing. In this case, the __dict__ attribute:

class Foo(object): pass
foo = Foo()
print (make_hash(foo.__dict__)) # 1209812346789
foo.a = 1
print (make_hash(foo.__dict__)) # -78956430974785

Alas, when you attempt to do the same thing with the class itself:

print (make_hash(Foo.__dict__)) # TypeError: unhashable type: 'dict_proxy'

The class __dict__ property is not a normal dictionary:

print (type(Foo.__dict__)) # type <'dict_proxy'>

Here is a similar mechanism as previous that will handle classes appropriately:

import copy

DictProxyType = type(object.__dict__)

def make_hash(o):

  """
  Makes a hash from a dictionary, list, tuple or set to any level, that 
  contains only other hashable types (including any lists, tuples, sets, and
  dictionaries). In the case where other kinds of objects (like classes) need 
  to be hashed, pass in a collection of object attributes that are pertinent. 
  For example, a class can be hashed in this fashion:

    make_hash([cls.__dict__, cls.__name__])

  A function can be hashed like so:

    make_hash([fn.__dict__, fn.__code__])
  """

  if type(o) == DictProxyType:
    o2 = {}
    for k, v in o.items():
      if not k.startswith("__"):
        o2[k] = v
    o = o2  

  if isinstance(o, (set, tuple, list)):

    return tuple([make_hash(e) for e in o])    

  elif not isinstance(o, dict):

    return hash(o)

  new_o = copy.deepcopy(o)
  for k, v in new_o.items():
    new_o[k] = make_hash(v)

  return hash(tuple(frozenset(sorted(new_o.items()))))

You can use this to return a hash tuple of however many elements you’d like:

# -7666086133114527897
print (make_hash(func.__code__))

# (-7666086133114527897, 3527539)
print (make_hash([func.__code__, func.__dict__]))

# (-7666086133114527897, 3527539, -509551383349783210)
print (make_hash([func.__code__, func.__dict__, func.__name__]))

NOTE: all of the above code assumes Python 3.x. Did not test in earlier versions, although I assume make_hash() will work in, say, 2.7.2. As far as making the examples work, I do know that

func.__code__ 

should be replaced with

func.func_code

回答 3

这是一个更清晰的解决方案。

def freeze(o):
  if isinstance(o,dict):
    return frozenset({ k:freeze(v) for k,v in o.items()}.items())

  if isinstance(o,list):
    return tuple([freeze(v) for v in o])

  return o


def make_hash(o):
    """
    makes a hash out of anything that contains only list,dict and hashable types including string and numeric types
    """
    return hash(freeze(o))  

Here is a clearer solution.

def freeze(o):
  if isinstance(o,dict):
    return frozenset({ k:freeze(v) for k,v in o.items()}.items())

  if isinstance(o,list):
    return tuple([freeze(v) for v in o])

  return o


def make_hash(o):
    """
    makes a hash out of anything that contains only list,dict and hashable types including string and numeric types
    """
    return hash(freeze(o))  

回答 4

下面的代码避免使用Python hash()函数,因为它不会提供在Python重新启动后保持一致的哈希值(请参阅Python 3.3中的哈希函数在会话之间返回不同的结果)。make_hashable()会将对象转换为嵌套元组,make_hash_sha256()还将转换为repr()以base64编码的SHA256哈希。

import hashlib
import base64

def make_hash_sha256(o):
    hasher = hashlib.sha256()
    hasher.update(repr(make_hashable(o)).encode())
    return base64.b64encode(hasher.digest()).decode()

def make_hashable(o):
    if isinstance(o, (tuple, list)):
        return tuple((make_hashable(e) for e in o))

    if isinstance(o, dict):
        return tuple(sorted((k,make_hashable(v)) for k,v in o.items()))

    if isinstance(o, (set, frozenset)):
        return tuple(sorted(make_hashable(e) for e in o))

    return o

o = dict(x=1,b=2,c=[3,4,5],d={6,7})
print(make_hashable(o))
# (('b', 2), ('c', (3, 4, 5)), ('d', (6, 7)), ('x', 1))

print(make_hash_sha256(o))
# fyt/gK6D24H9Ugexw+g3lbqnKZ0JAcgtNW+rXIDeU2Y=

The code below avoids using the Python hash() function because it will not provide hashes that are consistent across restarts of Python (see hash function in Python 3.3 returns different results between sessions). make_hashable() will convert the object into nested tuples and make_hash_sha256() will also convert the repr() to a base64 encoded SHA256 hash.

import hashlib
import base64

def make_hash_sha256(o):
    hasher = hashlib.sha256()
    hasher.update(repr(make_hashable(o)).encode())
    return base64.b64encode(hasher.digest()).decode()

def make_hashable(o):
    if isinstance(o, (tuple, list)):
        return tuple((make_hashable(e) for e in o))

    if isinstance(o, dict):
        return tuple(sorted((k,make_hashable(v)) for k,v in o.items()))

    if isinstance(o, (set, frozenset)):
        return tuple(sorted(make_hashable(e) for e in o))

    return o

o = dict(x=1,b=2,c=[3,4,5],d={6,7})
print(make_hashable(o))
# (('b', 2), ('c', (3, 4, 5)), ('d', (6, 7)), ('x', 1))

print(make_hash_sha256(o))
# fyt/gK6D24H9Ugexw+g3lbqnKZ0JAcgtNW+rXIDeU2Y=

回答 5

从2013年的回复中更新…

上述答案对我来说似乎都不可靠。原因是使用items()。据我所知,这是以机器相关的顺序出现的。

怎么样呢?

import hashlib

def dict_hash(the_dict, *ignore):
    if ignore:  # Sometimes you don't care about some items
        interesting = the_dict.copy()
        for item in ignore:
            if item in interesting:
                interesting.pop(item)
        the_dict = interesting
    result = hashlib.sha1(
        '%s' % sorted(the_dict.items())
    ).hexdigest()
    return result

Updated from 2013 reply…

None of the above answers seem reliable to me. The reason is the use of items(). As far as I know, this comes out in a machine-dependent order.

How about this instead?

import hashlib

def dict_hash(the_dict, *ignore):
    if ignore:  # Sometimes you don't care about some items
        interesting = the_dict.copy()
        for item in ignore:
            if item in interesting:
                interesting.pop(item)
        the_dict = interesting
    result = hashlib.sha1(
        '%s' % sorted(the_dict.items())
    ).hexdigest()
    return result

回答 6

保留键顺序,而不是 hash(str(dictionary))或者hash(json.dumps(dictionary))我宁愿快速和肮脏的解决方案:

from pprint import pformat
h = hash(pformat(dictionary))

即使对于像DateTimeJSON不可序列化的类型之类的类型,它也将起作用。

To preserve key order, instead of hash(str(dictionary)) or hash(json.dumps(dictionary)) I would prefer quick-and-dirty solution:

from pprint import pformat
h = hash(pformat(dictionary))

It will work even for types like DateTime and more that are not JSON serializable.


回答 7

您可以使用第三方frozendict模块冻结字典并使其可哈希化。

from frozendict import frozendict
my_dict = frozendict(my_dict)

要处理嵌套对象,可以使用:

import collections.abc

def make_hashable(x):
    if isinstance(x, collections.abc.Hashable):
        return x
    elif isinstance(x, collections.abc.Sequence):
        return tuple(make_hashable(xi) for xi in x)
    elif isinstance(x, collections.abc.Set):
        return frozenset(make_hashable(xi) for xi in x)
    elif isinstance(x, collections.abc.Mapping):
        return frozendict({k: make_hashable(v) for k, v in x.items()})
    else:
        raise TypeError("Don't know how to make {} objects hashable".format(type(x).__name__))

如果要支持更多类型,请使用functools.singledispatch(Python 3.7):

@functools.singledispatch
def make_hashable(x):
    raise TypeError("Don't know how to make {} objects hashable".format(type(x).__name__))

@make_hashable.register
def _(x: collections.abc.Hashable):
    return x

@make_hashable.register
def _(x: collections.abc.Sequence):
    return tuple(make_hashable(xi) for xi in x)

@make_hashable.register
def _(x: collections.abc.Set):
    return frozenset(make_hashable(xi) for xi in x)

@make_hashable.register
def _(x: collections.abc.Mapping):
    return frozendict({k: make_hashable(v) for k, v in x.items()})

# add your own types here

You could use the third-party frozendict module to freeze your dict and make it hashable.

from frozendict import frozendict
my_dict = frozendict(my_dict)

For handling nested objects, you could go with:

import collections.abc

def make_hashable(x):
    if isinstance(x, collections.abc.Hashable):
        return x
    elif isinstance(x, collections.abc.Sequence):
        return tuple(make_hashable(xi) for xi in x)
    elif isinstance(x, collections.abc.Set):
        return frozenset(make_hashable(xi) for xi in x)
    elif isinstance(x, collections.abc.Mapping):
        return frozendict({k: make_hashable(v) for k, v in x.items()})
    else:
        raise TypeError("Don't know how to make {} objects hashable".format(type(x).__name__))

If you want to support more types, use functools.singledispatch (Python 3.7):

@functools.singledispatch
def make_hashable(x):
    raise TypeError("Don't know how to make {} objects hashable".format(type(x).__name__))

@make_hashable.register
def _(x: collections.abc.Hashable):
    return x

@make_hashable.register
def _(x: collections.abc.Sequence):
    return tuple(make_hashable(xi) for xi in x)

@make_hashable.register
def _(x: collections.abc.Set):
    return frozenset(make_hashable(xi) for xi in x)

@make_hashable.register
def _(x: collections.abc.Mapping):
    return frozendict({k: make_hashable(v) for k, v in x.items()})

# add your own types here

回答 8

您可以使用地图库来做到这一点。具体来说就是maps.FrozenMap

import maps
fm = maps.FrozenMap(my_dict)
hash(fm)

要安装maps,只需执行以下操作:

pip install maps

它也处理嵌套的dict情况:

import maps
fm = maps.FrozenMap.recurse(my_dict)
hash(fm)

免责声明:我是maps图书馆的作者。

You can use the maps library to do this. Specifically, maps.FrozenMap

import maps
fm = maps.FrozenMap(my_dict)
hash(fm)

To install maps, just do:

pip install maps

It handles the nested dict case too:

import maps
fm = maps.FrozenMap.recurse(my_dict)
hash(fm)

Disclaimer: I am the author of the maps library.


回答 9

解决该问题的一种方法是对字典中的项进行元组化:

hash(tuple(my_dict.items()))

One way to approach the problem is to make a tuple of the dictionary’s items:

hash(tuple(my_dict.items()))

回答 10

我这样做是这样的:

hash(str(my_dict))

I do it like this:

hash(str(my_dict))

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