Python 获取字典名称

Get name of dictionary

提问by Gabriel

I find myself needing to iterate over a list made of dictionaries and I need, for every iteration, the name of which dictionary I'm iterating on.

我发现自己需要迭代一个由字典组成的列表，并且每次迭代我都需要我正在迭代的字典的名称。

Here's an MWE (the contents of the dicts are irrelevant for this example) :

这是一个 MWE（字典的内容与此示例无关）：

dict1 = {...}
dicta = {...}
dict666 = {...}

dict_list = [dict1, dicta, dict666]

for dc in dict_list:
    # Insert command that should replace ???
    print 'The name of the dictionary is: ', ???

If I just use dcwhere ???is, it will print the entire contents of the dictionary. How can I get the name of the dictionary being used?

如果我只使用dcwhere ???is，它将打印字典的全部内容。如何获得正在使用的字典的名称？

采纳答案by Adam Smith

Don't use a dict_list, use a dict_dictif you need their names. In reality, though, you should really NOT be doing this. Don't embed meaningful information in variable names. It's tough to get.

不要使用 a dict_list，dict_dict如果您需要他们的名字，请使用 a 。但实际上，您真的不应该这样做。不要在变量名中嵌入有意义的信息。很难得到。

dict_dict = {'dict1':dict1, 'dicta':dicta, 'dict666':dict666}

for name,dict_ in dict_dict.items():
    print 'the name of the dictionary is ', name
    print 'the dictionary looks like ', dict_

Alternatively make a dict_setand iterate over locals()but this is uglier than sin.

或者 make adict_set并迭代，locals()但这比罪更丑。

dict_set = {dict1,dicta,dict666}

for name,value in locals().items():
    if value in dict_set:
        print 'the name of the dictionary is ', name
        print 'the dictionary looks like ', value

Again: uglier than sin, but it DOES work.

再说一遍：比罪更丑，但它确实有效。

回答by Jeff Tsui

You should also consider adding a "name" key to each dictionary.

您还应该考虑为每个字典添加一个“名称”键。

The names would be:

名称将是：

for dc in dict_list:
    # Insert command that should replace ???
    print 'The name of the dictionary is: ', dc['name']

回答by martineau

Objects don't have names in Python and multiple names could be assigned to the same object.

对象在 Python 中没有名称，可以为同一个对象分配多个名称。

However, an object-oriented way to do what you want would be to subclass the built-indictdictionary class and add anameproperty to it. Instances of it would behave exactly like normal dictionaries and could be used virtually anywhere a normal one could be.

但是，做您想做的事情的面向对象的方法是将内置dict字典类子类化并name为其添加属性。它的实例的行为与普通词典完全一样，几乎可以在普通词典所在的任何地方使用。

class NamedDict(dict):
    def __init__(self, *args, **kwargs):
        try:
            self._name = kwargs.pop('name')
        except KeyError:
            raise KeyError('a "name" keyword argument must be supplied')
        super(NamedDict, self).__init__(*args, **kwargs)

    @classmethod
    def fromkeys(cls, name, seq, value=None):
        return cls(dict.fromkeys(seq, value), name=name)

    @property
    def name(self):
        return self._name

dict_list = [NamedDict.fromkeys('dict1', range(1,4)),
             NamedDict.fromkeys('dicta', range(1,4), 'a'),
             NamedDict.fromkeys('dict666', range(1,4), 666)]

for dc in dict_list:
    print 'the name of the dictionary is ', dc.name
    print 'the dictionary looks like ', dc

Output:

输出：

the name of the dictionary is  dict1
the dictionary looks like  {1: None, 2: None, 3: None}
the name of the dictionary is  dicta
the dictionary looks like  {1: 'a', 2: 'a', 3: 'a'}
the name of the dictionary is  dict666
the dictionary looks like  {1: 666, 2: 666, 3: 666}

回答by SummerEla

The following doesn't work on standard dictionaries, but does work just fine with collections dictionaries and counters:

以下不适用于标准字典，但适用于集合字典和计数器：

from collections import Counter

# instantiate Counter ditionary
test= Counter()

# create an empty name attribute field
test.name = lambda: None

# set the "name" attribute field to "name" = "test"
setattr(test.name, 'name', 'test')

# access the nested name field
print(test.name.name)

It's not the prettiest solution, but it is easy to implement and access.

这不是最漂亮的解决方案，但它易于实施和访问。

回答by Carl Brubaker

Here's my solution for a descriptive error message.

这是我针对描述性错误消息的解决方案。

def dict_key_needed(dictionary,key,msg='dictionary'):
    try:
        value = dictionary[key]
        return value
    except KeyError:
         raise KeyError(f"{msg} is missing key '{key}'")

回答by Habibur Rahman

If you want to read name and value

如果要读取名称和值

dictionary={"name1":"value1","name2":"value2","name3":"value3","name4":"value4"}
for name,value in dictionary.items():
    print(name)
    print(value)

If you want to read name only

如果您只想读取名称

dictionary={"name1":"value1","name2":"value2","name3":"value3","name4":"value4"}
for name in dictionary:
    print(name)

If you want to read value only

如果您只想读取值

dictionary={"name1":"value1","name2":"value2","name3":"value3","name4":"value4"}
for values in dictionary.values():
    print(values)

Here is your answer

这是你的答案

dic1 = {"dic":1}
dic2 = {"dic":2}
dic3 = {"dic":3}
dictionaries = [dic1,dic2,dic3]
for i in range(len(dictionaries)):
  my_var_name = [ k for k,v in locals().items() if v == dictionaries[i]][0]
  print(my_var_name)