Get name of dictionary

Question

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.

Here’s an MRE (Minimal Reproducible Example).
Contents of the dicts are irrelevant:

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 dc where ??? is, it will print the entire contents of the dictionary. How can I get the name of the dictionary being used?

Asked By: Gabriel

||

Source

Answer 1

Don’t use a dict_list, use a dict_dict if 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.

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_set and iterate over locals() but this is uglier than sin.

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.

Answered By: Adam Smith

Answer 2

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

The names would be:

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

Answered By: Jeff Tsui

Answer 3

Objects don’t have names in Python, a name is an identifier that can be assigned to an object, and multiple names could be assigned to the same one.

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

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:

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}

Answered By: martineau

Answer 4

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

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.

Answered By: SummerEla

Answer 5

Here’s my solution for a descriptive error message.

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

Answered By: Carl Brubaker

Answer 6

If you want to read name and value

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

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

If you want to read value only

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

Here is your answer

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)

Answered By: Habibur Rahman

Answer 7

While looking for a way to retrieve the name of a dict, couldn’t find anything simple; but on a whim tried this:

mydict = { None: 'mydict' }  # setup the dict and its name
mydict.update( {'a': 'A', 'b':'B' } )  # add items
print( f"ALL: {mydict}" )
print( f"Name: {mydict[None]}" )

produces:

ALL: {None: 'mydict', 'a': 'A', 'b': 'B'}
Name: mydict

To exclude the dict’s name when processing the contents:

for k,v in mydict.items():
    if k:
        print( v )

A
B

or with a list comprehension:

[ v for k,v in mydict.items() if k ]

['A', 'B']

Using a null string also works:

mydict = { '': 'mydict' }

Answered By: user21108062

Get name of dictionary

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