Convert a namedtuple into a dictionary

Question:

I have a named tuple class in python

class Town(collections.namedtuple('Town', [
    'name', 
    'population',
    'coordinates',
    'population', 
    'capital', 
    'state_bird'])):
    # ...

I’d like to convert Town instances into dictionaries. I don’t want it to be rigidly tied to the names or number of the fields in a Town.

Is there a way to write it such that I could add more fields, or pass an entirely different named tuple in and get a dictionary.

I can not alter the original class definition as its in someone else’s code. So I need to take an instance of a Town and convert it to a dictionary.

Answers:

TL;DR: there’s a method _asdict provided for this.

Here is a demonstration of the usage:

>>> fields = ['name', 'population', 'coordinates', 'capital', 'state_bird']
>>> Town = collections.namedtuple('Town', fields)
>>> funkytown = Town('funky', 300, 'somewhere', 'lipps', 'chicken')
>>> funkytown._asdict()
OrderedDict([('name', 'funky'),
             ('population', 300),
             ('coordinates', 'somewhere'),
             ('capital', 'lipps'),
             ('state_bird', 'chicken')])

This is a documented method of namedtuples, i.e. unlike the usual convention in python the leading underscore on the method name isn’t there to discourage use. Along with the other methods added to namedtuples, _make, _replace, _source, _fields, it has the underscore only to try and prevent conflicts with possible field names.


Note: For some 2.7.5 < python version < 3.5.0 code out in the wild, you might see this version:

>>> vars(funkytown)
OrderedDict([('name', 'funky'),
             ('population', 300),
             ('coordinates', 'somewhere'),
             ('capital', 'lipps'),
             ('state_bird', 'chicken')])

For a while the documentation had mentioned that _asdict was obsolete (see here), and suggested to use the built-in method vars. That advice is now outdated; in order to fix a bug related to subclassing, the __dict__ property which was present on namedtuples has again been removed by this commit.

Answered By: wim

There’s a built in method on namedtuple instances for this, _asdict.

As discussed in the comments, on some versions vars() will also do it, but it’s apparently highly dependent on build details, whereas _asdict should be reliable. In some versions _asdict was marked as deprecated, but comments indicate that this is no longer the case as of 3.4.

Answered By: Peter DeGlopper

On Ubuntu 14.04 LTS versions of python2.7 and python3.4 the __dict__ property worked as expected. The _asdict method also worked, but I’m inclined to use the standards-defined, uniform, property api instead of the localized non-uniform api.

$ python2.7

# Works on:
# Python 2.7.6 (default, Jun 22 2015, 17:58:13)  [GCC 4.8.2] on linux2
# Python 3.4.3 (default, Oct 14 2015, 20:28:29)  [GCC 4.8.4] on linux

import collections

Color = collections.namedtuple('Color', ['r', 'g', 'b'])
red = Color(r=256, g=0, b=0)

# Access the namedtuple as a dict
print(red.__dict__['r'])  # 256

# Drop the namedtuple only keeping the dict
red = red.__dict__
print(red['r'])  #256

Seeing as dict is the semantic way to get a dictionary representing soemthing, (at least to the best of my knowledge).


It would be nice to accumulate a table of major python versions and platforms and their support for __dict__, currently I only have one platform version and two python versions as posted above.

| Platform                      | PyVer     | __dict__ | _asdict |
| --------------------------    | --------- | -------- | ------- |
| Ubuntu 14.04 LTS              | Python2.7 | yes      | yes     |
| Ubuntu 14.04 LTS              | Python3.4 | yes      | yes     |
| CentOS Linux release 7.4.1708 | Python2.7 | no       | yes     |
| CentOS Linux release 7.4.1708 | Python3.4 | no       | yes     |
| CentOS Linux release 7.4.1708 | Python3.6 | no       | yes     |
Answered By: ThorSummoner

Python 3. Allocate any field to the dictionary as the required index for the dictionary, I used ‘name’.

import collections

Town = collections.namedtuple("Town", "name population coordinates capital state_bird")

town_list = []

town_list.append(Town('Town 1', '10', '10.10', 'Capital 1', 'Turkey'))
town_list.append(Town('Town 2', '11', '11.11', 'Capital 2', 'Duck'))

town_dictionary = {t.name: t for t in town_list}
Answered By: Andre Odendaal

Case #1: one dimension tuple

TUPLE_ROLES = (
    (912,"Role 21"),
    (913,"Role 22"),
    (925,"Role 23"),
    (918,"Role 24"),
)


TUPLE_ROLES[912]  #==> Error because it is out of bounce. 
TUPLE_ROLES[  2]  #==> will show Role 23.
DICT1_ROLE = {k:v for k, v in TUPLE_ROLES }
DICT1_ROLE[925] # will display "Role 23" 

Case #2: Two dimension tuple
Example: DICT_ROLES[961] # will show ‘Back-End Programmer’

NAMEDTUPLE_ROLES = (
    ('Company', ( 
            ( 111, 'Owner/CEO/President'), 
            ( 113, 'Manager'),
            ( 115, 'Receptionist'),
            ( 117, 'Marketer'),
            ( 119, 'Sales Person'),
            ( 121, 'Accountant'),
            ( 123, 'Director'),
            ( 125, 'Vice President'),
            ( 127, 'HR Specialist'),
            ( 141, 'System Operator'),
    )),
    ('Restaurant', ( 
            ( 211, 'Chef'), 
            ( 212, 'Waiter/Waitress'), 
    )),
    ('Oil Collector', ( 
            ( 211, 'Truck Driver'), 
            ( 213, 'Tank Installer'), 
            ( 217, 'Welder'),
            ( 218, 'In-house Handler'),
            ( 219, 'Dispatcher'),
    )),
    ('Information Technology', ( 
            ( 912, 'Server Administrator'),
            ( 914, 'Graphic Designer'),
            ( 916, 'Project Manager'),
            ( 918, 'Consultant'),
            ( 921, 'Business Logic Analyzer'),
            ( 923, 'Data Model Designer'),
            ( 951, 'Programmer'),
            ( 953, 'WEB Front-End Programmer'),
            ( 955, 'Android Programmer'),
            ( 957, 'iOS Programmer'),
            ( 961, 'Back-End Programmer'),
            ( 962, 'Fullstack Programmer'),
            ( 971, 'System Architect'),
    )),
)

#Thus, we need dictionary/set

T4 = {}
def main():
    for k, v in NAMEDTUPLE_ROLES:
        for k1, v1 in v:
            T4.update ( {k1:v1}  )
    print (T4[961]) # will display 'Back-End Programmer'
    # print (T4) # will display all list of dictionary

main()
Answered By: yongtaek jun

if no _asdict(), you can use this way:

def to_dict(model):
    new_dict = {}
    keys = model._fields
    index = 0
    for key in keys:
        new_dict[key] = model[index]
        index += 1

    return new_dict
Answered By: Ebubekir Tabak

Normally _asdict() returns a OrderedDict. this is how to convert from OrderedDict to a regular dict


town = Town('funky', 300, 'somewhere', 'lipps', 'chicken')
dict(town._asdict())

the output will be

{'capital': 'lipps',
 'coordinates': 'somewhere',
 'name': 'funky',
 'population': 300,
 'state_bird': 'chicken'}
Answered By: Simone