Python 将命名元组转换为字典
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Convert a namedtuple into a dictionary
提问by Without Me It Just Aweso
I have a named tuple class in python
我在 python 中有一个命名的元组类
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.
我想将 Town 实例转换为字典。我不希望它与城镇中的字段名称或数量严格绑定。
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.
有没有办法编写它，以便我可以添加更多字段，或者传入一个完全不同的命名元组并获取字典。
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.
我无法像在其他人的代码中那样更改原始类定义。所以我需要获取一个城镇的实例并将其转换为字典。
采纳答案by wim
TL;DR: there's a method _asdictprovided for this.
TL;DR：_asdict为此提供了一种方法。
Here is a demonstration of the usage:
下面是用法演示：
>>> 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 methodof 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.
这是namedtuples的文档化方法，即与python 中的通常约定不同，方法名称上的前导下划线不是为了阻止使用。与添加到命名元组的其他方法一起，_make, _replace, _source, _fields，它具有下划线只是为了尝试和防止与可能的字段名称发生冲突。
Note:For some 2.7.5 < python version < 3.5.0 code out in the wild, you might see this version:
注意：对于一些 2.7.5 < python version < 3.5.0 的代码，你可能会看到这个版本：
>>> vars(funkytown)
OrderedDict([('name', 'funky'),
('population', 300),
('coordinates', 'somewhere'),
('capital', 'lipps'),
('state_bird', 'chicken')])
For a while the documentation had mentioned that _asdictwas obsolete (see here), and suggested to use the built-in method vars. That advice is now outdated; in order to fix a bugrelated to subclassing, the __dict__property which was present on namedtuples has again been removed by this commit.
有一段时间，文档提到_asdict过时（请参阅此处），并建议使用内置方法vars。这个建议现在已经过时了。为了修复与子类化相关的错误__dict__，此提交再次删除了namedtuples 上存在的属性。
回答by Peter DeGlopper
There's a built in method on namedtupleinstances for this, _asdict.
namedtuple实例中有一个内置方法，_asdict.
As discussed in the comments, on some versions vars()will also do it, but it's apparently highly dependent on build details, whereas _asdictshould be reliable. In some versions _asdictwas marked as deprecated, but comments indicate that this is no longer the case as of 3.4.
正如评论中所讨论的，在某些版本上vars()也会这样做，但它显然高度依赖于构建细节，而_asdict应该是可靠的。在某些版本中_asdict被标记为已弃用，但评论表明从 3.4 开始不再如此。
回答by ThorSummoner
On Ubuntu 14.04 LTS versions of python2.7 and python3.4 the __dict__property worked as expected. The _asdictmethodalso worked, but I'm inclined to use the standards-defined, uniform, property api instead of the localized non-uniform api.
在 Ubuntu 14.04 LTS 版本的 python2.7 和 python3.4 上，该__dict__属性按预期工作。该_asdict方法也有效，但我倾向于使用标准定义的、统一的、属性 api 而不是本地化的非统一 api。
$ python2.7
$ 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 dictis the semantic way to get a dictionary representing soemthing, (at least to the best of my knowledge).
将dict视为获取表示某些事物的字典的语义方式（至少据我所知）。
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.
最好能积累一张主要的 Python 版本和平台以及它们对__dict__.
| 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 |
回答by Andre Odendaal
Python 3. Allocate any field to the dictionary as the required index for the dictionary, I used 'name'.
Python 3. 将任何字段分配给字典作为字典的必需索引，我使用了'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}
回答by yongtaek jun
Case #1: one dimension tuple
案例#1：一维元组
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'
案例#2：二维元组
示例：DICT_ROLES[961] # 将显示“后端程序员”
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()
回答by Ebubekir Tabak
if no _asdict(), you can use this way:
如果没有_asdict()，你可以这样使用：
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
