'Can I have a simple list of a dataclass field
Can I have easily a list of field from a dataclass ?
@dataclass
class C:
x: int
y: int
z: int
t: int
expected result:
[x,y,z]
Solution 1:[1]
The answer depends on whether or not you have access to an object of the class.
Just using the class
If you only have access to the class, then you can use dataclasses.fields(C) which returns a list of field objects (each of which has a .name property):
[field.name for field in dataclasses.fields(C)]
From an existing object
If you have a constructed object of the class, then you have two additional options:
- Use
dataclasses.fieldson the object:
[field.name for field in dataclasses.fields(obj)]
- Use
dataclasses.asdict(obj)(as pointed out by this answer) which returns a dictionary from field name to field value. It sounds like you are only interested in the.keys()of the dictionary:
dataclasses.asdict(obj).keys() # gives a dict_keys object
list(dataclasses.asdict(obj).keys()) # gives a list
list(dataclasses.asdict(obj)) # same
Full example
Here are all of the options using your example:
from dataclasses import dataclass, fields, asdict
@dataclass
class C:
x: int
y: int
z: int
t: int
# from the class
print([field.name for field in fields(C)])
# using an object
obj = C(1, 2, 3, 4)
print([field.name for field in fields(obj)])
print(asdict(obj).keys())
print(list(asdict(obj).keys()))
print(list(asdict(obj)))
Output:
['x', 'y', 'z', 't']
['x', 'y', 'z', 't']
dict_keys(['x', 'y', 'z', 't'])
['x', 'y', 'z', 't']
['x', 'y', 'z', 't']
Solution 2:[2]
You can use the asdict method of the dataclasses module. For example:
from dataclasses import dataclass, asdict
@dataclass
class Person:
age: int
name: str
adam = Person(25, 'Adam')
# if you want the keys
print(asdict(adam).keys()) # dict_keys(['age', 'name'])
# if you want the values
print(asdict(adam).values()) # dict_values([25, 'Adam'])
Both methods above return a View object which you can iterate on, or you can convert it to list using list(...).
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | teichert |
| Solution 2 |
