Python 怎么初始化一个参数很多的类来着

class Foo:
    def __init__(self, name, age, gender, family, score, ...):
        self.name = name
        self.age = age
        self.gender = gender
        ...

总之就是把参数列表里面所有的 args 和 kwargs 里的参数都变成类的属性变量。请问有什么简洁的办法吗？这样一个一个赋值太麻烦了。

Gender

age

参数

name

11 replies • 2020-02-07 10:41:22 +08:00

ericls

Feb 6, 2020 via iPhone

**kwargs 或者直接传个 dict 或者别的什么 iterable

watsy0007

Feb 6, 2020

data_classes

pengdirect

Feb 6, 2020 via iPhone

*arg

ipwx

Feb 6, 2020 via Android

优先选择 dataclass。不能用，就用 **kwargs, setattr

imn1

Feb 6, 2020

dataclass

676529483

Feb 6, 2020

虽然**kwargs 可以解决问题，但不觉得参数太多，应该抽象成单独的类传进去吗？

ClericPy

Feb 6, 2020

四种, https://paste.ubuntu.com/p/fMRyDqJPRY/

```python
# 1. use dataclass at python3.7+, recommended
from dataclasses import dataclass

@dataclass
class Data(object):
a: int
b: int
c: int
d: int

data = Data(1, 2, 3, 4)
print(data)
# Data(a=1, b=2, c=3, d=4)
print(data.a, data.b, data.c, data.d)
# 1 2 3 4

# 2. Use namedtuple

from typing import NamedTuple

class Data(NamedTuple):
a: int
b: int
c: int
d: int

data = Data(1, 2, 3, 4)
print(data)
# Data(a=1, b=2, c=3, d=4)
print(data.a, data.b, data.c, data.d)
# 1 2 3 4

# 3. Use __dict__ without __slots__

class Data(object):

def __init__(self, **kwargs):
super().__init__()
self.__dict__.update(kwargs)

data = Data(a=1, b=2, c=3, d=4)
print(data.a, data.b, data.c, data.d)
# 1 2 3 4

# 4. Use setattr with __slots__

class Data(object):
__slots__ = ('a', 'b', 'c', 'd')

def __init__(self, **kwargs):
super().__init__()
for k, v in kwargs.items():
setattr(self, k, v)

data = Data(a=1, b=2, c=3, d=4)
print(data.a, data.b, data.c, data.d)
# 1 2 3 4

```

作为一个程序员论坛, V 站貌似对代码支持的一塌糊涂

PTLin

Feb 6, 2020

class Foo:
def __init__(self, name, age):
self.__dict__.update(locals())

janxin

Feb 6, 2020

dataclass for 3.7+
attrs for 2.x/3.7-

mathzhaoliang

Feb 6, 2020

我现在这个代码要兼容 python3.5 python3.6
dataclass 看来用不了了

watsy0007

Feb 7, 2020

@mathzhaoliang 用不了用 NamedTuple