英文:
Conditional assignment of class variable in Python
问题
我可以帮你翻译以下代码部分：
# 使用默认值或者从kwargs中设置属性
default = 1
class foo():
def __init__(self, **kwargs):
self.X = kwargs['X'] if 'X' in kwargs else default
self.Y = kwargs['Y'] if 'Y' in kwargs else default
F = foo()
f = foo(X=2, Y=3)
这段代码会在foo类的初始化方法中根据传入的关键字参数kwargs来设置属性X和Y的值，如果kwargs中没有对应的键值，将使用默认值default。这种方法可以实现你想要的条件性设置属性的效果。
英文:
How can I conditionally set an attribute in the __init__ of a class, from a **kwargs parameter?
I know that I can do:
default = 1
class foo():
def __init__(self, X=None, Y=None):
self.X = X if X else default
self.Y = Y if Y else default
F = foo()
f = foo(X=2, Y=3)
but I want to make this work with a variable-keyword parameter (**kwargs), in order to understand them properly.
These attempts did not work:
default = 1
class foo():
def __init__(self, **kwargs):
self.X = default or kwargs['X']
self.Y = default or kwargs['Y']
F = foo()
f = foo(X=2, Y=3)
This way, only the default value is used even if other values are provided. If the order is switched (kwargs['X'] or default etc.), the code raises a KeyError if the keyword arguments are not provided.
default = 1
class foo():
def __init__(self, **kwargs):
self.X = value if kwargs["X"] is None else kwargs["X"]
self.Y = value if kwargs["Y"] is None else kwargs["Y"]
F = foo()
f = foo(X=2, Y=3)
This still raises a KeyError for missing keyword arguments.
答案1
得分: 0
类 foo():
def __init__(self, **kwargs):
self.X = kwargs.get('X', value) # get() 接受键名，以及默认值选项
self.Y = kwargs.get('Y', value)
# 或者使用带有默认值的普通参数
类 foo():
# 将字典改为任何类型
def __init__(self, x = None, y = None):
self.X = x or value
self.Y = y or value
# 或者更好地使用带有默认值的关键字参数，如果只有 x 和 y
类 foo():
# 将字典改为任何类型
def __init__(self, x: dict = None, y: dict = None):
self.X = x or value
self.Y = y or value
你还需要知道什么时候应该使用 **kwargs。
有两种常见情况：
第一种：你正在包装另一个接受许多关键字参数的函数，但你只是要将它们传递下去：
def my_wrapper(a, b, **kwargs):
先做些事情(a, b)
真正的函数(**kwargs)
第二种：你愿意接受任何关键字参数，例如在对象上设置属性：
类 OpenEndedObject:
def __init__(self, **kwargs):
for k, v in kwargs.items():
setattr(self, k, v)
foo = OpenEndedObject(a=1, foo='bar')
assert foo.a == 1
assert foo.foo == 'bar'
如果需要了解更多，请阅读以下内容：
为什么在Python中使用kwargs？有了实际应用场景，有何优势？
英文:
class foo():
def __init__(self, **kwargs):
self.X = kwargs.get('X', value) # get() take key_name, default value this opetions
self.Y = kwargs.get('Y', value)
# or use normal params with default value
class foo():
# change dict to any type
def __init__(self, x = None, y = None):
self.X = x or value
self.Y = y or value
# or better use Keyword Argument with default value if you have x y only
class foo():
# change dict to any type
def __init__(self, x: dict = None, y: dict = None):
self.X = x or value
self.Y = y or value
you need also know when should use **kwargs
There are two common cases:
First: You are wrapping another function which takes a number of keyword argument, but you are just going to pass them along:
def my_wrapper(a, b, **kwargs):
do_something_first(a, b)
the_real_function(**kwargs)
Second: You are willing to accept any keyword argument, for example, to set attributes on an object:
class OpenEndedObject:
def __init__(self, **kwargs):
for k, v in kwargs.items():
setattr(self, k, v)
foo = OpenEndedObject(a=1, foo='bar')
assert foo.a == 1
assert foo.foo == 'bar'
read this if you need know more
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