我希望课程的内容与以下相同:
class Player:
def __init__(self, **kwargs):
try:
self.last_name = kwargs['last_name']
except:
pass
try:
self.first_name = kwargs['first_name']
except:
pass
try:
self.score = kwargs['score']
except:
pass
但这看起来真的很草率。有没有更好的方法来定义 __ init __ 方法?我希望所有的关键字参数都是可选的。
答案 0 :(得分:11)
如果你只有3个关键字args,那么这会更好。
class Player:
def __init__(self, last_name=None, first_name=None, score=None):
self.last_name = last_name
self.first_name = first_name
self.score = score
答案 1 :(得分:10)
如果你只有3个参数,那么Bhargav Rao的解决方案更合适,但如果你有很多潜在的论点,那么试试:
class Player:
def __init__(self, **kwargs):
self.last_name = kwargs.get('last_name')
# .. etc.
kwargs.get('xxx')
将返回xxx
密钥(如果存在),如果不存在则返回None。 .get
采用可选的第二个参数,如果xxx
不在kwargs
(而非None
),则返回该参数,例如要将属性设置为空字符串,请使用kwargs.get('xxx', "")
。
如果您确实希望该属性未定义(如果该属性不在kwargs
中,那么这将执行此操作:
class Player:
def __init__(self, **kwargs):
for k, v in kwargs.items():
setattr(self, k, v)
这将是令人惊讶的行为所以我建议不要这样做。
答案 2 :(得分:4)
这是实现此目标的一种方式,可以轻松更改:
class Player:
_VALID_KEYWORDS = {'last_name', 'first_name', 'score'}
def __init__(self, **kwargs):
for keyword, value in kwargs.items():
if keyword in self._VALID_KEYWORDS:
setattr(self, keyword, value)
else:
raise ValueError(
"Unknown keyword argument: {!r}".format(keyword))
样本用法:
Player(last_name="George", attitude="snarky")
结果:
Traceback (most recent call last):
File "keyword_checking.py", line 13, in <module>
Player(last_name="George", attitude="snarky")
File "keyword_checking.py", line 11, in __init__
raise ValueError("Unknown keyword argument: {!r}".format(keyword))
ValueError: Unknown keyword argument: 'attitude'
答案 3 :(得分:3)
您可以使用keyword arguments:
class Player:
def __init__(self, last_name=None, first_name=None, score=None):
self.last_name = last_name
self.first_name = first_name
self.score = score
obj = Player('Max', 'Jhon')
print obj.first_name, obj.last_name
Jhon Max
使用参数 **kwargs
class Player:
def __init__(self, **args):
self.last_name = args.get('last_name')
self.first_name = args.get('first_name')
self.score = args.get('score', 0) # 0 is the default score.
obj = Player(first_name='Max', last_name='Jhon')
print obj.first_name, obj.last_name, obj.score
Max Jhon 0
答案 4 :(得分:0)
我可以尝试这种方式:
#!/usr/bin/python
class Player(object):
def __init__(self, **kwargs):
for key, val in kwargs.items():
self.__dict__[key] = val
obj = Player(first_name='First', last_name='Last')
print obj.first_name
print obj.last_name
newobj = Player(first_name='First')
print newobj.first_name
输出:
First
Last
First
答案 5 :(得分:0)
这取决于你想要的最终结果。如果要创建一个在字典中定义属性的类,可以使用setattr。
class Player:
def __init__(self, **kwargs):
for key, value in kwargs.items():
setattr(self, key, value)
In [1]: player = Player(first_name='john', last_name='wayne', score=100)
In [2]: player.first_name
Out[2]: 'john'
In [3]: player.last_name
Out[3]: 'wayne'
In [4]: player.score
Out[4]: 100
In [5]: player.address
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
<ipython-input-6-0f7ee474d904> in <module>()
----> 1 player.address
AttributeError: 'Player' object has no attribute 'address'