Python函数中的默认参数在哪里-WhereisthedefaultparameterinPythonfunction

10  

As others already said, the default values are stored in the function object.

正如其他人已经说过的那样,默认值存储在函数对象中。

For example, in CPython you can do this:

例如,在CPython中,您可以这样做:

>>> def f(a=[]):
...     pass
...
>>> f.func_defaults
([],)
>>> f.func_code.co_varnames
('a',)
>>>

However, co_varnames may contain more than names of args so it needs further processing and these attributes might not even be there in other Python implementations. Therefore you should use the inspect module instead which takes care of all implementation details for you:

但是,co_varnames可能包含多个args名称,因此需要进一步处理,这些属性可能甚至不存在于其他Python实现中。因此,您应该使用inspect模块,它会为您处理所有实现细节:

>>> import inspect
>>> spec = inspect.getargspec(f)
>>> spec
ArgSpec(args=['a'], varargs=None, keywords=None, defaults=([],))
>>>

The ArgSpec is a named tuple so you can access all values as attributes:

ArgSpec是一个命名元组,因此您可以将所有值作为属性访问:

>>> spec.args
['a']
>>> spec.defaults
([],)
>>>

As the documentation says, the defaults tuple always corresponds to the n last arguments from args. This gives you your mapping.

正如文档所说,默认元组总是对应于args中的n个最后一个参数。这为您提供了映射。

To create a dict you could do this:

要创建一个dict,你可以这样做:

>>> dict(zip(spec.args[-len(spec.defaults):], spec.defaults))
{'a': []}
>>>

4  

It's in the function object, in the func_defaults:

它位于函数对象的func_defaults中:

def f(a=[]): a.append(3)

print f.func_defaults # ([],)

f()

print f.func_defaults # ([3],)

4  

It's attached to the function object, see foo.func_defaults:

它附加到函数对象,请参阅foo.func_defaults:

>>> foo()
>>> foo.func_defaults
([3],)
>>> foo()
>>> foo.func_defaults
([3, 3],)

In case if you want to get the mapping of a onto [], you can access foo.func_code:

如果你想获得一个到[]的映射,你可以访问foo.func_code:

defaults = foo.func_defaults
# the args are locals in the beginning:
args = foo.func_code.co_varnames[:foo.func_code.co_argcount] 
def_args = args[-len(defaults):]  # the args with defaults are in the end
print dict(zip(def_args, defaults)) # {'a': []}

(But, apparently, the yak's version is better.)

(但是,显然,牦牛的版本更好。)

2  

It is stored in the func_defaults attribute of the function object.

它存储在函数对象的func_defaults属性中。

>>> foo.func_defaults
([],)
>>> foo()
([3],)

1  

I have found an interesting situation: in python 2.5.2 version, try the function 'foo()'

我发现了一个有趣的情况:在python 2.5.2版本中,尝试函数'foo()'

>>> foo()
[1]
>>> foo()
[1]
>>> foo()
[1]

Because the objects of the function called are different:

因为调用的函数的对象是不同的:

>>> id(foo())
4336826757314657360
>>> id(foo())
4336826757314657008
>>> id(foo())
4336826757314683160

In 2.7.2 version:

在2.7.2版本中:

>>> foo()
[1]
>>> foo()
[1, 1]
>>> foo()
[1, 1, 1]

In this case, the object is the same each time calling the function:

在这种情况下,每次调用函数时对象都是相同的:

>>> id(foo())
29250192
>>> id(foo())
29250192
>>> id(foo())
29250192

Is it a problem of different versions?

这是不同版本的问题吗?