扩展dict类 [英] Extending the dict class
问题描述
Hello Pythoneers：
我需要将一个命名参数列表传递给给定的
订单中的函数，
并确保使用
中的keys（）检索这些命名参数。例如：
keyargs = {}
keyargs [''one''] = 1
keyargs [''two' '] = 2
keyargs [''three''] = 3
myfunc（** keyargs）
-myfunc将使用keys（）以相同的顺序检索关键参数
，因为它们已设置，即keyargs.keys（）== [''one''，''two''，''three' '）
为实现这一目标，我将dict子类化并在__init __（），__ setitem __（）和keys（）中添加了所需的行
。然后我将dict分配给了我的新课程，但是只获得了字典所需的行为
创建如下：
d = dict（）
但不适用于创建的字典：
d = {}
或
myfunc（** keyargs）
是否可以在这些情况下强制字典创建使用
我自己的dict类而不是默认类？
我想我们可以把它作为一个更通用的问题来表达：如果我想要修改字典类的行为，那么
是否有任何方法可以在解释器范围内进行那么特殊的dict构造函数
就像上面提到的那样使用修改后的版本？
感谢您的帮助
Hello Pythoneers:
I need to pass a list of named arguments to a function in a given
order,
and make sure these named arguments are retrieved using keys() in the
same order they were given. Example:
keyargs={}
keyargs[''one'']=1
keyargs[''two'']=2
keyargs[''three'']=3
myfunc(**keyargs)
-myfunc would retrieve key arguments with keys() in the same order
as they were set, i.e. keyargs.keys() == [''one'', ''two'', ''three'']
To achieve that, I subclassed dict and added the required lines
in __init__(), __setitem__() and keys(). I then assigned dict to
my new class but only get the desired behaviour for dictionaries
created like:
d=dict()
but not for dictionaries created like:
d={}
or
myfunc(**keyargs)
Is it possible to force dictionary creation in these case to use
my own dict class instead of the default one?
I guess we can formulate this as a more generic question: if I
want to modify the behaviour of the dictionary class, is there
any way to do it interpreter-wide so that special dict constructors
like those mentioned above use the modified version?
Thanks for helping
推荐答案
chosenchu写道：
chosechu wrote:
是否可以在这些情况下强制使用字典创建
我自己的dict类而不是默认类？
Is it possible to force dictionary creation in these case to use
my own dict class instead of the default one?
否
No
我想我们可以将此表示为一个更通用的问题：如果我
想要修改字典类的行为，有没有
任何方式来解释整个范围，以便特殊字典构造函数
如上所述使用修改后的版本？
I guess we can formulate this as a more generic question: if I
want to modify the behaviour of the dictionary class, is there
any way to do it interpreter-wide so that special dict constructors
like those mentioned above use the modified version?
并非没有修改和重新编译解释器。
Not without modifying and recompiling the interpreter.
Duncan Booth写道：
Duncan Booth wrote:
是否有可能强制字典创建在这种情况下使用
我自己的dict类代替默认的一个？
Is it possible to force dictionary creation in these case to use
my own dict class instead of the default one?
否
No
哎哟。我期待这样的事情，感谢你确认。
如果我可能：这似乎与我不一致。我创建了一个增强的
版本的类，没有可能扩展3个合法的
构造函数中的2个：{}和func（** d）。虽然我同意更改基类
行为是一个坏主意，因为我们有可能将
dict重新分配给另一个类，我们应该能够一致地执行它，即
也重新给予两位特殊建设者。
只是关于这个主题的2c ...无论如何我找到了解决办法。
Ouch. I was expecting something like that, thanks for confirming it.
If I may: this seems inconsistent to me. I have created an augmented
version of the class with no possibility to extend 2 of the 3 legal
constructors: {} and func(**d). While I agree changing base class
behaviour is a bad idea, since we have the possibility to re-assign
dict to another class we should be able to do it consistently, i.e.
re-assing the two special constructors too.
Just my 2c on the topic... I have found workarounds anyway.
chosenchu写道：
chosechu wrote:
Hello Pythoneers：
我需要将命名参数列表传递给给定的
命令中的函数，
并确保使用
$中的keys（）检索这些命名参数b $ b给出了相同的订单。例如：
keyargs = {}
keyargs [''one''] = 1
keyargs [''two' '] = 2
keyargs [''three''] = 3
myfunc（** keyargs）
-myfunc将使用keys（）以相同的顺序检索关键参数
，因为它们已设置，即keyargs.keys（）== [''one''，''two''，''three' ']
Hello Pythoneers:
I need to pass a list of named arguments to a function in a given
order,
and make sure these named arguments are retrieved using keys() in the
same order they were given. Example:
keyargs={}
keyargs[''one'']=1
keyargs[''two'']=2
keyargs[''three'']=3
myfunc(**keyargs)
-myfunc would retrieve key arguments with keys() in the same order
as they were set, i.e. keyargs.keys() == [''one'', ''two'', ''three'']
我不确定你为什么要这样做 - 也许你可以
告诉你更多关于你的实际用途案例？
无论如何，由于它不是直接可能的，可能的解决方法
可以传递一系列（键，值）元组相反（可能是
* args）。这当然要求你可以修改实现
of myfunc（）。
-
bruno desthuilliers
python -c" print''@''。join（[''。''。join（[w [:: - 1] for w in p.split（''。''）]）for br /'
p in''o **** @ xiludom.gro''。split（''@''）]）"
I''m not sure to understand why you want to do so - perhaps you could
tell more about your real use case ?
Anyway, and since it''s not directly possible, a possible workaround
could be to pass a sequence of (key, value) tuples instead (possibly as
*args). This of course requires that you can modify the implementation
of myfunc().
--
bruno desthuilliers
python -c "print ''@''.join([''.''.join([w[::-1] for w in p.split(''.'')]) for
p in ''o****@xiludom.gro''.split(''@'')])"
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