为什么没有python dict.update（）返回对象？(Why doesn't a python dict.update() return the object?)

为什么没有python dict.update（）返回对象？(Why doesn't a python dict.update() return the object?)

我想做：

award_dict = { "url" : "http://facebook.com", "imageurl" : "http://farm4.static.flickr.com/3431/3939267074_feb9eb19b1_o.png", "count" : 1, } def award(name, count, points, desc_string, my_size, parent) : if my_size > count : a = { "name" : name, "description" : desc_string % count, "points" : points, "parent_award" : parent, } a.update(award_dict) return self.add_award(a, siteAlias, alias).award

但是，如果在功能上感觉真的很麻烦，我会宁愿做的：

return self.add_award({ "name" : name, "description" : desc_string % count, "points" : points, "parent_award" : parent, }.update(award_dict), siteAlias, alias).award

为什么不更新返回对象，以便可以链接？

JQuery做这个链接。 为什么python不能接受？

I 'm trying to do :

award_dict = { "url" : "http://facebook.com", "imageurl" : "http://farm4.static.flickr.com/3431/3939267074_feb9eb19b1_o.png", "count" : 1, } def award(name, count, points, desc_string, my_size, parent) : if my_size > count : a = { "name" : name, "description" : desc_string % count, "points" : points, "parent_award" : parent, } a.update(award_dict) return self.add_award(a, siteAlias, alias).award

But if felt really cumbersome in the function, and I would have rather done :

return self.add_award({ "name" : name, "description" : desc_string % count, "points" : points, "parent_award" : parent, }.update(award_dict), siteAlias, alias).award

Why doesn't update return the object so you can chain?

JQuery does this to do chaining. Why isn't it acceptable in python?

最满意答案

Python主要实现了一个务实的命令查询分离的风格：mutator返回None （以实际的方式引发的例外，如pop ;-)，所以它们不可能与访问器混淆（同样，分配不是表达式，语句表达式分离在那里，等等）。

这并不意味着，当你真正想要的时候，并没有很多方法来合并，例如， dict(a, **award_dict)使得一个新的命令就像你似乎希望的那样。为什么不使用THAT如果你真的觉得很重要？

编辑 ：btw，没有必要，在你的具体情况下，创建a路途，或者：

dict(name=name, description=desc % count, points=points, parent_award=parent, **award_dict)

创建一个与您的a.update(award_dict)具有完全相同语义的单个dict（包括在冲突的情况下）， award_dict中的条目会覆盖您明确给出的事实;获取其他语义，即具有明确的条目“获胜”这样的冲突，通过作为唯一的位置 arg， 在关键字之前 ，并且没有**形式 - dict(award_dict, name=name等）的dict(award_dict, name=name 。

Python's mostly implementing a pragmatically tinged flavor of command-query separation: mutators return None (with pragmatically induced exceptions such as pop;-) so they can't possibly be confused with accessors (and in the same vein, assignment is not an expression, the statement-expression separation is there, and so forth).

That doesn't mean there aren't a lot of ways to merge things up when you really want, e.g., dict(a, **award_dict) makes a new dict much like the one you appear to wish .update returned -- so why not use THAT if you really feel it's important?

Edit: btw, no need, in your specific case, to create a along the way, either:

dict(name=name, description=desc % count, points=points, parent_award=parent, **award_dict)

creates a single dict with exactly the same semantics as your a.update(award_dict) (including, in case of conflicts, the fact that entries in award_dict override those you're giving explicitly; to get the other semantics, i.e., to have explicit entries "winning" such conflicts, pass award_dict as the sole positional arg, before the keyword ones, and bereft of the ** form -- dict(award_dict, name=name etc etc).