打印字典名称 [英] print names of dictionaries

问题描述

还是新的。学习属性和功能等等。


对不起，如果这是显而易见的，但是如果我为某些

词典定义一个函数，我怎么能只打印字典的名称？


例如假设应用程序，目录，网站是模块中定义的词典，

如何在与其他人做其他事情之前打印他们的名字。


def printdict（ dicts = [app，dirs，sites]）：

dict in dicts：

打印???


谢谢


rpd

Still new. Learning attributes and functions and so on.

Sorry if this is obvious, but if I''m defining a function for some
dictionaries, how can I print just the names of the dictionaries?

E.g. assuming apps, dirs, sites are dictionaries defined in the module,
how can I print just their names before doing other things with them.

def printdict(dicts=[apps, dirs, sites]):
for dict in dicts:
print ???

Thank you

rpd

推荐答案

BartlebyScrivener写道：

BartlebyScrivener wrote:

仍然新。学习属性和功能等等。
很抱歉，如果这很明显，但如果我正在为某些词典定义一个函数，我怎样才能打印出词典的名称？

Still new. Learning attributes and functions and so on.
Sorry if this is obvious, but if I''m defining a function for some
dictionaries, how can I print just the names of the dictionaries?

简答：你不能。

http://pyfaq.infogami.com/how-can-my...e-of-an-object


....

Jay Graves

Short answer: You can''t.

http://pyfaq.infogami.com/how-can-my...e-of-an-object

....
Jay Graves

这里这可能是你想做的事情：

Here''s an OO way that may do what you want:

class MyD（dict）：
... .def __init __（self，dic，rep）：

.... dict .__ init __（self，dic）

.... self.rep = rep

.... def __repr __（自我）：

....返回self.rep

.... apps = MyD（{' '阿尔法' '：1'， '测试' '：2}'， '应用程序''）<无线电通信/> apps
apps apps.keys（）

class MyD(dict): .... def __init__(self,dic,rep):
.... dict.__init__(self,dic)
.... self.rep = rep
.... def __repr__(self):
.... return self.rep
.... apps = MyD({''alpha'':1,''beta'':2},''apps'')
apps apps apps.keys()

[''alpha''，''beta'']


当然，最简单的方法就是使用元组（字典，字符串）。


作为旁注，因为dict是内置类型，功能，它可能不是

将它用作循环变量的好方式。


THN

[''alpha'', ''beta'']

Of course, the easiest way is just to use a tuple (dict,string).

As a side note, since dict is a builtin type and function, it might not
be good style to use it as a loop variable.

THN

哇，


这是值得深思的。谢谢。


我看到他们对改变方法的意思。我只需在每个字典中输入一个密钥，每个字典名称为er，名称及其名称值。


谢谢！


rick

Wow,

That''s food for thought. Thanks.

I see what they mean about change of approach. I''ll just stick a key in
each dictionary called, er, name with its name value.

Thank you!

rick

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