Python:将字典中的变量加载到名称空间中 [英] Python: load variables in a dict into namespace

问题描述

我想在函数外部使用一堆在函数中定义的局部变量.因此，我在返回值中传递了x=locals().

I want to use a bunch of local variables defined in a function, outside of the function. So I am passing x=locals() in the return value.

如何将该词典中定义的所有变量加载到函数外部的名称空间中，这样我可以简单地使用variable，而不是使用x['variable']访问值.

How can I load all the variables defined in that dictionary into the namespace outside the function, so that instead of accessing the value using x['variable'], I could simply use variable.

推荐答案

考虑Bunch替代方法:

class Bunch(object):
  def __init__(self, adict):
    self.__dict__.update(adict)

因此，如果您有字典d，并且想要使用语法x.foo而不是笨拙的d['foo']访问(读取)其值，则只需

so if you have a dictionary d and want to access (read) its values with the syntax x.foo instead of the clumsier d['foo'], just do

x = Bunch(d)

这在内部和外部函数中均有效-与将d注入globals()相比，它非常更干净，更安全！记住Python Zen中的最后一行...:

this works both inside and outside functions -- and it's enormously cleaner and safer than injecting d into globals()! Remember the last line from the Zen of Python...:

>>> import this
The Zen of Python, by Tim Peters
   ...
Namespaces are one honking great idea -- let's do more of those!

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