使用locals()的Python词典理解给出KeyError [英] Python dictionary comprehension using locals() gives KeyError

问题描述

>>> a = 1
>>> print { key: locals()[key] for key in ["a"] }
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 1, in <dictcomp>
KeyError: 'a'

如何创建具有这种理解力的字典?

How can I create a dictionary with a comprehension like this?

推荐答案

dict理解具有其自己的名称空间，而该名称空间中的locals()没有a.从技术上讲，除了最外面的可迭代对象(这里是["a"])的初始可迭代对象之外的所有东西，几乎都是作为嵌套函数运行的，最外面的可迭代对象作为参数传递.

A dict comprehension has its own namespace, and locals() in that namespace has no a. Technically speaking, everything but the initial iterable for the outermost iterable (here ["a"]) is run almost as a nested function with the outermost iterable passed in as an argument.

如果您改用globals()或创建对dict理解的locals()词典外的引用，您的代码将起作用:

Your code works if you used globals() instead, or created a reference to the locals() dictionary outside of the dict comprehension:

l = locals()
print { key: l[key] for key in ["a"] }

演示:

>>> a = 1
>>> l = locals()
>>> { key: l[key] for key in ["a"] }
{'a': 1}
>>> { key: globals()[key] for key in ["a"] }
{'a': 1}

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