Python dictionary comprehension using locals() gives KeyError

Question

>>> 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'

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

It should result in {"a": 1}.

Answer 1

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.

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"] }

Demo:

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

Answer 2

You can try using globals() instead:

print {key : globals()[key] for key in ["a"]}

since a is not defined in the scope of the dict comprehension (as @MartijnPieters said).