using locals() inside dictionary comprehension

Question

The following code doesn't work, I assume because the locals() variable inside the comprehension will refer to the nested block where comprehension is evaluated:

def f():
    a = 1
    b = 2
    list_ = ['a', 'b']
    dict_ = {x : locals()[x] for x in list_}

I could use globals() instead, and it seems to work, but that may come with some additional problems (e.g., if there was a variable from a surrounding scope that happens to have the same name).

Is there anything that would make the dictionary using the variables precisely in the scope of function f?

Note: I am doing this because I have many variables that I'd like to put in a dictionary later, but don't want to complicate the code by writing dict_['a'] instead of a in the meantime.

Answer 1

You could perhaps do this:

def f(): 
    a = 1 
    b = 2 
    list_ = ['a', 'b'] 
    locals_ = locals()
    dict_ = dict((x, locals_[x]) for x in list_)

However, I would strongly discourage the use of locals() for this purpose.

Answer 2

I believe that you're right: the locals() inside the dict comprehension will refer to the comprehension's namespace.

One possible solution (if it hasn't already occurred to you):

f_locals = locals()
dict_ = {x : f_locals[x] for x in list_}