In python's loops, how to load pickles to variable's string content instead of the variable itself?

Question

I have a directory full of python pickles containing dicts and I want to open the pickles in memory with their base name. If I have them in memory then I do not have to open and close the same dict(s) multiple times within other nested loops.

So list_of_pickles = os.listdir(path/to/pickles)

but

for i in list_of_pickles:
    pickle_base = i.rsplit('.')[0]
    name_pickle = f'{path/to/pickles}/{i}'

    with open(name_pickle,'rb') as current_pickle:
        pickle_base = pickle.load(current_pickle)

fails to assign the pickles to their base names and assign them to pickle_base literally. Being able to assign to the pickles' base name would make things easier.

Answer 1

I did it with locals()[{variable_name}]. So, if one wants something assigned to the variable's string content instead of the variable's literal name use locals()[{variable_name}].

For example, in the above code:

for i in list_of_pickles:
    pickle_base = i.rsplit('.')[0]
    name_pickle = f'{path/to/pickles}/{i}'

    with open(name_pickle,'rb') as current_pickle:
        locals()[pickle_base] = pickle.load(current_pickle)