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As I understood:

• something is needed to be loaded
• it is needed to be loaded often, because file with code which uses this something is edited often
• you don't want to wait until it will be loaded every time

Maybe such solution will be okay for you.

You can write script loader file in such way (tested on Python 3.8):

import importlib.util, traceback, sys, gc
# Example data
import pickle
something = pickle.loads(pickle.dumps([123]))
if __name__ == '__main__':
 try:
 mod_path = sys.argv[1]
 except IndexError:
 print('Usage: python3', sys.argv[0], 'PATH_TO_SCRIPT')
 exit(1)
 modules_before = list(sys.modules.keys())
 argv = sys.argv[1:]
 while True:
 MOD_NAME = '__main__'
 spec = importlib.util.spec_from_file_location(MOD_NAME, mod_path)
 mod = importlib.util.module_from_spec(spec)
 # Change to needed global name in the target module
 mod.something = something
 
 sys.modules[MOD_NAME] = mod
 sys.argv = argv
 try:
 spec.loader.exec_module(mod)
 except:
 traceback.print_exc()
 del mod, spec
 modules_after = list(sys.modules.keys())
 for k in modules_after:
 if k not in modules_before:
 del sys.modules[k]
 gc.collect()
 print('Press enter to re-run, CTRL-C to exit')
 sys.stdin.readline()

Example of module:

# Change 1 to some different number when first script is running and press enter
something[0] += 1 
print(something)

Should work. And should reduce the reload time of pickle close to zero 🌝

UPD Add a possibility to accept script name with command line arguments