First of all, let me say that I read the many threads with similar topics on creating dynamically named variables, but they mostly relate to Python 2 or they assume you are
On the exec/eval/locals question
At least on the CPython implementation modifications to the locals() dictionary do not actually change the names in the local scope, which is why it's meant to be used read-only. You can change it, and you can see your changes in the dictionary object, but the actual local scope is not changed.
exec() takes two optional dictionary arguments, a global scope and a local scope. It defaults to globals() and locals(), but since changes to locals() aren't "real" outside of the dictionary, exec() only affects the "real" local scope when globals() is locals(), i.e. in a module outside of any function. (So in your case it's failing because it's inside a function scope).
The "better" way to use exec() in this case is to pass in your own dictionary, then operate on the values in that.
def foo():
exec_scope = {}
exec("y = 2", exec_scope)
print(exec_scope['y'])
foo()
In this case, exec_scope is used as the global and local scope for the exec, and after the exec it will contain {'y': 2, '__builtins__': __builtins__} (the builtins are inserted for you if not present)
If you want access to more globals you could do exec_scope = dict(globals()).
Passing in different global and local scope dictionaries can produce "interesting" behavior.
If you pass the same dictionary into successive calls to exec or eval, then they have the same scope, which is why your eval worked (it implicitly used the locals() dictionary).
On dynamic variable names
If you set the name from a string, what's so wrong about getting the value as a string (i.e. what a dictionary does)? In other words, why would you want to set locals()['K'] and then access K? If K is in your source it's not really a dynamically set name... hence, dictionaries.