How to modify the local namespace in python

Question

How can I modify the local namespace of a function in python? I know that locals() returns the local namespace of the function when called inside it, but I want to do something like this (I have a reason why I want to do this where g is not accessible to f, but it's quicker to give a trivial, stupid example to illustrate the problem):

def g():
   pass

def f():
    g()

f.add_to_locals({'g':g})

Answer 1

You've a couple of options. First, note that g in your example isn't actually a local to the function (ie. not assigned within it), it's a global (ie hasn't been assigned to a local variable). This means that it will be looked up in the module the function is defined in. This is fortunate, as there's no way of altering locals externally (short of patching the bytecode), as they get assigned when the function runs, not before.

One option is simply to inject your function into the function's module's namespace. This will work, but will affect every function in that module that accesses the variable, rather than just the one function.

To affect just the one function, you need to instead point that func_globals somewhere else. Unfortunately, this is a read-only property, but you can do what you want by recreating the function with the same body, but a different global namespace:

import new
f = new.function(f.func_code, {'g': my_g_function}, f.func_name, f.func_defaults, f.func_closure)

f will now be indentical, except that it will look for globals in the provided dict. Note that this rebinds the whole global namespace - if there are variables there that f does look up, make sure you provide them too. This is also fairly hacky though, and may not work on versions of python other than cpython.