python - How to capture the current namespace inside nested functions? -

suppose 1 had following:

def outer(foo, bar, baz):     frobozz = foo + bar + baz      def inner(x, y, z):         return dict(globals().items() + locals().items())      return inner(7, 8, 9) 

the value returned inner dictionary obtained merging dictionaries returned globals() , locals(), shown. in general¹, returned value not contain entries foo, bar, baz, , frobozz, though these variables visible within inner, , therefore, arguably belong in inner's namespace.

one way facilitate capturing of inner's namespace following kluge:

def outer(foo, bar, baz):     frobozz = foo + bar + baz      def inner(x, y, z, _kluge=locals().items()):         return dict(globals().items() + _kluge + locals().items())      return inner(7, 8, 9) 

is there better way capture inner's namespace sort of kluge?

---

^{1 unless, is, variables having names happen present in current global namespace.}

this isn't dynamic, , it's not best way, could access variables inside inner add them "namespace":

def outer(foo, bar, baz):     frobozz = foo + bar + baz      def inner(x, y, z):         foo, bar, baz, frobozz         return dict(globals().items() + locals().items())      return inner(7, 8, 9) 

you store outer function's locals variable, , use variable inside inner's return value:

def outer(foo, bar, baz):     frobozz = foo + bar + baz     outer_locals = locals()     def inner(x, y, z):         return dict(outer_locals.items() + globals().items() + locals().items())     return inner(7, 8, 9)