Python locals() for containing scope

Question

TL;DR: I want a locals() that looks in a containing scope.

Hi, all.

I'm teaching a course on Python programming to some chemist friends, and I want to be sure I really understand scope.

Consider:

def a():
    x = 1
    def b():
        print(locals())
        print(globals())
    b()

Locals prints an empty environment, and globals prints the usual globals. How do I get access to the environment where x is stored? Clearly the interpreter knows about it because I can refer to it.

Related: When does scoping happen? The following nameErrors on a = x+2 only if x=3 is included:

def a():
    x = 1
    def b():
        a = x+2
        x = 3
    b()

If you comment out x=3, the code works. Does this mean that python makes a lexical-scope pass over the code before it interprets it?

Answer 1

What is happening in your code is that when python see's the x=3 line in your b() method, it is recreating x with a scope within the b function instead of using the x with it's scope in the a function.

because your code then goes:

    a = x+2
    x = 3

it is saying that you need to define the inner scoped x before referencing it.

however, when you do not have the assigning of x within the b function python does not attempt to make a lower scoped x and will not throw any errors.

The following code will illustrate this:

def a():
    x = 1
    def b():
        x = 3
        print (x)

    def c():
        print (x)

    b()
    c()
    print (x)

a()

However if you were to declare x as global you could use it within a function, like so:

def a():
    global x
    x = 1

    def d():
        global x
        x += 2

    print (x)
    d()
    print (x)

a()

Python 3 has also added a nonlocal keyword that will let you access a variable from an enclosing scope, usage is like:

def a():
    x = 1

    def d():
        nonlocal x
        x += 2

    print (x)
    d()
    print (x)

a()

Will print the same results as the global example.

---

In case I miss-read the question:

As per the answer to Get locals from calling namespace in Python:

you can use:

import inspect

def a():
    x = 1

    def d():
        frame = inspect.currentframe()
        try:
            print (frame.f_back.f_locals)
        finally:
            del frame
    d()

a()

to get the local scope of the functions caller.