Python nested functions variable scoping [duplicate]

I've read almost all the other questions about the topic, but my code still doesn't work.

I think I'm missing something about python variable scope.

Here is my code:

PRICE_RANGES = {
                64:(25, 0.35),
                32:(13, 0.40),
                16:(7, 0.45),
                8:(4, 0.5)
                }

def get_order_total(quantity):
    global PRICE_RANGES
    _total = 0
    _i = PRICE_RANGES.iterkeys()
    def recurse(_i):
        try:
            key = _i.next()
            if quantity % key != quantity:
                _total += PRICE_RANGES[key][0]
            return recurse(_i) 
        except StopIteration:
            return (key, quantity % key)

    res = recurse(_i)

And I get

"global name '_total' is not defined"

I know the problem is on the _total assignment, but I can't understand why. Shouldn't recurse() have access to the parent function's variables?

Can someone explain to me what I'm missing about python variable scope?

In Python 3, you can use the nonlocal statement to access non-local, non-global scopes.

The nonlocal statement causes a variable definition to bind to a previously created variable in the nearest scope. Here are some examples to illustrate:

def sum_list_items(_list):
    total = 0

    def do_the_sum(_list):
        for i in _list:
            total += i

    do_the_sum(_list)

    return total

sum_list_items([1, 2, 3])

The above example will fail with the error: UnboundLocalError: local variable 'total' referenced before assignment

Using nonlocal we can get the code to work:

def sum_list_items(_list):
    total = 0

    def do_the_sum(_list):

        # Define the total variable as non-local, causing it to bind
        # to the nearest non-global variable also called total.
        nonlocal total

        for i in _list:
            total += i

    do_the_sum(_list)

    return total

sum_list_items([1, 2, 3])

But what does "nearest" mean? Here is another example:

def sum_list_items(_list):

    total = 0

    def do_the_sum(_list):

        # The nonlocal total binds to this variable.
        total = 0

        def do_core_computations(_list):

            # Define the total variable as non-local, causing it to bind
            # to the nearest non-global variable also called total.
            nonlocal total

            for i in _list:
                total += i

        do_core_computations(_list)

    do_the_sum(_list)

    return total

sum_list_items([1, 2, 3])

In the above example, total will bind to the variable defined inside the do_the_sum function, and not the outer variable defined in the sum_list_items function, so the code will return 0.

def sum_list_items(_list):

    # The nonlocal total binds to this variable.
    total = 0

    def do_the_sum(_list):

        def do_core_computations(_list):

            # Define the total variable as non-local, causing it to bind
            # to the nearest non-global variable also called total.
            nonlocal total

            for i in _list:
                total += i

        do_core_computations(_list)

    do_the_sum(_list)

    return total

sum_list_items([1, 2, 3])

In the above example the nonlocal assignment traverses up two levels before it locates the total variable that is local to sum_list_items.

Here's an illustration that gets to the essence of David's answer.

def outer():
    a = 0
    b = 1

    def inner():
        print a
        print b
        #b = 4

    inner()

outer()

With the statement b = 4 commented out, this code outputs 0 1, just what you'd expect.

But if you uncomment that line, on the line print b, you get the error

UnboundLocalError: local variable 'b' referenced before assignment

It seems mysterious that the presence of b = 4 might somehow make b disappear on the lines that precede it. But the text David quotes explains why: during static analysis, the interpreter determines that b is assigned to in inner, and that it is therefore a local variable of inner. The print line attempts to print the b in that inner scope before it has been assigned.

When I run your code I get this error:

UnboundLocalError: local variable '_total' referenced before assignment

This problem is caused by this line:

_total += PRICE_RANGES[key][0]

The documentation about Scopes and Namespaces says this:

A special quirk of Python is that – if no global statement is in effect – assignments to names always go into the innermost scope. Assignments do not copy data — they just bind names to objects.

So since the line is effectively saying:

_total = _total + PRICE_RANGES[key][0]

it creates _total in the namespace of recurse(). Since _total is then new and unassigned you can't use it in the addition.

Rather than declaring a special object or map or array, one can also use a function attribute. This makes the scoping of the variable really clear.

def sumsquares(x,y):
  def addsquare(n):
    sumsquares.total += n*n

  sumsquares.total = 0
  addsquare(x)
  addsquare(y)
  return sumsquares.total

Of course this attribute belongs to the function (defintion), and not to the function call. So one must be mindful of threading and recursion.