Python nested function variable scope

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.

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.

Python nested functions variable scopingStack Overflow