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Why doesn't python delete a function's variables when it it out of scope?

In python, I know that functions run in the context that they are called from, not the context that they are defined in. However, in the code below, where a function is returned from another function, it still has access to the variables (specifically the list 'cache') of the function it was defined in, even though that function is now out of scope (I think), and so I would have though it's variables would have been deleted. But it can also access global variables from the current context?
def wrapper(func):
cache = []
def func_new(n):
cache.append(func(n))
print(cache)
return func_new
def add1(n):
return(n+1)
x = wrapper(add1)
x(5)
x(2)
print(cache)

It doesn't?
>>> x = wrapper(add1)
>>> x(5)
[6]
>>> x(2)
[6, 3]
>>> print(cache)
Traceback (most recent call last):
File "<pyshell#7>", line 1, in <module>
print(cache)
NameError: name 'cache' is not defined
As for the cache variable being defined within the wrapper, that matches behavior elsewhere; the function has access to all variables where it is defined. The opposite is not true; variables defined in the function are not accessible in the context scope unless explicitly a global.
Does this example help?
>>> x = 4
>>> def test():
print(x)
>>> test()
4

how to change an existing variable to a global variable inside of an if condition

I'm trying to make a little choose your own adventure game for fun and can't find a way to make this work. I want to be able to have a consistent pick up method where a specific action adds one to a variable and unlocks lines of dialogue under other actions. What I've got right now is something like this.
item = 0
while True:
act = input()
if "action" in act:
if(item != 1):
print("You don't have that item.")
continue
else:
print("You use the item.")
continue
if "take item" in act:
print("You take the item.")
global item
item = item + 1
continue
The issue I'm finding is that when I try to set the global variable, the program claims a syntax warning:
main.py:107: SyntaxWarning: name 'item' is assigned to before global declaration global item
since the variable is stated before the global, but if I don't state the variable before hand, the system I have in place to prevent the wrong dialogue from printing won't work. How do I have a conditional global variable that updates an existing local variable?

TLDR
You don't need the global keyword. Remove that line and your code will work.
Short Explanation
In Python, everything is "global" (not accurate, I will explain in long explanation) so you rarely need to use the global keyword. You can try the following code in the interactive Python REPL:
>>> def test(y):
... return x, y
...
>>> test(3)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in test
NameError: name 'x' is not defined
>>> x = 4
>>> test(3)
(4, 3)
Here, I'm using a function as an example, if and while/for loops are exactly the same.
I'm declaring a function called test that takes in a parameter y, but it will return the tuple (x, y). Here, x is undefined. You would expect that it returns an error, but it actually doesn't until you run it.
When we run test(3), Python will complain that x is not defined so it cannot run the function. However, if we define x = 4 even after our function declaration, the function will work. Hence test(3) will return (4, 3). Here, x is treated as a global variable.
Long Explanation
>>> x = 3
>>> y = 4
>>> def test2():
... x = 1
... y = 2
... return x, y
...
>>> test2()
(1, 2)
>>> x
3
>>> y
4
>>>
Notice that y is not touched, even though it's clearly being assigned a different value in the function test2.
How is this possible? Actually, Python functions technically do have a scope of their own, and changes to function variables are isolated and do not change the variables of the same name outside the function. Loops and if statements don't have their own scope, so the variables inside if statements are the exact variables outside them.
>>> x = 3
>>> y = 4
>>> if True:
... x = 1
... y = 2
...
>>> x
1
>>> y
2
When you reference a variable in functions, Python will check if a value is assigned to a variable of that name in that function (either as a parameter or inside the function block). If there is a variable with that name, Python will use the variable in the function (within the scope of the function). If not, Python will try to look for a function that's in the scope outside (either another function or the global scope). If Python searched till the global scope and there is no variable of that name, it will finally produce an error.
>>> x = 3
>>> y = 4
>>> def function1():
... x = 4 # this x is being used
... def function2():
... print(x)
... function2()
...
>>> function1()
4
>>> def test3():
... x = 1
... y = 2
... return (x, y)
...
>>> test3()
(1, 2)
>>> x
3
>>> y
4
This means, only when you are trying to change the value of a variable outside of the current scope, you will use the global keyword. However, this is BAD PRACTICE. Whenever you find yourself doing this, you need to think about whether you should refactor your code to using classes. It's a very bad practice to change the value of global variables inside functions no matter the language you are using.
>>> x = 3
>>> y = 4
>>> def test4():
... global x, y # refers to the global x and y
... x = 1
... y = 2
... return (x, y)
...
>>> test4()
(1, 2)
>>> x
1
>>> y
2

python: name resolution clarification?

I'm reading the python reference name resolution, which reads
A class definition is an executable statement that may use and define names. These references follow the normal rules for name resolution with an exception that unbound local variables are looked up in the global namespace.
Based on that, I would expect the following code
x = 10
def f():
x = 5
class Test:
y = x
return Test
print(f().y)
to print 10, however it prints 5. is this a mistake in the reference, or am I misunderstanding something?

In this case, 'normal' rules apply:
x = 'global'
def f():
x = 'defined in f'
class Test:
print(x) # not local, normal rules apply
f()
# defined in f
In this second case, we would expect an UnboundLocalError: local variable 'x' referenced before assignment if we were inside a function:
x = 'global'
def f():
x = 'defined in f'
class Test:
print(x) # unbound local at this time
x = 'assigned in Test'
print(x)
But for the first print(x), x will be taken from the global namespace:
f()
# global
# assigned in Test

I think #khelwood gave the answer. The value in the variable:
Test.y
is an integer that you define, but you never give the x = 10 to the function.
Maybe what you want is actually:
x = 10
def f(x=5):
class Test:
y = x
return Test
print(f().y) #print default 5
print(f(x).y)
The last line print 10 since x is given to the function, and therefore the class set y as x

You can see the exception by wrapping f in another function that defines a local variable x:
x = 10
def g():
x = 7
def f():
class Test:
y = x
return Test
return f()
print(g().y)
Now the output is 7, as the lookup of x in the class statement bypasses the local scope of g to find the global value of x.
In your code, because class establishes a new namespace, but not a new scope, the value of x is taken from the current local scope, that of f.

Must I always reference a variable as global in my function if I were to use it?

Firstly, it's related to this function:
a = 3
def b():
a = a+1
return a
returns an error.
Isn't it when the local environment of b couldn't find 'a', then it goes to the parent environment where b is defined, that is, where 'a' is found to be 3?
Why must I reference it with a global a in order for it to work? Meaning, do I have to do this only when the immediate parent environment is the global environment? Because the below function seems to be able to reference the parent environment:
def b():
def c():
print c
return c()

This is such a common issue that is is actually in the Python FAQ:
[...] when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope. Since the last statement in foo assigns a new value to x, the compiler recognizes it as a local variable.
So, Python sees you're assigning to a local variable with a = .. and thinks, hey, a local variable, I'm fine with this. But, when it actually bumps into the content of that assignment (a + 1) and sees the same name a it complains because you're trying to reference it before assigning it.
This is why the following doesn't raise an error:
def b():
a = 20
a = a + 2
return a
You first make an assignment, a is treated as local to the function b and when a = a + 2 is encountered, Python knows a is a local variable, no confusion is present here.
In the other case you added:
def b():
def c():
print c
return c()
You are only referencing a name enclosed in the scope of function b, referencing. If you changed this to a similar assignment as before, you'll get the same UnboundLocalError:
def b():
def c():
c = c + 20
return c()

Because you are doing an assignment and the scope has to be clarified.
Your case:
a = 3
def b():
a = a + 1
return a
print(b())
Output:
Traceback (most recent call last):
File "./tests/test1.py", line 12, in <module>
print(b())
File "./tests/test1.py", line 9, in b
a = a + 1
UnboundLocalError: local variable 'a' referenced before assignment
When you do the assignement a = ... you have created always a local variable unless it has been declared global which is not the case. Local resolution takes precedence.
Your example would obviously work with:
a = 3
def b():
global a
a = a + 1
return a
print(b())
Which outputs:
4
But using globals, imho, should not be the way and hence passing a as a parameter to function b would be it:
a = 3
def b(a):
a = a + 1
return a
print(b(a))
Which agains produces the expected 4.
EDIT:
Following the edition in the OP: you can always "reference" the global scope (and parent scope). But if you make assignments they take place in the local scope.
If your code (which is wronly indented and contains a new c which isnt' anywhere) is meant to do this:
a = 3
def b():
def c():
print(a)
return c()
b()
It works ... and prints 3 because there is NO assignment and only referencing. As soon as you executing an assignment a would again be created in the local scope.

Assignment will create a new variable in Python.
Here may help you.

Variable scope in nested functions

Could someone explain why the following program fails:
def g(f):
for _ in range(10):
f()
def main():
x = 10
def f():
print x
x = x + 1
g(f)
if __name__ == '__main__':
main()
with the message:
Traceback (most recent call last):
File "a.py", line 13, in <module>
main()
File "a.py", line 10, in main
g(f)
File "a.py", line 3, in g
f()
File "a.py", line 8, in f
print x
UnboundLocalError: local variable 'x' referenced before assignment
But if I simply change the variable x to an array, it works:
def g(f):
for _ in range(10):
f()
def main():
x = [10]
def f():
print x[0]
x[0] = x[0] + 1
g(f)
if __name__ == '__main__':
main()
with the output
10
11
12
13
14
15
16
17
18
19
The reason I am confused is, if from f() it can't access x, why it becomes accessible if x is an array?
Thanks.

But this answer says the problem is with assigning to x. If that's it,
then printing it should work just fine, shouldn't it?
You have to understand the order in which things happen. Before your python code is even compiled and executed, something called a parser reads through the python code and checks the syntax. Another thing the parser does is mark variables as being local. When the parser sees an assignment in the code in a local scope, the variable on the lefthand side of the assignment is marked as local. At that point, nothing has even been compiled yet--let alone executed, and therefore no assignment takes place; the variable is merely marked as a local variable.
After the parser is finished, the code is compiled and executed. When execution reaches the print statement:
def main():
x = 10 #<---x in enclosing scope
def f():
print x #<-----
x = x + 1 #<-- x marked as local variable inside the function f()
the print statement looks like it should go ahead and print the x in the enclosing scope (the 'E' in the LEGB lookup process). However, because the parser previously marked x as a local variable inside f(), python does not proceed past the local scope (the 'L' in the LEGB lookup process) to lookup x. Because x has not been assigned to in the local scope at the time 'print x' executes, python spits out an error.
Note that even if the code where an assignment occurs will NEVER execute, the parser still marks the variable on the left of an assignment as a local variable. The parser has no idea about how things will execute, so it blindly searches for syntax errors and local variables throughout your file--even in code that can never execute. Here are some examples of that:
def dostuff ():
x = 10
def f():
print x
if False: #The body of the if will never execute...
a b c #...yet the parser finds a syntax error here
return f
f = dostuff()
f()
--output:--
File "1.py", line 8
a b c
^
SyntaxError: invalid syntax
The parser does the same thing when marking local variables:
def dostuff ():
x = 10
def f():
print x
if False: #The body of the if will never execute...
x = 0 #..yet the parser marks x as a local variable
return f
f = dostuff()
f()
Now look what happens when you execute that last program:
Traceback (most recent call last):
File "1.py", line 11, in <module>
f()
File "1.py", line 4, in f
print x
UnboundLocalError: local variable 'x' referenced before assignment
When the statement 'print x' executes, because the parser marked x as a local variable the lookup for x stops at the local scope.
That 'feature' is not unique to python--it happens in other languages too.
As for the array example, when you write:
x[0] = x[0] + 1
that tells python to go lookup up an array named x and assign something to its first element. Because there is no assignment to anything named x in the local scope, the parser does not mark x as a local variable.

The reason is in first example you used an assignment operation, x = x + 1, so when the functions was defined python thought that x is local variable. But when you actually called the function python failed to find any value for the x on the RHS locally, So raised an Error.
In your second example instead of assignment you simply changed a mutable object, so python will never raise any objection and will fetch x[0]'s value from the enclosing scope(actually it looks for it firstly in the enclosing scope, then global scope and finally in the builtins, but stops as soon as it was found).
In python 3x you can handle this using the nonlocal keyword and in py2x you can either pass the value to the inner function or use a function attribute.
Using function attribute:
def main():
main.x = 1
def f():
main.x = main.x + 1
print main.x
return f
main()() #prints 2
Passing the value explicitly:
def main():
x = 1
def f(x):
x = x + 1
print x
return x
x = f(x) #pass x and store the returned value back to x
main() #prints 2
Using nonlocal in py3x:
def main():
x = 1
def f():
nonlocal x
x = x + 1
print (x)
return f
main()() #prints 2

The problem is that the variable x is picked up by closure. When you try to assign to a variable that is picked up from the closure, python will complain unless you use the global or nonlocal1 keywords. In the case where you are using a list, you're not assigning to the name -- You can modify an object which got picked up in the closure, but you can't assign to it.
Basically, the error occurs at the print x line because when python parses the function, It sees that x is assigned to so it assumes x must be a local variable. When you get to the line print x, python tries to look up a local x but it isn't there. This can be seen by using dis.dis to inspect the bytecode. Here, python uses the LOAD_FAST instruction which is used for local variables rather than the LOAD_GLOBAL instruction which is used for non-local variables.
Normally, this would cause a NameError, but python tries to be a little more helpful by looking for x in func_closure or func_globals 2. If it finds x in one of those, it raises an UnboundLocalError instead to give you a better idea about what is happening -- You have a local variable which couldn't be found (isn't "bound").
1python3.x only
2python2.x -- On python3.x, those attributes have changed to __closure__ and __globals__ respectively

The problem is in the line
x = x + 1
This is the first time x being assigned in function f(), telling the compiler that x is a local name. It conflicts with the previous line print x, which can't find any previous assignment of the local x.
That's where your error UnboundLocalError: local variable 'x' referenced before assignment comes from.
Note that the error happens when compiler tries to figure out which object the x in print x refers to. So the print x doesn't executes.
Change it to
x[0] = x[0] + 1
No new name is added. So the compiler knows you are referring to the array outside f().