How can I reassing a global variables in local namespaces in Python?

Question

I have a problem of not being able to reassing global variables in local namespaces. For example you can't change global x's value in example:
global x
x = 5
def f():
x = 10

I thought I could have fixed that by:
global x
x = 5
def f():
nonlocal x
x = 10

But it results in syntax errors.
Is there a correct way to fix that problem?

Answer 1

Seb TheS apologizes, I didn't know that and thank you for letting me know.
And I just read about nonlocal variable, it says:
Nonlocal variable are used in nested function whose local scope is not defined. This means, the variable can be neither in the local nor the global scope.
And a change in nonlocal variable changes the value of local variable.

Answer 2

Yes syntax error 'cause there's no keyword called nonlocal in Python, lol.
In the first code, in the function's scope, it cannot find that x is a global variable but it sees x as a local variable within the function with a value of 10. And hence upon printing x outside the function, it'll have a value of 5 and not 10.
You must declare x as global inside the function.
def f():
global x
x=10

Answer 3

Seb TheS The nonlocal keyword allows to assign to variables in an outer scope, but not a global scope. So you can't use
nonlocal in your function because then it would try to assign to a global scope. That's why you have got a SyntaxError. You can use nonlocal in a nested function only

To change the global variable write global x inside the function:
x = 5
def f():
global x
x = 10
print(x) #5
f()
print(x) #10

Answer 4

M. Watney Actually that works, but help("keywords") says that "nonlocal" is a keyword in Python, so I thought it would have some meaning within the problem.

Local and Global (and __builtins__) were already quite clear.

Answer 5

You need to declare global variable inside funtion:
x = 5
def f():
global x
x = 10