Globals in Functions | Python Fiddle

"""
If you define a variable outside of a function before running the function,
it does not get updated by the function when used. They are totally separate.
The only way to update it is to explicitly define its value outside of the function.
This contrasts with C, wherein a variable can be defined outside of a function,
and then be updated within one.

HOWEVER, a variable can be defined as global, and then it will work. But that's considered
bad practice in most situations, so don't...

ALSO, methods (such as list.append(1) or using list[0]=1) will work on lists outside of a function,
without declaring that list to be global. This is because these do not rebind the list, but instead
find the list in global space and act on it.
"""

# TYPICAL CASE
x = 1

def fun1():
    x = 2

fun1() #x is still 1 because not global

print x

def fun2():
    x = 2
    return x

fun2() #the output of fun2 was not saved anywhere; the value of x is still 1

print x

x = fun1() #fun1 has no returned output, so x is now nothing

print x
 
x = fun2() #FINALLY, x is now 2 because it has been directly updated

print x

# GLOBAL CASE
x = 1

print x

def fun3():
    global x
    x = 2
    
fun3() #because x was declared global within the function, it can update it outside of itself

print x

# LIST METHOD CASE
y = [1,2,3,1]
a=0

def fun4():
    y[3] = 4 #changing a list entry is like calling a member function on the list; it will affect it
    a = 1 #directly rebinding the variable "a" creates a local binding separate from the global one;
          #unless "global" is used, of course
    
fun4()
print y #will have new value at index 3
print a #will be the same old value of "a"

def fun5():
    y.append(5) #like changing a list entry, this will find the list in global space and affect it

fun5()
print y

Python Fiddle

Python Cloud IDE