In Python what is a global statement?

Solution 1:

Every "variable" in python is limited to a certain scope. The scope of a python "file" is the module-scope. Consider the following:

#file test.py
myvariable = 5  # myvariable has module-level scope

def func():
    x = 3       # x has "local" or function level scope.

Objects with local scope die as soon as the function exits and can never be retrieved (unless you return them), but within a function, you can access variables in the module level scope (or any containing scope):

myvariable = 5
def func():
    print(myvariable)  # prints 5

def func2():
    x = 3
    def func3():
        print(x)       # will print 3 because it picks it up from `func2`'s scope

    func3()

However, you can't use assignment on that reference and expect that it will be propagated to an outer scope:

myvariable = 5
def func():
    myvariable = 6     # creates a new "local" variable.  
                       # Doesn't affect the global version
    print(myvariable)  # prints 6

func()
print(myvariable)      # prints 5

Now, we're finally to global. The global keyword is the way that you tell python that a particular variable in your function is defined at the global (module-level) scope.

myvariable = 5
def func():
    global myvariable
    myvariable = 6    # changes `myvariable` at the global scope
    print(myvariable) # prints 6

func()
print(myvariable)  # prints 6 now because we were able 
                   # to modify the reference in the function

In other words, you can change the value of myvariable in the module-scope from within func if you use the global keyword.

As an aside, scopes can be nested arbitrarily deep:

def func1():
    x = 3
    def func2():
        print("x=",x,"func2")
        y = 4
        def func3():
            nonlocal x  # try it with nonlocal commented out as well.  See the difference.
            print("x=",x,"func3")
            print("y=",y,"func3")
            z = 5
            print("z=",z,"func3")
            x = 10

        func3()

    func2()
    print("x=",x,"func1")

func1()

Now in this case, none of the variables are declared at the global scope, and in python2, there is no (easy/clean) way to change the value of x in the scope of func1 from within func3. That's why the nonlocal keyword was introduced in python3.x . nonlocal is an extension of global that allows you to modify a variable that you picked up from another scope in whatever scope it was pulled from.

Solution 2:

mgilson did a good job but I'd like to add some more.

list1 = [1]
list2 = [1]

def main():
    list1.append(3)
    #list1 = [9]
    list2 = [222]

    print list1, list2


print "before main():", list1, list2
>>> [1] [1]
main()
>>> [1,3] [222]
print list1, list2    
>>> [1, 3] [1]

Inside a function, Python assumes every variable as local variable unless you declare it as global, or you are accessing a global variable.

list1.append(2) 

was possible because you are accessing the 'list1' and lists are mutable.

list2 = [222]

was possible because you are initializing a local variable.

However if you uncomment #list1 = [9], you will get

UnboundLocalError: local variable 'list1' referenced before assignment

It means you are trying to initialize a new local variable 'list1' but it was already referenced before, and you are out of the scope to reassign it.

To enter the scope, declare 'list1' as global.

I strongly recommend you to read this even though there is a typo in the end.