Global variables are accessible inside and outside of functions. Local variables are only accessible inside the function. In the example below, the function can access both the global and the local variable. However, trying to access the local variable outside the function produces an error.
global_var = 'foo' def ex1(): local_var = 'bar' print global_var print local_var ex1() print global_var print local_var # this gives an error
foo bar foo Traceback (most recent call last): File "nested_scope.py", line 12, in
print local_var # this gives an error NameError: name 'local_var' is not defined
*Setting* a global variable from
within a function is not as simple. If I set a variable in a function with
the same name as a global variable, I am actually creating a new local
variable. In the example below,
even after the function is called.
var = 'foo' def ex2(): var = 'bar' print 'inside the function var is ', var ex2() print 'outside the function var is ', var
inside the function var is bar outside the function var is foo
To set the global variable
inside a function, I
need to use the
global statement. This declares the inner
variable to have module scope. Now
after the function is called.
var = 'foo' def ex3(): global var var = 'bar' print 'inside the function var is ', var ex3() print 'outside the function var is ', var
inside the function var is bar outside the function var is bar
Scoping for nested functions works similarly.
In the example below, the inner function can access both
var_inner. However, the outer function cannot access
var_inner. Side note: the inner function is considered a
closure if it makes
reference to a non-global outside variable.
def ex4(): var_outer = 'foo' def inner(): var_inner = 'bar' print var_outer print var_inner inner() print var_outer print var_inner # this gives an error ex4()
foo bar foo Traceback (most recent call last): File "nested_scope.py", line 53, in
ex3() File "nested_scope.py", line 51, in ex3 print var_inner # this gives an error NameError: global name 'var_inner' is not defined
Like Example 2, setting a variable in the
inner function creates a new local variable instead of modifying the
outer variable. In the example below,
var in the outer
function does not get changed to
def ex5(): var = 'foo' def inner(): var = 'bar' print 'inside inner, var is ', var inner() print 'inside outer function, var is ', var ex5()
inside inner, var is bar inside outer function, var is foo
However, using the
keyword won't work in this case.
global cause a variable to
have module scope, but I want my variable to have the scope of the outer
function. Per the
Python 3000 Status Update, Python 3000 will have a
keyword to solve this problem. See
PEP 3104 for more information about
nonlocal and nested
scopes. In the example below,
var is still not changed
'bar' in the outer function.
def ex6(): var = 'foo' def inner(): global var var = 'bar' print 'inside inner, var is ', var inner() print 'inside outer function, var is ', var ex6()
inside inner, var is bar inside outer function, var is foo
A workaround is to create an empty class to use as an additional
namespace. Now the variable in the outer function can be set to
class Namespace: pass def ex7(): ns = Namespace() ns.var = 'foo' def inner(): ns.var = 'bar' print 'inside inner, ns.var is ', ns.var inner() print 'inside outer function, ns.var is ', ns.var ex7()
inside inner, ns.var is bar inside outer function, ns.var is bar
Update 2010-03-01: According to Alexander's comment below, this is not a good way to do things.
I learned about this method from Nihiliad's comment on my recursion example. To me, this seems like a more elegant alternative to the solution in Example 7.
def ex8(): ex8.var = 'foo' def inner(): ex8.var = 'bar' print 'inside inner, ex8.var is ', ex8.var inner() print 'inside outer function, ex8.var is ', ex8.var ex8()
inside inner, ex8.var is bar inside outer function, ex8.var is bar
Core Python Programming, Second Edition, Ch 11
Many thanks for this post! I could not find the solution to setting global variables anywhere in all my python documentation.
Another workaround is to wrap the variable in a mutable:
... var = ['foo']
... def inner():
... var = 'bar'
... print 'inside inner, var is ', var
... print 'inside outer function, var is ', var
inside inner, var is ['bar']
inside outer function, var is ['bar']
Can anybody tell me what's wrong in the below code:
argref_list = 
for entry in arg_l: # processing #nnnn($Fmm)
the output is:
note that 'NONE' is printed (strange) and secondly throws the error that NONETYPE has no append function.
I am fade up with PYTHON!
Nice post, told me exactly what I needed to know!
Thanks for the clear and concise examples. The python tutorial left me confused about variable scope. This cleared it up.
This confused me for a long time, but your explanation is understandable. Thanks very much!
It's clear explanation about variables scope.that python official documentation aborted these. thanks,thanks,for your sharing.
Thanks a lot. I did wrote this code in my programm a wonder why he didn't work (I have hear that python support closure. my code was :
def main(): count = 0 def counting(): for i in range(10): count += 1 counting() return count
which throw an exception. if first change it to :
def main(): count = 0 def counting(count): for i in range(10): count += 1 counting() return count
but thanks to you, i update it to :
def main(): main.count = 0 def counting(): for i in range(10): main.count += 1 counting() return main.count
Nice Post and keep posting stuff like this.
Sorry but your editor didn't keep the formatting
roger, Thanks for the comment. I fixed the formatting for your code. To format code blocks, put 4 spaces before each line and a blank line before and after the block.
firstly great article, really really useful.
I was wondering... in a code I am writing, I assign a global variable a value from a function, but I then want to delete that global variable from another function in which identically named local variables are defined. For this reason, "del variable" deletes the local not the global. How would I go about deleting the global variable?
Thanks, this was helpful.
I do have a question about scope within a class. I am using Python 2.5
def class foo(object): def test(self): test.var = 1 print test.var f = foo() f.test()
Result is "global name test is not defined"
I can get around this by using a "namespace" or mutable wrapper as described but I was just curious why this didn't work
I think you get this error message because
test is a method of
foo and not a global. However, the following code gives an error as well:
class foo(object): def test(self): self.test.var = 1 print self.test f = foo() f.test()
AttributeError: 'instancemethod' object has no attribute 'var'
A variation of Example 7:
def ex7b(): class ns: pass ns.var = 'foo' def inner(): ns.var = 'bar' print 'inside inner, ns.var is ', ns.var inner() print 'inside outer function, ns.var is ', ns.var ex7b()
akaihola, thanks. it does seem to make more sense to put the namespace within the outer function. btw, nice blog-- i see we have some similar interests (twisted, scribus, emacs, django)
I think the snippet you posted is trying to access 'test', a method of 'foo', before its declaration is complete (within the same block), thus the error. Did you mean
self.var = 3
foo.var = 3
Can someone help me understand what is wrong with this example?
class T: A = range(2) B = range(4) s = sum(i*j for i in A for j in B)
It produces the exception:
<type 'exceptions.NameError'>: global name 'j' is not defined
The exception above is especially confusing when the following similar example (I just replaced the generator by an explicit array) works:
class T: A = range(2) B = range(4) s = sum([(i*j) for i in A for j in B])
(BTW, the class scope declarations are intentional).
Example 8 is not a very good way to do things. ex8's properties are effectively global variables. That means that something like http://www.paulgraham.com/accgen.html won't work this way. (inner functions defined this way are not closures)
Leo: sorry, I've run in to this also, but don't know the answer.
Alexander: thank you for alerting me to this. I added a quick note to example 8 above.
As for the question in #17, see: http://docs.python.org/reference/executionmodel.html#naming-and-binding
"A scope defines the visibility of a name within a block. If a local variable is defined in a block, its scope includes that block. If the definition occurs in a function block, the scope extends to any blocks contained within the defining one, unless a contained block introduces a different binding for the name. The scope of names defined in a class block is limited to the class block; it does not extend to the code blocks of methods – this includes generator expressions since they are implemented using a function scope. This means that the following will fail:"
class A: a = 42 b = list(a + i for i in range(10))
#18: ex8 vars are indeed effectively global, but you can make it work if you use attributes of the inner function. Here is Paul Graham's accgen:
def foo(n): def inner(i): inner.n += i return inner.n inner.n = n return inner >>> f = foo(10) >>> print f(1), f(2), f(3) 11 13 16 >>> f = foo(20) >>> print f(1), f(2), f(3) 21 23 26
Thank you! This explained my closure woes.
Nice post.. gave me very useful info, not quite clear from python error messages.. But I find it incomplete as it's missing description of how scope works in classes. Thanks none the less..