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Python cannot read global variable in function even though I used global keyword

I want to access the variable number_of_messages in class A but I get "number_of_messages" is not defined error even though I used global keyword. Here is a code sample:
class A:
number_of_messages=0;
def inc(self):
global number_of_messages
number_of_messages+=1
print(A().inc())

Use the class attribute instead:
class A:
def ___init__(self):
self.number_of_messages=0
def inc(self):
self.number_of_messages+=1
a = A()
print(a.inc())
print(a.number_of_messages)
but you can also:
number_of_messages = 0
class A():
def inc(self):
global number_of_messages
number_of_messages+=1
a = A()
a.inc()
print(number_of_messages)
you just forgot to declare the variable in the global scope

That's not a global. That's a class attribute. Write
def inc(self):
A.number_of_messages += 1
You don't need the global statement.

Calling variables in another module python

I'm trying to access variables I created in one function inside another module in Python to plot a graph however, Python can't find them.
Heres some example code:
class1:
def method1
var1 = []
var2 = []
#Do something with var1 and var2
print var1
print var2
return var1,var2
sample = class1()
sample.method1
here is class 2
from class1 import *
class number2:
sample.method1()
This does as intended and prints var1 and var2 but I can't call var1 or var2 inside class number 2
FIXED EDIT:
Incase anyone else has this issue, I fixed it by importing this above class two
from Module1 import Class1,sample
And then inside class2
var1,var2 = smaple.method1()

The code you posted is full of syntax errors as Francesco sayed in his comment. Perhaps you could paste the correct one.
You don't import from a class but from a package or a module. Plus you don't "call" a variable unless it's a callable.
In your case you could just have :
file1.py :
class class1:
def __init__(self): # In your class's code, self is the current instance (= this for othe languages, it's always the first parameter.)
self.var = 0
def method1(self):
print(self.var)
sample = class1()
file2.py :
from file1 import class1, sample
class class2(class1):
def method2(self):
self.var += 1
print(self.var)
v = class2() # create an instance of class2 that inherits from class1
v.method1() # calls method inherited from class1 that prints the var instance variable
sample.method1() # same
print(v.var) # You can also access it from outside the class definition.
v.var += 2 # You also can modify it.
print(v.var)
v.method2() # Increment the variable, then print it.
v.method2() # same.
sample.method1() # Print var from sample.
#sample.method2() <--- not possible because sample is an instance of class1 and not of class2
Note that to have method1() in class2, class2 must inherit from class1. But you can still import variables from other packages/modules.
Note also that var is unique for each instance of the class.

Share variables between modules

I have this situation:
library_file1.py:
class A:
def foo(self):
print("bar")
def baz(self):
pass
project_file.py:
from library_file1 import A
class B(A):
def baz(self):
print(the_variable)
library_file2.py:
from project_file import B
the_variable = 7
b = B()
b.foo() # prints "bar"
b.baz() # I want this to print "7", but I don't know how
How do I allow code to be written in project_file.py that can access variables from library_file2.py? The only solution I can think of is this:
project_file.py:
from library_file1 import A
class B(A):
def baz(self, the_variable):
print(the_variable)
library_file2.py:
from project_file import B
the_variable = 7
b = B()
b.foo()
b.baz(the_variable)
but this feels awkward and doesn't scale to many variables like the_variable.

Quite easy: you need the variable to be in project_file.py instead of library_file2.py.
Change project_file to:
from library_file1 import A
the_variable = None
class B(A):
def baz(self):
print(the_variable)
And then in library_file2:
import project_file
from project_file import B
project_file.the_variable = 7
b = B()
b.foo() # prints "bar"
b.baz() # prints "7"
Using an argument is also a good solution (you should avoid globals as much as you can).
There is no such a thing as a truly global variable in python. Variables are always attached to a scope. We usually say that a variable is "global" when it's in the module scope.
The widest scope is the built-in scope, so it would be theoretically possible for you to add something as a built-in. However this is a really bad practice.
Also it doesn't complete fix the problem, because all files could access that new built-in, but they couldn't re-assign it without explicitly mentioning the scope.

Why can't Python access a subfunction from outside?

def A():
def B():
#do something
a = A()
a.B()
Why isn't the above (such simple code) possible in Python? Is there a 'pythonic' (legible, unsurprising, non-hacky) workaround that does not turn A() into a class?
Edit 1: The above was explained to me that B is local to A, thus it only exists as long as A is being evaluated. So if we make it global (and be sure not to have it overriden), then why doesn't this work?
def A():
def B():
#do something
return A()
a = A()
a.B()
It says it's returning a 'NoneType' object.

Because a function definition just creates a name in the local namespace. What you are doing is no different than:
def f():
a = 2
and then asking why you can't access a from outside the function. Names bound inside a function are local to the function.
In addition, your proposed code is strange. when you do a = f(), you are setting a to the return value of the function. Your function returns nothing, so you can't hope to access anything through the return value. It is possible to return the inner function directly:
def f():
def g():
return "blah"
return g
>>> func = f()
>>> func()
'blah'
And this can indeed be useful. But there isn't a generic way to access things inside the function from outside except by modifying global variables (which is usually a bad idea) or returning the values. That's how functions work: they take inputs and return outputs; they don't make their innards available to the outside word.

To call B with the syntax you want, use:
def A():
def B():
print("I'm B")
A.B = B
return A
a = A()
a.B()
A.B()

What is the Python equivalent of static variables inside a function?

What is the idiomatic Python equivalent of this C/C++ code?
void foo()
{
static int counter = 0;
counter++;
printf("counter is %d\n", counter);
}
specifically, how does one implement the static member at the function level, as opposed to the class level? And does placing the function into a class change anything?

A bit reversed, but this should work:
def foo():
foo.counter += 1
print "Counter is %d" % foo.counter
foo.counter = 0
If you want the counter initialization code at the top instead of the bottom, you can create a decorator:
def static_vars(**kwargs):
def decorate(func):
for k in kwargs:
setattr(func, k, kwargs[k])
return func
return decorate
Then use the code like this:
#static_vars(counter=0)
def foo():
foo.counter += 1
print "Counter is %d" % foo.counter
It'll still require you to use the foo. prefix, unfortunately.
(Credit: #ony)

You can add attributes to a function, and use it as a static variable.
def myfunc():
myfunc.counter += 1
print myfunc.counter
# attribute must be initialized
myfunc.counter = 0
Alternatively, if you don't want to setup the variable outside the function, you can use hasattr() to avoid an AttributeError exception:
def myfunc():
if not hasattr(myfunc, "counter"):
myfunc.counter = 0 # it doesn't exist yet, so initialize it
myfunc.counter += 1
Anyway static variables are rather rare, and you should find a better place for this variable, most likely inside a class.

One could also consider:
def foo():
try:
foo.counter += 1
except AttributeError:
foo.counter = 1
Reasoning:
much pythonic ("ask for forgiveness not permission")
use exception (thrown only once) instead of if branch (think StopIteration exception)

Many people have already suggested testing 'hasattr', but there's a simpler answer:
def func():
func.counter = getattr(func, 'counter', 0) + 1
No try/except, no testing hasattr, just getattr with a default.

Other answers have demonstrated the way you should do this. Here's a way you shouldn't:
>>> def foo(counter=[0]):
... counter[0] += 1
... print("Counter is %i." % counter[0]);
...
>>> foo()
Counter is 1.
>>> foo()
Counter is 2.
>>>
Default values are initialized only when the function is first evaluated, not each time it is executed, so you can use a list or any other mutable object to store static values.

Python doesn't have static variables but you can fake it by defining a callable class object and then using it as a function. Also see this answer.
class Foo(object):
# Class variable, shared by all instances of this class
counter = 0
def __call__(self):
Foo.counter += 1
print Foo.counter
# Create an object instance of class "Foo," called "foo"
foo = Foo()
# Make calls to the "__call__" method, via the object's name itself
foo() #prints 1
foo() #prints 2
foo() #prints 3
Note that __call__ makes an instance of a class (object) callable by its own name. That's why calling foo() above calls the class' __call__ method. From the documentation:
Instances of arbitrary classes can be made callable by defining a __call__() method in their class.

Here is a fully encapsulated version that doesn't require an external initialization call:
def fn():
fn.counter=vars(fn).setdefault('counter',-1)
fn.counter+=1
print (fn.counter)
In Python, functions are objects and we can simply add, or monkey patch, member variables to them via the special attribute __dict__. The built-in vars() returns the special attribute __dict__.
EDIT: Note, unlike the alternative try:except AttributeError answer, with this approach the variable will always be ready for the code logic following initialization. I think the try:except AttributeError alternative to the following will be less DRY and/or have awkward flow:
def Fibonacci(n):
if n<2: return n
Fibonacci.memo=vars(Fibonacci).setdefault('memo',{}) # use static variable to hold a results cache
return Fibonacci.memo.setdefault(n,Fibonacci(n-1)+Fibonacci(n-2)) # lookup result in cache, if not available then calculate and store it
EDIT2: I only recommend the above approach when the function will be called from multiple locations. If instead the function is only called in one place, it's better to use nonlocal:
def TheOnlyPlaceStaticFunctionIsCalled():
memo={}
def Fibonacci(n):
nonlocal memo # required in Python3. Python2 can see memo
if n<2: return n
return memo.setdefault(n,Fibonacci(n-1)+Fibonacci(n-2))
...
print (Fibonacci(200))
...

Use a generator function to generate an iterator.
def foo_gen():
n = 0
while True:
n+=1
yield n
Then use it like
foo = foo_gen().next
for i in range(0,10):
print foo()
If you want an upper limit:
def foo_gen(limit=100000):
n = 0
while n < limit:
n+=1
yield n
If the iterator terminates (like the example above), you can also loop over it directly, like
for i in foo_gen(20):
print i
Of course, in these simple cases it's better to use xrange :)
Here is the documentation on the yield statement.

Other solutions attach a counter attribute to the function, usually with convoluted logic to handle the initialization. This is inappropriate for new code.
In Python 3, the right way is to use a nonlocal statement:
counter = 0
def foo():
nonlocal counter
counter += 1
print(f'counter is {counter}')
See PEP 3104 for the specification of the nonlocal statement.
If the counter is intended to be private to the module, it should be named _counter instead.

Using an attribute of a function as static variable has some potential drawbacks:
Every time you want to access the variable, you have to write out the full name of the function.
Outside code can access the variable easily and mess with the value.
Idiomatic python for the second issue would probably be naming the variable with a leading underscore to signal that it is not meant to be accessed, while keeping it accessible after the fact.
Using closures
An alternative would be a pattern using lexical closures, which are supported with the nonlocal keyword in python 3.
def make_counter():
i = 0
def counter():
nonlocal i
i = i + 1
return i
return counter
counter = make_counter()
Sadly I know no way to encapsulate this solution into a decorator.
Using an internal state parameter
Another option might be an undocumented parameter serving as a mutable value container.
def counter(*, _i=[0]):
_i[0] += 1
return _i[0]
This works, because default arguments are evaluated when the function is defined, not when it is called.
Cleaner might be to have a container type instead of the list, e.g.
def counter(*, _i = Mutable(0)):
_i.value += 1
return _i.value
but I am not aware of a builtin type, that clearly communicates the purpose.

A little bit more readable, but more verbose (Zen of Python: explicit is better than implicit):
>>> def func(_static={'counter': 0}):
... _static['counter'] += 1
... print _static['counter']
...
>>> func()
1
>>> func()
2
>>>
See here for an explanation of how this works.

_counter = 0
def foo():
global _counter
_counter += 1
print 'counter is', _counter
Python customarily uses underscores to indicate private variables. The only reason in C to declare the static variable inside the function is to hide it outside the function, which is not really idiomatic Python.

def staticvariables(**variables):
def decorate(function):
for variable in variables:
setattr(function, variable, variables[variable])
return function
return decorate
#staticvariables(counter=0, bar=1)
def foo():
print(foo.counter)
print(foo.bar)
Much like vincent's code above, this would be used as a function decorator and static variables must be accessed with the function name as a prefix. The advantage of this code (although admittedly anyone might be smart enough to figure it out) is that you can have multiple static variables and initialise them in a more conventional manner.

After trying several approaches I ended up using an improved version of #warvariuc's answer:
import types
def func(_static=types.SimpleNamespace(counter=0)):
_static.counter += 1
print(_static.counter)

The idiomatic way is to use a class, which can have attributes. If you need instances to not be separate, use a singleton.
There are a number of ways you could fake or munge "static" variables into Python (one not mentioned so far is to have a mutable default argument), but this is not the Pythonic, idiomatic way to do it. Just use a class.
Or possibly a generator, if your usage pattern fits.

A static variable inside a Python method
class Count:
def foo(self):
try:
self.foo.__func__.counter += 1
except AttributeError:
self.foo.__func__.counter = 1
print self.foo.__func__.counter
m = Count()
m.foo() # 1
m.foo() # 2
m.foo() # 3

Another (not recommended!) twist on the callable object like https://stackoverflow.com/a/279598/916373, if you don't mind using a funky call signature, would be to do
class foo(object):
counter = 0;
#staticmethod
def __call__():
foo.counter += 1
print "counter is %i" % foo.counter
>>> foo()()
counter is 1
>>> foo()()
counter is 2

Soulution n +=1
def foo():
foo.__dict__.setdefault('count', 0)
foo.count += 1
return foo.count

A global declaration provides this functionality. In the example below (python 3.5 or greater to use the "f"), the counter variable is defined outside of the function. Defining it as global in the function signifies that the "global" version outside of the function should be made available to the function. So each time the function runs, it modifies the value outside the function, preserving it beyond the function.
counter = 0
def foo():
global counter
counter += 1
print("counter is {}".format(counter))
foo() #output: "counter is 1"
foo() #output: "counter is 2"
foo() #output: "counter is 3"

Using a decorator and a closure
The following decorator can be used create static function variables. It replaces the declared function with the return from itself. This implies that the decorated function must return a function.
def static_inner_self(func):
return func()
Then use the decorator on a function that returns another function with a captured variable:
#static_inner_self
def foo():
counter = 0
def foo():
nonlocal counter
counter += 1
print(f"counter is {counter}")
return foo
nonlocal is required, otherwise Python thinks that the counter variable is a local variable instead of a captured variable. Python behaves like that because of the variable assignment counter += 1. Any assignment in a function makes Python think that the variable is local.
If you are not assigning to the variable in the inner function, then you can ignore the nonlocal statement, for example, in this function I use to indent lines of a string, in which Python can infer that the variable is nonlocal:
#static_inner_self
def indent_lines():
import re
re_start_line = re.compile(r'^', flags=re.MULTILINE)
def indent_lines(text, indent=2):
return re_start_line.sub(" "*indent, text)
return indent_lines
P.S. There is a deleted answer that proposed the same. I don't know why the author deleted it.
https://stackoverflow.com/a/23366737/195417

Prompted by this question, may I present another alternative which might be a bit nicer to use and will look the same for both methods and functions:
#static_var2('seed',0)
def funccounter(statics, add=1):
statics.seed += add
return statics.seed
print funccounter() #1
print funccounter(add=2) #3
print funccounter() #4
class ACircle(object):
#static_var2('seed',0)
def counter(statics, self, add=1):
statics.seed += add
return statics.seed
c = ACircle()
print c.counter() #1
print c.counter(add=2) #3
print c.counter() #4
d = ACircle()
print d.counter() #5
print d.counter(add=2) #7
print d.counter() #8    
If you like the usage, here's the implementation:
class StaticMan(object):
def __init__(self):
self.__dict__['_d'] = {}
def __getattr__(self, name):
return self.__dict__['_d'][name]
def __getitem__(self, name):
return self.__dict__['_d'][name]
def __setattr__(self, name, val):
self.__dict__['_d'][name] = val
def __setitem__(self, name, val):
self.__dict__['_d'][name] = val
def static_var2(name, val):
def decorator(original):
if not hasattr(original, ':staticman'):
def wrapped(*args, **kwargs):
return original(getattr(wrapped, ':staticman'), *args, **kwargs)
setattr(wrapped, ':staticman', StaticMan())
f = wrapped
else:
f = original #already wrapped
getattr(f, ':staticman')[name] = val
return f
return decorator

Instead of creating a function having a static local variable, you can always create what is called a "function object" and give it a standard (non-static) member variable.
Since you gave an example written C++, I will first explain what a "function object" is in C++. A "function object" is simply any class with an overloaded operator(). Instances of the class will behave like functions. For example, you can write int x = square(5); even if square is an object (with overloaded operator()) and not technically not a "function." You can give a function-object any of the features that you could give a class object.
# C++ function object
class Foo_class {
private:
int counter;
public:
Foo_class() {
counter = 0;
}
void operator() () {
counter++;
printf("counter is %d\n", counter);
}
};
Foo_class foo;
In Python, we can also overload operator() except that the method is instead named __call__:
Here is a class definition:
class Foo_class:
def __init__(self): # __init__ is similair to a C++ class constructor
self.counter = 0
# self.counter is like a static member
# variable of a function named "foo"
def __call__(self): # overload operator()
self.counter += 1
print("counter is %d" % self.counter);
foo = Foo_class() # call the constructor
Here is an example of the class being used:
from foo import foo
for i in range(0, 5):
foo() # function call
The output printed to the console is:
counter is 1
counter is 2
counter is 3
counter is 4
counter is 5
If you want your function to take input arguments, you can add those to __call__ as well:
# FILE: foo.py - - - - - - - - - - - - - - - - - - - - - - - - -
class Foo_class:
def __init__(self):
self.counter = 0
def __call__(self, x, y, z): # overload operator()
self.counter += 1
print("counter is %d" % self.counter);
print("x, y, z, are %d, %d, %d" % (x, y, z));
foo = Foo_class() # call the constructor
# FILE: main.py - - - - - - - - - - - - - - - - - - - - - - - - - - - -
from foo import foo
for i in range(0, 5):
foo(7, 8, 9) # function call
# Console Output - - - - - - - - - - - - - - - - - - - - - - - - - -
counter is 1
x, y, z, are 7, 8, 9
counter is 2
x, y, z, are 7, 8, 9
counter is 3
x, y, z, are 7, 8, 9
counter is 4
x, y, z, are 7, 8, 9
counter is 5
x, y, z, are 7, 8, 9

This answer builds on #claudiu 's answer.
I found that my code was getting less clear when I always had
to prepend the function name, whenever I intend to access a static variable.
Namely, in my function code I would prefer to write:
print(statics.foo)
instead of
print(my_function_name.foo)
So, my solution is to :
add a statics attribute to the function
in the function scope, add a local variable statics as an alias to my_function.statics
from bunch import *
def static_vars(**kwargs):
def decorate(func):
statics = Bunch(**kwargs)
setattr(func, "statics", statics)
return func
return decorate
#static_vars(name = "Martin")
def my_function():
statics = my_function.statics
print("Hello, {0}".format(statics.name))
Remark
My method uses a class named Bunch, which is a dictionary that supports
attribute-style access, a la JavaScript (see the original article about it, around 2000)
It can be installed via pip install bunch
It can also be hand-written like so:
class Bunch(dict):
def __init__(self, **kw):
dict.__init__(self,kw)
self.__dict__ = self

I personally prefer the following to decorators. To each their own.
def staticize(name, factory):
"""Makes a pseudo-static variable in calling function.
If name `name` exists in calling function, return it.
Otherwise, saves return value of `factory()` in
name `name` of calling function and return it.
:param name: name to use to store static object
in calling function
:type name: String
:param factory: used to initialize name `name`
in calling function
:type factory: function
:rtype: `type(factory())`
>>> def steveholt(z):
... a = staticize('a', list)
... a.append(z)
>>> steveholt.a
Traceback (most recent call last):
...
AttributeError: 'function' object has no attribute 'a'
>>> steveholt(1)
>>> steveholt.a
[1]
>>> steveholt('a')
>>> steveholt.a
[1, 'a']
>>> steveholt.a = []
>>> steveholt.a
[]
>>> steveholt('zzz')
>>> steveholt.a
['zzz']
"""
from inspect import stack
# get scope enclosing calling function
calling_fn_scope = stack()[2][0]
# get calling function
calling_fn_name = stack()[1][3]
calling_fn = calling_fn_scope.f_locals[calling_fn_name]
if not hasattr(calling_fn, name):
setattr(calling_fn, name, factory())
return getattr(calling_fn, name)

Building on Daniel's answer (additions):
class Foo(object):
counter = 0
def __call__(self, inc_value=0):
Foo.counter += inc_value
return Foo.counter
foo = Foo()
def use_foo(x,y):
if(x==5):
foo(2)
elif(y==7):
foo(3)
if(foo() == 10):
print("yello")
use_foo(5,1)
use_foo(5,1)
use_foo(1,7)
use_foo(1,7)
use_foo(1,1)
The reason why I wanted to add this part is , static variables are used not only for incrementing by some value, but also check if the static var is equal to some value, as a real life example.
The static variable is still protected and used only within the scope of the function use_foo()
In this example, call to foo() functions exactly as(with respect to the corresponding c++ equivalent) :
stat_c +=9; // in c++
foo(9) #python equiv
if(stat_c==10){ //do something} // c++
if(foo() == 10): # python equiv
#add code here # python equiv
Output :
yello
yello
if class Foo is defined restrictively as a singleton class, that would be ideal. This would make it more pythonic.

I write a simple function to use static variables:
def Static():
### get the func object by which Static() is called.
from inspect import currentframe, getframeinfo
caller = currentframe().f_back
func_name = getframeinfo(caller)[2]
# print(func_name)
caller = caller.f_back
func = caller.f_locals.get(
func_name, caller.f_globals.get(
func_name
)
)
class StaticVars:
def has(self, varName):
return hasattr(self, varName)
def declare(self, varName, value):
if not self.has(varName):
setattr(self, varName, value)
if hasattr(func, "staticVars"):
return func.staticVars
else:
# add an attribute to func
func.staticVars = StaticVars()
return func.staticVars
How to use:
def myfunc(arg):
if Static().has('test1'):
Static().test += 1
else:
Static().test = 1
print(Static().test)
# declare() only takes effect in the first time for each static variable.
Static().declare('test2', 1)
print(Static().test2)
Static().test2 += 1

Miguel Angelo's self-redefinition solution is even possible without any decorator:
def fun(increment=1):
global fun
counter = 0
def fun(increment=1):
nonlocal counter
counter += increment
print(counter)
fun(increment)
fun() #=> 1
fun() #=> 2
fun(10) #=> 12
The second line has to be adapted to get a limited scope:
def outerfun():
def innerfun(increment=1):
nonlocal innerfun
counter = 0
def innerfun(increment=1):
nonlocal counter
counter += increment
print(counter)
innerfun(increment)
innerfun() #=> 1
innerfun() #=> 2
innerfun(10) #=> 12
outerfun()
The plus of the decorator is that you don't have to pay extra attention to the scope of your construction.

Sure this is an old question but I think I might provide some update.
It seems that the performance argument is obsolete.
The same test suite appears to give similar results for siInt_try and isInt_re2.
Of course results vary, but this is one session on my computer with python 3.4.4 on kernel 4.3.01 with Xeon W3550.
I have run it several times and the results seem to be similar.
I moved the global regex into function static, but the performance difference is negligible.
isInt_try: 0.3690
isInt_str: 0.3981
isInt_re: 0.5870
isInt_re2: 0.3632
With performance issue out of the way, it seems that try/catch would produce the most future- and cornercase- proof code so maybe just wrap it in function