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Why can I change class attributes for an instance without changing the class value in Python?

In Java, there's no way to change a static variable (what I thought was more or less the same as a Python class attribute) for just an instance.
In Python, however, I tried the following code in an interactive session:
>>> class TestClass:
... x = 0
...
>>> a = TestClass()
>>> a.x += 1
>>> a.x
1
>>> TestClass.x
0
>>> TestClass.x = 2
>>> a.x
1
>>> TestClass.x
2
Further, I checked the IDs of both a.x and TestClass.x and determined that they weren't equal, so I guessed that there's an instance-level x and a class-level x. Can anyone explain why this is?
I figure the best way to modify a class attribute through an instance through a method that modifies TestClass.x, but I also can't figure out quite why this behavior exists as is.

Basically, when looking up a variable, python will first look at the instance level to see if the variable exists there, and then at the class level. When you do
a.x += 1
What's happening is that it's setting a.x to the class level x variable plus 1.
And later on when you do a.x, it's looking to see if x is available at the instance level (and it is), and so it's using that value.
But if you did
b = TestClass() and then b.x without setting x on the instance level, then python would look up x on the instance level (does not exist), and then on the class level (does exist), and use that value

Python 3 REGEX .finditer: Usefulness of hex address in "callable_iterator object at 0x..."?

The following Python 3.5 prints add_date_iter == <callable_iterator object at 0xb78e218c>.
import re
date_added_attrs = re.compile(r'( +ADD_DATE=)("(\d+)")')
add_date_iter = date_added_attrs.finditer(test_string)
print("add_date_iter ==", add_date_iter)
So far, so good. BUT of what use is 0xb78e218c? It appears to be a hexadecimal memory or object address. Whatever it is, why / how if at all might a Python 3 program make use of it?
EDIT: My question is NOT about REGEX. The REGEX works fine. My question is, what's the purpose / benefit of the hexadecimal value returned by the .finditer operation?

what's the purpose / benefit of the hexadecimal value returned by the .finditer operation?
When you write print("add_date_iter ==", add_date_iter), you are simply converting the iterator object to a string and printing it. That is, <callable_iterator object at 0xb78e218c> is the return value of the object's internal __str__ method.*
The hexadecimal address is telling you where in memory the iterator is stored. It is the same value you would get if you ran hex(id(add_date_iter)). It is generally only useful when you're trying to figure out the internals of how Python is managing memory during a certain process, or if you want to check whether two variables are holding a reference to the same object. When comparing objects, you can think of id(a) == id(b) as a long way of writing a is b.
More detail...
For instance, if you had this code:
class A:
def __init__(self):
self.val = 0
a = A()
b = A()
print(id(a), id(b))
print(a is b)
b = a
print(id(a), id(b))
print(a is b)
You would get an output like this:
140665126149392 140665230088528
False
140665126149392 140665126149392
True
In the first case, even though the instance variables val have the same value, the objects themselves are different. After writing b = a, though, both variables now refer to the same object.
One place you can get tripped up with this is with integers, which is why you should always use == instead of is unless you really know what you're doing (or you're checking is None, since None is actually a singleton object):
a = 5
b = 5
a is b # True
a = 300
b = 300
a is b # False
One final point: since Python has first class functions (i.e. functions are treated as objects), everything discussed above works with functions, too. Just reference the function without parentheses, like id(print).
* Note: If you write add_date_iter without print in the interactive shell, it will call the __repr__ method instead

Assign results of function call in one line in python

How can I assign the results of a function call to multiple variables when the results are stored by name (not index-able), in python.
For example (tested in Python 3),
import random
# foo, as defined somewhere else where we can't or don't want to change it
def foo():
t = random.randint(1,100)
# put in a dummy class instead of just "return t,t+1"
# because otherwise we could subscript or just A,B = foo()
class Cat(object):
x = t
y = t + 1
return Cat()
# METHOD 1
# clearly wrong; A should be 1 more than B; they point to fields of different objects
A,B = foo().x, foo().y
print(A,B)
# METHOD 2
# correct, but requires two lines and an implicit variable
t = foo()
A,B = t.x, t.y
del t # don't really want t lying around
print(A,B)
# METHOD 3
# correct and one line, but an obfuscated mess
A,B = [ (t.x,t.y) for t in (foo(),) ][0]
print(A,B)
print(t) # this will raise an exception, but unless you know your python cold it might not be obvious before running
# METHOD 4
# Conforms to the suggestions in the links below without modifying the initial function foo or class Cat.
# But while all subsequent calls are pretty, but we have to use an otherwise meaningless shell function
def get_foo():
t = foo()
return t.x, t.y
A,B = get_foo()
What we don't want to do
If the results were indexable ( Cat extended tuple/list, we had used a namedtuple, etc.), we could simply write A,B = foo() as indicated in the comment above the Cat class. That's what's recommended here , for example.
Let's assume we have a good reason not to allow that. Maybe we like the clarity of assigning from the variable names (if they're more meaningful than x and y) or maybe the object is not primarily a container. Maybe the fields are properties, so access actually involves a method call. We don't have to assume any of those to answer this question though; the Cat class can be taken at face value.
This question already deals with how to design functions/classes the best way possible; if the function's expected return value are already well defined and does not involve tuple-like access, what is the best way to accept multiple values when returning?

I would strongly recommend either using multiple statements, or just keeping the result object without unpacking its attributes. That said, you can use operator.attrgetter for this:
from operator import attrgetter
a, b, c = attrgetter('a', 'b', 'c')(foo())

Get array name in function it is passed to [duplicate]

I already read How to get a function name as a string?.
How can I do the same for a variable? As opposed to functions, Python variables do not have the __name__ attribute.
In other words, if I have a variable such as:
foo = dict()
foo['bar'] = 2
I am looking for a function/attribute, e.g. retrieve_name() in order to create a DataFrame in Pandas from this list, where the column names are given by the names of the actual dictionaries:
# List of dictionaries for my DataFrame
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [retrieve_name(d) for d in list_of_dicts]

With Python 3.8 one can simply use f-string debugging feature:
>>> foo = dict()
>>> f'{foo=}'.split('=')[0]
'foo'
One drawback of this method is that in order to get 'foo' printed you have to add f'{foo=}' yourself. In other words, you already have to know the name of the variable. In other words, the above code snippet is exactly the same as just
>>> 'foo'

Even if variable values don't point back to the name, you have access to the list of every assigned variable and its value, so I'm astounded that only one person suggested looping through there to look for your var name.
Someone mentioned on that answer that you might have to walk the stack and check everyone's locals and globals to find foo, but if foo is assigned in the scope where you're calling this retrieve_name function, you can use inspect's current frame to get you all of those local variables.
My explanation might be a little bit too wordy (maybe I should've used a "foo" less words), but here's how it would look in code (Note that if there is more than one variable assigned to the same value, you will get both of those variable names):
import inspect
x, y, z = 1, 2, 3
def retrieve_name(var):
callers_local_vars = inspect.currentframe().f_back.f_locals.items()
return [var_name for var_name, var_val in callers_local_vars if var_val is var]
print(retrieve_name(y))
If you're calling this function from another function, something like:
def foo(bar):
return retrieve_name(bar)
foo(baz)
And you want the baz instead of bar, you'll just need to go back a scope further. This can be done by adding an extra .f_back in the caller_local_vars initialization.
See an example here: ideone

The only objects in Python that have canonical names are modules, functions, and classes, and of course there is no guarantee that this canonical name has any meaning in any namespace after the function or class has been defined or the module imported. These names can also be modified after the objects are created so they may not always be particularly trustworthy.
What you want to do is not possible without recursively walking the tree of named objects; a name is a one-way reference to an object. A common or garden-variety Python object contains no references to its names. Imagine if every integer, every dict, every list, every Boolean needed to maintain a list of strings that represented names that referred to it! It would be an implementation nightmare, with little benefit to the programmer.

TL;DR
Use the Wrapper helper from python-varname:
from varname.helpers import Wrapper
foo = Wrapper(dict())
# foo.name == 'foo'
# foo.value == {}
foo.value['bar'] = 2
For list comprehension part, you can do:
n_jobs = Wrapper(<original_value>)
users = Wrapper(<original_value>)
queues = Wrapper(<original_value>)
priorities = Wrapper(<original_value>)
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [d.name for d in list_of_dicts]
# ['n_jobs', 'users', 'queues', 'priorities']
# REMEMBER that you have to access the <original_value> by d.value
I am the author of the python-varname package. Please let me know if you have any questions or you can submit issues on Github.
The long answer
Is it even possible?
Yes and No.
We are retrieving the variable names at runtime, so we need a function to be called to enable us to access the previous frames to retrieve the variable names. That's why we need a Wrapper there. In that function, at runtime, we are parsing the source code/AST nodes in the previous frames to get the exact variable name.
However, the source code/AST nodes in the previous frames are not always available, or they could be modified by other environments (e.g: pytest's assert statement). One simple example is that the codes run via exec(). Even though we are still able to retrieve some information from the bytecode, it needs too much effort and it is also error-prone.
How to do it?
First of all, we need to identify which frame the variable is given. It's not always simply the direct previous frame. For example, we may have another wrapper for the function:
from varname import varname
def func():
return varname()
def wrapped():
return func()
x = wrapped()
In the above example, we have to skip the frame inside wrapped to get to the right frame x = wrapped() so that we are able to locate x. The arguments frame and ignore of varname allow us to skip some of these intermediate frames. See more details in the README file and the API docs of the package.
Then we need to parse the AST node to locate where the variable is assigned value (function call) to. It's not always just a simple assignment. Sometimes there could be complex AST nodes, for example, x = [wrapped()]. We need to identify the correct assignment by traversing the AST tree.
How reliable is it?
Once we identify the assignment node, it is reliable.
varname is all depending on executing package to look for the node. The node executing detects is ensured to be the correct one (see also this).
It partially works with environments where other AST magics apply, including pytest, ipython, macropy, birdseye, reticulate with R, etc. Neither executing nor varname is 100% working with those environments.
Do we need a package to do it?
Well, yes and no, again.
If your scenario is simple, the code provided by #juan Isaza or #scohe001 probably is enough for you to work with the case where a variable is defined at the direct previous frame and the AST node is a simple assignment. You just need to go one frame back and retrieve the information there.
However, if the scenario becomes complicated, or we need to adopt different application scenarios, you probably need a package like python-varname, to handle them. These scenarios may include to:
present more friendly messages when the source code is not available or AST nodes are not accessible
skip intermediate frames (allows the function to be wrapped or called in other intermediate frames)
automatically ignores calls from built-in functions or libraries. For example: x = str(func())
retrieve multiple variable names on the left-hand side of the assignment
etc.
How about the f-string?
Like the answer provided by #Aivar Paalberg. It's definitely fast and reliable. However, it's not at runtime, meaning that you have to know it's foo before you print the name out. But with varname, you don't have to know that variable is coming:
from varname import varname
def func():
return varname()
# In external uses
x = func() # 'x'
y = func() # 'y'
Finally
python-varname is not only able to detect the variable name from an assignment, but also:
Retrieve variable names directly, using nameof
Detect next immediate attribute name, using will
Fetch argument names/sources passed to a function using argname
Read more from its documentation.
However, the final word I want to say is that, try to avoid using it whenever you can.
Because you can't make sure that the client code will run in an environment where the source node is available or AST node is accessible. And of course, it costs resources to parse the source code, identify the environment, retrieve the AST nodes and evaluate them when needed.

On python3, this function will get the outer most name in the stack:
import inspect
def retrieve_name(var):
"""
Gets the name of var. Does it from the out most frame inner-wards.
:param var: variable to get name from.
:return: string
"""
for fi in reversed(inspect.stack()):
names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
if len(names) > 0:
return names[0]
It is useful anywhere on the code. Traverses the reversed stack looking for the first match.

I don't believe this is possible. Consider the following example:
>>> a = []
>>> b = a
>>> id(a)
140031712435664
>>> id(b)
140031712435664
The a and b point to the same object, but the object can't know what variables point to it.

def name(**variables):
return [x for x in variables]
It's used like this:
name(variable=variable)

>> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v == my_var][0]
>> my_var_name
'my_var'
In case you get an error if myvar points to another variable, try this (suggested by #mherzog)-
>> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v is my_var][0]
>> my_var_name
'my_var'
locals() - Return a dictionary containing the current scope's local variables.
by iterating through this dictionary we can check the key which has a value equal to the defined variable, just extracting the key will give us the text of variable in string format.
from (after a bit changes)
https://www.tutorialspoint.com/How-to-get-a-variable-name-as-a-string-in-Python

I wrote the package sorcery to do this kind of magic robustly. You can write:
from sorcery import dict_of
columns = dict_of(n_jobs, users, queues, priorities)
and pass that to the dataframe constructor. It's equivalent to:
columns = dict(n_jobs=n_jobs, users=users, queues=queues, priorities=priorities)

Here's one approach. I wouldn't recommend this for anything important, because it'll be quite brittle. But it can be done.
Create a function that uses the inspect module to find the source code that called it. Then you can parse the source code to identify the variable names that you want to retrieve. For example, here's a function called autodict that takes a list of variables and returns a dictionary mapping variable names to their values. E.g.:
x = 'foo'
y = 'bar'
d = autodict(x, y)
print d
Would give:
{'x': 'foo', 'y': 'bar'}
Inspecting the source code itself is better than searching through the locals() or globals() because the latter approach doesn't tell you which of the variables are the ones you want.
At any rate, here's the code:
def autodict(*args):
get_rid_of = ['autodict(', ',', ')', '\n']
calling_code = inspect.getouterframes(inspect.currentframe())[1][4][0]
calling_code = calling_code[calling_code.index('autodict'):]
for garbage in get_rid_of:
calling_code = calling_code.replace(garbage, '')
var_names, var_values = calling_code.split(), args
dyn_dict = {var_name: var_value for var_name, var_value in
zip(var_names, var_values)}
return dyn_dict
The action happens in the line with inspect.getouterframes, which returns the string within the code that called autodict.
The obvious downside to this sort of magic is that it makes assumptions about how the source code is structured. And of course, it won't work at all if it's run inside the interpreter.

This function will print variable name with its value:
import inspect
def print_this(var):
callers_local_vars = inspect.currentframe().f_back.f_locals.items()
print(str([k for k, v in callers_local_vars if v is var][0])+': '+str(var))
***Input & Function call:***
my_var = 10
print_this(my_var)
***Output**:*
my_var: 10

>>> locals()['foo']
{}
>>> globals()['foo']
{}
If you wanted to write your own function, it could be done such that you could check for a variable defined in locals then check globals. If nothing is found you could compare on id() to see if the variable points to the same location in memory.
If your variable is in a class, you could use className.dict.keys() or vars(self) to see if your variable has been defined.

I have a method, and while not the most efficient...it works! (and it doesn't involve any fancy modules).
Basically it compares your Variable's ID to globals() Variables' IDs, then returns the match's name.
def getVariableName(variable, globalVariables=globals().copy()):
""" Get Variable Name as String by comparing its ID to globals() Variables' IDs
args:
variable(var): Variable to find name for (Obviously this variable has to exist)
kwargs:
globalVariables(dict): Copy of the globals() dict (Adding to Kwargs allows this function to work properly when imported from another .py)
"""
for globalVariable in globalVariables:
if id(variable) == id(globalVariables[globalVariable]): # If our Variable's ID matches this Global Variable's ID...
return globalVariable # Return its name from the Globals() dict

In Python, the def and class keywords will bind a specific name to the object they define (function or class). Similarly, modules are given a name by virtue of being called something specific in the filesystem. In all three cases, there's an obvious way to assign a "canonical" name to the object in question.
However, for other kinds of objects, such a canonical name may simply not exist. For example, consider the elements of a list. The elements in the list are not individually named, and it is entirely possible that the only way to refer to them in a program is by using list indices on the containing list. If such a list of objects was passed into your function, you could not possibly assign meaningful identifiers to the values.
Python doesn't save the name on the left hand side of an assignment into the assigned object because:
It would require figuring out which name was "canonical" among multiple conflicting objects,
It would make no sense for objects which are never assigned to an explicit variable name,
It would be extremely inefficient,
Literally no other language in existence does that.
So, for example, functions defined using lambda will always have the "name" <lambda>, rather than a specific function name.
The best approach would be simply to ask the caller to pass in an (optional) list of names. If typing the '...','...' is too cumbersome, you could accept e.g. a single string containing a comma-separated list of names (like namedtuple does).

I think it's so difficult to do this in Python because of the simple fact that you never will not know the name of the variable you're using. So, in his example, you could do:
Instead of:
list_of_dicts = [n_jobs, users, queues, priorities]
dict_of_dicts = {"n_jobs" : n_jobs, "users" : users, "queues" : queues, "priorities" : priorities}

Many of the answers return just one variable name. But that won't work well if more than one variable have the same value. Here's a variation of Amr Sharaki's answer which returns multiple results if more variables have the same value.
def getVariableNames(variable):
results = []
globalVariables=globals().copy()
for globalVariable in globalVariables:
if id(variable) == id(globalVariables[globalVariable]):
results.append(globalVariable)
return results
a = 1
b = 1
getVariableNames(a)
# ['a', 'b']

just another way to do this based on the content of input variable:
(it returns the name of the first variable that matches to the input variable, otherwise None. One can modify it to get all variable names which are having the same content as input variable)
def retrieve_name(x, Vars=vars()):
for k in Vars:
if isinstance(x, type(Vars[k])):
if x is Vars[k]:
return k
return None

If the goal is to help you keep track of your variables, you can write a simple function that labels the variable and returns its value and type. For example, suppose i_f=3.01 and you round it to an integer called i_n to use in a code, and then need a string i_s that will go into a report.
def whatis(string, x):
print(string+' value=',repr(x),type(x))
return string+' value='+repr(x)+repr(type(x))
i_f=3.01
i_n=int(i_f)
i_s=str(i_n)
i_l=[i_f, i_n, i_s]
i_u=(i_f, i_n, i_s)
## make report that identifies all types
report='\n'+20*'#'+'\nThis is the report:\n'
report+= whatis('i_f ',i_f)+'\n'
report+=whatis('i_n ',i_n)+'\n'
report+=whatis('i_s ',i_s)+'\n'
report+=whatis('i_l ',i_l)+'\n'
report+=whatis('i_u ',i_u)+'\n'
print(report)
This prints to the window at each call for debugging purposes and also yields a string for the written report. The only downside is that you have to type the variable twice each time you call the function.
I am a Python newbie and found this very useful way to log my efforts as I program and try to cope with all the objects in Python. One flaw is that whatis() fails if it calls a function described outside the procedure where it is used. For example, int(i_f) was a valid function call only because the int function is known to Python. You could call whatis() using int(i_f**2), but if for some strange reason you choose to define a function called int_squared it must be declared inside the procedure where whatis() is used.

Maybe this could be useful:
def Retriever(bar):
return (list(globals().keys()))[list(map(lambda x: id(x), list(globals().values()))).index(id(bar))]
The function goes through the list of IDs of values from the global scope (the namespace could be edited), finds the index of the wanted/required var or function based on its ID, and then returns the name from the list of global names based on the acquired index.

Whenever I have to do it, mostly while communicating json schema and constants with the frontend I define a class as follows
class Param:
def __init__(self, name, value):
self.name = name
self.value = value
Then define the variable with name and value.
frame_folder_count = Param({'name':'frame_folder_count', 'value':10})
Now you can access the name and value using the object.
>>> frame_folder_count.name
'frame_folder_count'

>>> def varname(v, scope=None):
d = globals() if not scope else vars(scope); return [k for k in d if d[k] == v]
...
>>> d1 = {'a': 'ape'}; d2 = {'b': 'bear'}; d3 = {'c': 'cat'}
>>> ld = [d1, d2, d3]
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3']]
>>> d5 = d3
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]
>>> def varname(v, scope=None):
d = globals() if not scope else vars(scope); return [k for k in d if d[k] is v]
...
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]
As you see and is noted here, there can be multiple variables with the same value or even address, so using a wrapper to keep the names with the data is best.

Following method will not return the name of variable but using this method you can create data frame easily if variable is available in global scope.
class CustomDict(dict):
def __add__(self, other):
return CustomDict({**self, **other})
class GlobalBase(type):
def __getattr__(cls, key):
return CustomDict({key: globals()[key]})
def __getitem__(cls, keys):
return CustomDict({key: globals()[key] for key in keys})
class G(metaclass=GlobalBase):
pass
x, y, z = 0, 1, 2
print('method 1:', G['x', 'y', 'z']) # Outcome: method 1: {'x': 0, 'y': 1, 'z': 2}
print('method 2:', G.x + G.y + G.z) # Outcome: method 2: {'x': 0, 'y': 1, 'z': 2}
A = [0, 1]
B = [1, 2]
pd.DataFrame(G.A + G.B) # It will return a data frame with A and B columns

Some of the previous cases would fail if there are two variables with the same value. So it is convenient to alert it:
Defining function:
# Variable to string of variable name
def var_name(variable,i=0):
results = []
for name in globals():
if eval(name) == variable:
results.append(name)
if len(results) > 1:
print('Warning:' )
print(' var_name() has found',len(results), 'possible outcomes.')
print(' Please choose the suitable parameter "i". Where "i" is the index')
print(' that matches your choice from the list below.')
print(' ',results) ; print('')
return results[i]
Use:
var_1 = 10
var_name(var_1) # Output will be "var_1"
If you have 2 variables with the same value like var_1 = 8 and var_2 = 8, then a warning will appear.
var_1 = 8
var_2 = 8
var_name(var_2) # Output will be "var_1" too but Warning will appear

You can get your variable as kwargs and return it as string:
var=2
def getVarName(**kwargs):
return list(kwargs.keys())[0]
print (getVarName(var = var))
Note: variable name must be equal to itself.

I try to get name from inspect locals, but it cann't process var likes a[1], b.val.
After it, I got a new idea --- get var name from the code, and I try it succ!
code like below:
#direct get from called function code
def retrieve_name_ex(var):
stacks = inspect.stack()
try:
func = stacks[0].function
code = stacks[1].code_context[0]
s = code.index(func)
s = code.index("(", s + len(func)) + 1
e = code.index(")", s)
return code[s:e].strip()
except:
return ""

You can try the following to retrieve the name of a function you defined (does not work for built-in functions though):
import re
def retrieve_name(func):
return re.match("<function\s+(\w+)\s+at.*", str(func)).group(1)
def foo(x):
return x**2
print(retrieve_name(foo))
# foo

When finding the name of a variable from its value,
you may have several variables equal to the same value,
for example var1 = 'hello' and var2 = 'hello'.
My solution:
def find_var_name(val):
dict_list = []
global_dict = dict(globals())
for k, v in global_dict.items():
dict_list.append([k, v])
return [item[0] for item in dict_list if item[1] == val]
var1 = 'hello'
var2 = 'hello'
find_var_name('hello')
Outputs
['var1', 'var2']

Compressed version of iDilip's answer:
import inspect
def varname(x):
return [k for k,v in inspect.currentframe().f_back.f_locals.items() if v is x][0]
hi = 123
print(varname(hi))

It's totally possible to get the name of an instance variable, so long as it is the property of a class.
I got this from Effective Python by Brett Slatkin. Hope it helps someone:
The class must implement the get, set, and set_name dunder methods, which are part of the "Descriptor Protocol"
This worked when I ran it:
class FieldThatKnowsItsName():
def __init__(self):
self.name = None
self._value= None
self.owner = None
def __set_name__(self, owner, name):
self.name = name
self.owner = owner
self.owner.fields[self.name] = self
def __get__(self, instance, instance_type):
return self
def __set__(self, instance, value):
self = value
class SuperTable:
fields = {}
field_1=FieldThatKnowsItsName()
field_2=FieldThatKnowsItsName()
table = SuperTable()
print(table.field_1.name)
print(table.field_2.name)
You can then add methods and or extend your datatype as you like.
As a bonus, the set_name(self, owner, name) dunder also passes the parent instance, so the Field class instance can register itself with the parent.
I got this from Effective Python by Brett Slatkin. It took a while to figure out how to implement.

How can I do the same for a variable? As opposed to functions, Python variables do not have the __name__ attribute.
The problem comes up because you are confused about terminology, semantics or both.
"variables" don't belong in the same category as "functions". A "variable" is not a thing that takes up space in memory while the code is running. It is just a name that exists in your source code - so that when you're writing the code, you can explain which thing you're talking about. Python uses names in the source code to refer to (i.e., give a name to) values. (In many languages, a variable is more like a name for a specific location in memory where the value will be stored. But Python's names actually name the thing in question.)
In Python, a function is a value. (In some languages, this is not the case; although there are bytes of memory used to represent the actual executable code, it isn't a discrete chunk of memory that your program logic gets to interact with directly.) In Python, every value is an object, meaning that you can assign names to it freely, pass it as an argument, return it from a function, etc. (In many languages, this is not the case.) Objects in Python have attributes, which are the things you access using the . syntax. Functions in Python have a __name__ attribute, which is assigned when the function is created. Specifically, when a def statement is executed (in most languages, creation of a function works quite differently), the name that appears after def is used as a value for the __name__ attribute, and also, independently, as a variable name that will get the function object assigned to it.
But most objects don't have an attribute like that.
In other words, if I have a variable such as:
That's the thing: you don't "have" the variable in the sense that you're thinking of. You have the object that is named by that variable. Anything else depends on the information incidentally being stored in some other object - such as the locals() of the enclosing function. But it would be better to store the information yourself. Instead of relying on a variable name to carry information for you, explicitly build the mapping between the string name you want to use for the object, and the object itself.

Python: update value of an instance and use it in a class from imported module

I have been studying Python for three months and now I have a question that I could not solve by using google, but luckily I am able to simplify it here:
I have a var_class module:
#var_class.py
class A(object):
def __init__(self, x):
self.x = x+2
zz = A(10)
class B():
b = 0
def __init__(self):
pass
def update(self):
B.b = zz.x
and in main program I have:
#main.py
from var_class import *
b_class = B()
b_class.b # I get 0 as expected
zz = A(100)
b_class.update()
b_class.b # Instead of 102, I get 12 unexpectedly
You see my goal is to change "zz" frequently for every data input and then update a class variable 'b', and the reason I wrote zz = A(10) in var_class.py is that otherwise when I import it, module 'var_class' is missing 'zz' in 'class B', it would give error "global name zz is not defined".
However, as I write it like this now, looks like the value '10' is stuck to the class, and I am not able to change it in the main program. Don't know how to overcome this.
Thanks in advance for any help.
Answer:
alKid wrote the whole answer first, have to thanks to alexvassel and Mr. C too, would like to know if there is way to thanks them, others helped me with the knowledge, also appreciate a lot.

Easy understanding:
You can't do that, you're inside var_class module, so zz is A(10).
How about passing zz as a parameter? Like this!
class A(object):
def __init__(self, x):
self.x = x+2
zz = A(10)
class B():
b = 0
def __init__(self):
pass
def update(self, zz):
B.b = zz.x
Hope this helps!

When you do B.b = zz.x (update method) you are inside the module var_class, so zz is A(10)

python runtime finds variables by namespace. Namespace is something like scope. When B.b = zz.x executes, the runtime first searches the local namespace(the function namespace -- update), it sees no zz. Then, it goes to the module space(var_class), well we get the variable and stop searching.
The python namespace search order:
1. local namespace, the function scope
2. global namespace, the module scope
3. built-in namespace
Better not use global variables around.
your code may like this:
class B():
b = 0
def __init__(self):
pass
def update(self, zz):
B.b = zz.x

That is a matter of scope! you are using zz = A(100) in your main.py. But when you call b_class.update(), b_class has to get "some2 variable called "zz" - and the one which is available is the one you defined in your class.py - and this one still has the value A(10)!!!
To work around this, you have different options.
The dirty hack (but closest to your solution) would be to define a global variable and use ist (google "python globals" for more info on this approach).
I would suggest passing A (or A.x) to your "update" method b_class.update(A.x). This avoids globals and is more readable. Of course, you'll have to adjust B.update for the new parameter.

when you import a variable from a module, you get a copy of the variable, not the original. you need to write to the original directly.
from a.b import c
from a.b.c import var
a.b.c.var = 1
var = 2
a.b.c.var
1
var
2
Edit: So, more correctly, in python, there are objects in memory, and then there are names for the objects. When you import a module, you create two separate names, but they both point to the same object - i.e. they have a reference to the same object. It's very similar to the below code, which doesn't require any imports at all:
>>> a = 4
>>> b = a
>>> b
4
>>> a
4
>>> b = 'something different'
>>> b
'something different'
>>> a
4
Why did changing b not also change a? The sequence is like this: First, we create an object (the 4) and point the name 'a' at it. Then, we create a name 'b', and we point it at the same object. So, now looking up those names returns the same object. Now, we then go back and point the name 'b' at a different object, a string. The object 'a' is pointing to still exists, and has not changed.
New users more often trip up on this the other way, with objects like lists:
>>> a = [1,2,3,4,5]
>>> b = a
>>> b
[1,2,3,4,5]
>>> a
[1,2,3,4,5]
>>> b.append(6)
>>> b
[1,2,3,4,5,6]
>>> a
[1,2,3,4,5,6]
>>> b = b[1:3]
>>> b
[2,3]
>>> a
[1,2,3,4,5,6]
What's going on here? Well, first we create a list object, and then point the name 'a' at it. Then we create the name 'b' and point it to the same object. So, 'a' and 'b' both point to the same list. Then, we use the reference b to get the object and modify it. Note that we haven't changed what 'b' points to in this case - we grabbed the reference, and then modified the object it points to directly. So, in this case, both 'a' and 'b' will see the change - they both point to the modified object. Then, we extract a slice, and assign it to 'b'. Now, this actually creates a new object, and points 'b' at it - 'b' is no longer pointing to the original object. So now 'a' and 'b' point to different objects, and now updates to one are no longer reflected in the other.
The import case is just a special case of this.

maybe you can do this
class B():
b = 0
def __init__(self):
pass
def update(self,value):
self.b =self.b+value