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Why does my module not behave like a singleton?

I have a JSON file that I am using as a datastore in a small game I am using as a way to learn Python.
I am proficient in a number of other languages.
I have several classes that want read access to the JSON so I want to load the JSON from the file into a variable and then allow the other classes to access the variable via getters and setters, because each class wants different parts of the JSON.
This sounds like a job for a Singleton. I understood that a Python Module behaves like a singleton.
However, when I import the Module into my classes the variable resets?
Here is a very cut down example:
Module:- state_manager
x=45
def set_x(value):
x=value
def get_x():
return x
Class:- Game
import Player
import state_manager
value = state_manager.get_x()
Class:- Player
import state_manager
state_manager.set_x(12)
By setting breakpoints I can see that when Player is imported by Game that Player sets the value of x in state_manager to 12.
But when I look at the value of x returned to Game using state_manager.get_x() I get 45.
Why is this?
What is the correct way in Python to create a Module or Object that can be shared among other classes?
I realise I can construct a Singleton myself but I thought I'd use the features of Python.

By setting breakpoints I can see that when Player is imported by Game that Player sets the value of x in state_manager to 12.
I am fairly sure that you're doing something wrong in your inspection, because the set_x function, at least as you quoted it...
x=45
def set_x(value):
x=value
...does not do what you think it does. Since x is being assigned to in the scope of set_x, it does not refer to the global (module-level) variable x, but to a local variable x that is immediately discarded as part of the stack frame when set_x returns. The existence of static assignments is effectively how local variables are declared in Python. The fix is to declare x as referring to the global variable:
x=45
def set_x(value):
global x
x=value

You need to declare x global in any function that attempts to set it globally:
def set_x(value):
global x
x=value
Without the global declaration, x is just a function-local variable.
In general, if a function assigns to a variable, anywhere in the function, then that variable is local unless it is explicitly declared global (or nonlocal). If a function only reads a variable, without setting it, then the variable is taken from a higher scope (e.g., a global, or an up-level reference).

Limit Python function scope to local variables only

Is there a way to limit function so that it would only have access to local variable and passed arguments?
For example, consider this code
a = 1
def my_fun(x):
print(x)
print(a)
my_fun(2)
Normally the output will be
2
1
However, I want to limit my_fun to local scope so that print(x) would work but throw an error on print(a). Is that possible?

I feel like I should preface this with: Do not actually do this.
You (sort of) can with functions, but you will also disable calls to all other global methods and variables, which I do not imagine you would like to do.
You can use the following decorator to have the function act like there are no variables in the global namespace:
import types
noglobal = lambda f: types.FunctionType(f.__code__, {})
And then call your function:
a = 1
#noglobal
def my_fun(x):
print(x)
print(a)
my_fun(2)
However this actually results in a different error than you want, it results in:
NameError: name 'print' is not defined
By not allowing globals to be used, you cannot use print() either.
Now, you could pass in the functions that you want to use as parameters, which would allow you to use them inside the function, but this is not a good approach and it is much better to just keep your globals clean.
a = 1
#noglobal
def my_fun(x, p):
p(x)
p(a)
my_fun(2, print)
Output:
2
NameError: name 'a' is not defined

Nope. The scoping rules are part of a language's basic definition. To change this, you'd have to alter the compiler to exclude items higher on the context stack, but still within the user space. You obviously don't want to limit all symbols outside the function's context, as you've used one in your example: the external function print. :-)

assignment of variable after function definition in Python

I am new to Python so I am unsure about the difference in variable assignment before or after the function definition.
Specifically, the first example was adopted from Lutz's book.
def tester(start):
print("inside tester")
def nested(label):
print("inside nested")
print(label,nested.state)
nested.state += 1
print("done with nested")
nested.state = start
print("done with tester")
return nested
F = tester(0)
F('spam')
F.state
F.state
The objective of the code is to store the state information without using nonlocal.
I am unsure what nested.state means here. I am unsure because nested.state is used inside nested() function (i.e. nested.state +=1) and outside nested() function (i.e. nested.state = start).
I modified the code above to see whether Python accepts assigning variable after function declaration for nested() and to see whether there is any concept I am missing relating to function.variable call (i.e. nested.state call).
def tester(start):
def nested(label):
print(label, state)
state += 1 #replaced 'nested.state' with 'state' here
state = start #replaced 'nested.state' with 'state' here
return nested
F=tester(0)
F('spam')
F('ham')
Unfortunately, above code generates error local variable 'state' referenced before assignment. This tells me that I am missing some concept about function.variable (i.e. nested.state).
Can someone please help me understand three things:
I. why it is that the code with nested.state doesn't generate any error but state does?
II. what does nested.state mean? If nested.state is a mechanism to access function's variables, why is it that the call inside nested() function also uses nested.state and not state?
III. If nested.state is a mechanism to access variable inside function, then why is it that PyCharm fails to show state under dropdown when I type nested.?
I'd appreciate any help. I research SO, and couldn't find any explanation on such problems.

The reason the first code example worked is because it was assigning and referencing an attribute of the nested function object. The key concept to understand here, is that Python allows you to assign new, arbitrary attributes to objects - including functions:
>>> def func(a, b):
return a + b
>>> func(1, 2)
3
>>> func.attr = 5
>>> func.attr
5
The first code example takes advantage of this fact by using the nested function object to store the necessary state. This is the same concept as using any other object to store the state. It's particularly convenient, however, to use a function since it's readily available.
In the second example, a normal variable is used. Because of this, normal scoping rules apply which means simply that the state variable defined in tester is not the state variable being referenced in nested. Thus, an error is raised.

Actually, I think you're asking a question about scope in Python, ignoring your code, check this:
def scope_level_1():
variable = 'Scope_level_1'
def scope_level_2():
variable = 'Scope_level_2'
def core():
nonlocal variable
variable += '_TOUCHED_AND_MODIFIED_BY_CORE'
print(variable)
return core()
return scope_level_2()
scope_level_1()
# 'Scope_level_2_TOUCHED_AND_MODIFIED_BY_CORE'
Don't worry about the keyword nonlocal, treat it just as a declaring to make code more readable.

First, remember a += b is the same as a = a + b. So a must exist before getting to the +=.
Simply put, in the first example the function nested has an attribute called state (accessed by nested.state). It is an attribute, which means that once you tell nested that it has an attribute called state (you are doing this in line 9 when nested.state = start) it keep that attribute. So, in the first example nested.state exists when you get to the +=.
In the second example, you are declaring a variable called state in tester, and another variable called state in nested. The one in nested could be called potato for all that matters, because it is not the same variable. Therefore when you arrive to the +=, the variable state does not exist!

Update locals from inside a function

I would like to write a function which receives a local namespace dictionary and update it. Something like this:
def UpdateLocals(local_dict):
d = {'a':10, 'b':20, 'c':30}
local_dict.update(d)
When I call this function from the interactive python shell it works all right, like this:
a = 1
UpdateLocals(locals())
# prints 20
print a
However, when I call UpdateLocals from inside a function, it doesn't do what I expect:
def TestUpdateLocals():
a = 1
UpdateLocals(locals())
print a
# prints 1
TestUpdateLocals()
How can I make the second case work like the first?
UPDATE:
Aswin's explanation makes sense and is very helpful to me. However I still want a mechanism to update the local variables. Before I figure out a less ugly approach, I'm going to do the following:
def LoadDictionary():
return {'a': 10, 'b': 20, 'c': 30}
def TestUpdateLocals():
a = 1
for name, value in LoadDictionary().iteritems():
exec('%s = value' % name)
Of course the construction of the string statements can be automated, and the details can be hidden from the user.

You have asked a very good question. In fact, the ability to update local variables is very important and crucial in saving and loading datasets for machine learning or in games. However, most developers of Python language have not come to a realization of its importance. They focus too much on conformity and optimization which is nevertheless important too.
Imagine you are developing a game or running a deep neural network (DNN), if all local variables are serializable, saving the entire game or DNN can be simply put into one line as print(locals()), and loading the entire game or DNN can be simply put into one line as locals().update(eval(sys.stdin.read())).
Currently, globals().update(...) takes immediate effect but locals().update(...) does not work because Python documentation says:
The default locals act as described for function locals() below:
modifications to the default locals dictionary should not be
attempted. Pass an explicit locals dictionary if you need to see
effects of the code on locals after function exec() returns.
Why they design Python in such way is because of optimization and conforming the exec statement into a function:
To modify the locals of a function on the fly is not possible without
several consequences: normally, function locals are not stored in a
dictionary, but an array, whose indices are determined at compile time
from the known locales. This collides at least with new locals added
by exec. The old exec statement circumvented this, because the
compiler knew that if an exec without globals/locals args occurred in
a function, that namespace would be "unoptimized", i.e. not using the
locals array. Since exec() is now a normal function, the compiler does
not know what "exec" may be bound to, and therefore can not treat is
specially.
Since global().update(...) works, the following piece of code will work in root namespace (i.e., outside any function) because locals() is the same as globals() in root namespace:
locals().update({'a':3, 'b':4})
print(a, b)
But this will not work inside a function.
However, as hacker-level Python programmers, we can use sys._getframe(1).f_locals instead of locals(). From what I have tested so far, on Python 3, the following piece of code always works:
def f1():
sys._getframe(1).f_locals.update({'a':3, 'b':4})
print(a, b)
f1()
However, sys._getframe(1).f_locals does not work in root namespace.

The locals are not updated here because, in the first case, the variable declared has a global scope. But when declared inside a function, the variable loses scope outside it.
Thus, the original value of the locals() is not changed in the UpdateLocals function.
PS: This might not be related to your question, but using camel case is not a good practice in Python. Try using the other method.
update_locals() instead of UpdateLocals()
Edit To answer the question in your comment:
There is something called a System Stack. The main job of this system stack during the execution of a code is to manage local variables, make sure the control returns to the correct statement after the completion of execution of the called function etc.,
So, everytime a function call is made, a new entry is created in that stack,
which contains the line number (or instruction number) to which the control has to return after the return statement, and a set of fresh local variables.
The local variables when the control is inside the function, will be taken from the stack entry. Thus, the set of locals in both the functions are not the same. The entry in the stack is popped when the control exits from the function. Thus, the changes you made inside the function are erased, unless and until those variables have a global scope.

How can one create new scopes in python

In many languages (and places) there is a nice practice of creating local scopes by creating a block like this.
void foo()
{
... Do some stuff ...
if(TRUE)
{
char a;
int b;
... Do some more stuff ...
}
... Do even more stuff ...
}
How can I implement this in python without getting the unexpected indent error and without using some sort of if True: tricks

Why do you want to create new scopes in python anyway?
The normal reason for doing it in other languages is variable scoping, but that doesn't happen in python.
if True:
a = 10
print a

In Python, scoping is of three types : global, local and class. You can create specialized 'scope' dictionaries to pass to exec / eval(). In addition you can use nested scopes
(defining a function within another). I found these to be sufficient in all my code.
As Douglas Leeder said already, the main reason to use it in other languages is variable scoping and that doesn't really happen in Python. In addition, Python is the most readable language I have ever used. It would go against the grain of readability to do something like if-true tricks (Which you say you want to avoid). In that case, I think the best bet is to refactor your code into multiple functions, or use a single scope. I think that the available scopes in Python are sufficient to cover every eventuality, so local scoping shouldn't really be necessary.

If you just want to create temp variables and let them be garbage collected right after using them, you can use
del varname
when you don't want them anymore.
If its just for aesthetics, you could use comments or extra newlines, no extra indentation, though.

Python has exactly two scopes, local and global. Variables that are used in a function are in local scope no matter what indentation level they were created at. Calling a nested function will have the effect that you're looking for.
def foo():
a = 1
def bar():
b = 2
print a, b #will print "1 2"
bar()
Still like everyone else, I have to ask you why you want to create a limited scope inside a function.

variables in list comprehension (Python 3+) and generators are local:
>>> i = 0
>>> [i+1 for i in range(10)]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> i
0
but why exactly do you need this?

A scope is a textual region of a
Python program where a namespace is
directly accessible. “Directly
accessible” here means that an
unqualified reference to a name
attempts to find the name in the
namespace...
Please, read the documentation and clarify your question.
btw, you don't need if(TRUE){} in C, a simple {} is sufficient.

As mentioned in the other answers, there is no analogous functionality in Python to creating a new scope with a block, but when writing a script or a Jupyter Notebook, I often (ab)use classes to introduce new namespaces for similar effect. For example, in a notebook where you might have a model "Foo", "Bar" etc. and related variables you might want to create a new scope to avoid having to reuse names like
model = FooModel()
optimizer = FooOptimizer()
...
model = BarModel()
optimizer = BarOptimizer()
or suffix names like
model_foo = ...
optimizer_foo = ...
model_bar = ...
optimizer_bar= ...
Instead you can introduce new namespaces with
class Foo:
model = ...
optimizer = ...
loss = ....
class Bar:
model = ...
optimizer = ...
loss = ...
and then access the variables as
Foo.model
Bar.optimizer
...
I find that using namespaces this way to create new scopes makes code more readable and less error-prone.

While the leaking scope is indeed a feature that is often useful,
I have created a package to simulate block scoping (with selective leaking of your choice, typically to get the results out) anyway.
from scoping import scoping
a = 2
with scoping():
assert(2 == a)
a = 3
b = 4
scoping.keep('b')
assert(3 == a)
assert(2 == a)
assert(4 == b)
https://pypi.org/project/scoping/

I would see this as a clear sign that it's time to create a new function and refactor the code. I can see no reason to create a new scope like that. Any reason in mind?

def a():
def b():
pass
b()
If I just want some extra indentation or am debugging, I'll use if True:

Like so, for arbitrary name t:
### at top of function / script / outer scope (maybe just big jupyter cell)
try: t
except NameError:
class t
pass
else:
raise NameError('please `del t` first')
#### Cut here -- you only need 1x of the above -- example usage below ###
t.tempone = 5 # make new temporary variable that definitely doesn't bother anything else.
# block of calls here...
t.temptwo = 'bar' # another one...
del t.tempone # you can have overlapping scopes this way
# more calls
t.tempthree = t.temptwo; del t.temptwo # done with that now too
print(t.tempthree)
# etc, etc -- any number of variables will fit into t.
### At end of outer scope, to return `t` to being 'unused'
del t
All the above could be in a function def, or just anyplace outside defs along a script.
You can add or del new elements to an arbitrary-named class like that at any point. You really only need one of these -- then manage your 'temporary' namespace as you like.
The del t statement isn't necessary if this is in a function body, but if you include it, then you can copy/paste chunks of code far apart from each other and have them work how you expect (with different uses of 't' being entirely separate, each use starting with the that try: t... block, and ending with del t).
This way if t had been used as a variable already, you'll find out, and it doesn't clobber t so you can find out what it was.
This is less error prone then using a series of random=named functions just to call them once -- since it avoids having to deal with their names, or remembering to call them after their definition, especially if you have to reorder long code.
This basically does exactly what you want: Make a temporary place to put things you know for sure won't collide with anything else, and which you are responsible for cleaning up inside as you go.
Yes, it's ugly, and probably discouraged -- you will be directed to decompose your work into a set of smaller, more reusable functions.

As others have suggested, the python way to execute code without polluting the enclosing namespace is to put it in a class or function. This presents a slight and usually harmless problem: defining the function puts its name in the enclosing namespace. If this causes harm to you, you can name your function using Python's conventional temporary variable "_":
def _():
polluting_variable = foo()
...
_() # Run the code before something overwrites the variable.
This can be done recursively as each local definition masks the definition from the enclosing scope.
This sort of thing should only be needed in very specific circumstances. An example where it is useful is when using Databricks' %run magic, which executes the contents of another notebook in the current notebook's global scope. Wrapping the child notebook's commands in temporary functions prevents them from polluting the global namespace.