Develop Reference

Will Python automatically detect that the function was never called but defined?

True or False
If a function is defined but never called, then Python automatically detects that and issues a warning

One of the issues with this is that functions in Python are first class objects. So their name can be reassigned. For example:
def myfunc():
pass
a = myfunc
myfunc = 42
a()
We also have closures, where a function is returned by another function and the original name goes out of scope.
Unfortunately it is also perfectly legal to define a function with the same name as an existing one. For example:
def myfunc(): # <<< This code is never called
pass
def myfunc():
pass
myfunc()
So any tracking must include the function's id, not just its name - although that won't help with closures, since the id could get reused. It also won't help if the __name__ attribute of the function is reassigned.
You could track function calls using a decorator. Here I have used the name and the id - the id on its own would not be readable.
import functools
globalDict = {}
def tracecall(f):
#functools.wraps(f)
def wrapper(*args, **kwargs):
global globalDict
key = "%s (%d)" % (f.__name__, id(f))
# Count the number of calls
if key in globalDict:
globalDict[key] += 1
else:
globalDict[key] = 1
return f(*args, **kwargs)
return wrapper
#tracecall
def myfunc1():
pass
myfunc1()
myfunc1()
#tracecall
def myfunc1():
pass
a = myfunc1
myfunc1 = 42
a()
print(globalDict)
Gives:
{'myfunc1 (4339565296)': 2, 'myfunc1 (4339565704)': 1}
But that only gives the functions that have been called, not those that have not!
So where to go from here? I hope you can see that the task is quite difficult given the dynamic nature of python. But I hope the decorator I show above could at least allow you to diagnose the way the code is used.

No it is not. Python is not detect this. If you want to detect which functions are called or not during the run time you can use global set in your program. Inside each function add function name to set. Later you can print your set content and check if the the function is called or not.

False. Ignoring the difficulty and overhead of doing this, there's no reason why it would be useful.
A function that is defined in a module (i.e. a Python file) but not called elsewhere in that module might be called from a different module, so that doesn't deserve a warning.
If Python were to analyse all modules that get run over the course of a program, and print a warning about functions that were not called, it may be that a function was not called because of the input in this particular run e.g. perhaps in a calculator program there is a "multiply" function but the user only asked to sum some numbers.
If Python were to analyse all modules that make up a program and note and print a warning about functions that could not possibly be called (this is impossible but stay with me here) then it would warn about functions that were intended for use in other programs. E.g. if you have two calculator programs, a simple one and an advanced one, maybe you have a central calc.py with utility functions, and then advanced functions like exp and log could not possibly be called when that's used as part of simple program, but that shouldn't cause a warning because they're needed for the advanced program.

Python basic nested statements and scope concept

def func():
def nested():
global x
x = 1
x = 2
func()
print(x)
The correct answer would be '2' and the reason why is because func() is not defined. But when I read this it seems that func() is defined as nested(). I would think that when you call func() that would then automatically call nested(). I'm struggling to grasp this and understand why I shouldn't read it that way.

You're defining nested inside func, but you're not calling nested() anywhere in func, so when you call func() it's effectively doing nothing.
To do what you want try defining func as:
def func():
def nested():
global x
x = 1
nested()
UPDATE: After StevenRumbalski's comment, I think a small addition about what exactly is going on in that function can help clarify things around.
Python's functions are themselves objects which can respond to the operator ().
When you define a new function, what you're actually doing is instantiating a function object and giving it a name. In the example above, def func() creates a function instance and gives it name func, so that when you apply operator () to func (i.e. when you call the function with func()) the code of the function associated to that name is executed.
Let's now take one step further and look at what happens with nested.
Nested is defined inside func's scope, so when you exit func's scope the name nested is not defined anymore.
If, however, you return the function object to the caller of func, you can use that object to run nested's code.
A small example:
def returning_func():
def nested():
print("I am nested!")
return nested # note the lack of the () operator!
In this case, if we do
my_reference_to_nested = func()
nothing gets printed, because nested is defined but not executed.
If we call it, however:
my_reference_to_nested()
we execute nested's code and print to the output I am nested!

Definition of a function that does nothing

I need to define a function depending on a condition, so that it either does something or nothing.
e.g.
# Code block that *defines* bar
if condition:
bar = do_stuff
else:
bar = # ?
# End of the definition of bar
bar()
do_stuff is a function defined somewhere else that does some useful computation. I want bar() to do be defined to do nothing if the condition above is not met. Any ideas?
Note: I designed the if-else block above for illustration purposes. I know that I could just call do_stuff() in the if body and pass in the else body. The code above is just to illustrate a code block after which I need bar to be defined either to do something or to do nothing. The block just defines the function and is not meant to call it.

To set the variable bar to be a function that does nothing, there are two options:
Using a lambda:
bar = lambda: None
Using a function definition:
def bar():
pass
You could also use return None instead of pass for the function definition, the behavior is equivalent.

You can make a function that does nothing with:
def func():
pass

Python custom function

I have been working at learning Python over the last week and it has been going really well, however I have now been introduced to custom functions and I sort of hit a wall. While I understand the basics of it, such as:
def helloworld():
print("Hello World!")
helloworld()
I know this will print "Hello World!".
However, when it comes to getting information from one function to another, I find that confusing. ie: function1 and function2 have to work together to perform a task. Also, when to use the return command.
Lastly, when I have a list or a dictionary inside of a function. I'll make something up just as an example.
def my_function():
my_dict = {"Key1":Value1,
"Key2":Value2,
"Key3":Value3,
"Key4":Value4,}
How would I access the key/value and be able to change them from outside of the function? ie: If I had a program that let you input/output player stats or a character attributes in a video game.
I understand bits and pieces of this, it just confuses me when they have different functions calling on each other.
Also, since this was my first encounter with the custom functions. Is this really ambitious to pursue and this could be the reason for all of my confusion? Since this is the most complex program I have seen yet.

Functions in python can be both, a regular procedure and a function with a return value. Actually, every Python's function will return a value, which might be None.
If a return statement is not present, then your function will be executed completely and leave normally following the code flow, yielding None as a return value.
def foo():
pass
foo() == None
>>> True
If you have a return statement inside your function. The return value will be the return value of the expression following it. For example you may have return None and you'll be explicitly returning None. You can also have return without anything else and there you'll be implicitly returning None, or, you can have return 3 and you'll be returning value 3. This may grow in complexity.
def foo():
print('hello')
return
print('world')
foo()
>>>'hello'
def add(a,b):
return a + b
add(3,4)
>>>7
If you want a dictionary (or any object) you created inside a function, just return it:
def my_function():
my_dict = {"Key1":Value1,
"Key2":Value2,
"Key3":Value3,
"Key4":Value4,}
return my_dict
d = my_function()
d['Key1']
>>> Value1
Those are the basics of function calling. There's even more. There are functions that return functions (also treated as decorators. You can even return multiple values (not really, you'll be just returning a tuple) and a lot a fun stuff :)
def two_values():
return 3,4
a,b = two_values()
print(a)
>>>3
print(b)
>>>4
Hope this helps!

The primary way to pass information between functions is with arguments and return values. Functions can't see each other's variables. You might think that after
def my_function():
my_dict = {"Key1":Value1,
"Key2":Value2,
"Key3":Value3,
"Key4":Value4,}
my_function()
my_dict would have a value that other functions would be able to see, but it turns out that's a really brittle way to design a language. Every time you call my_function, my_dict would lose its old value, even if you were still using it. Also, you'd have to know all the names used by every function in the system when picking the names to use when writing a new function, and the whole thing would rapidly become unmanageable. Python doesn't work that way; I can't think of any languages that do.
Instead, if a function needs to make information available to its caller, return the thing its caller needs to see:
def my_function():
return {"Key1":"Value1",
"Key2":"Value2",
"Key3":"Value3",
"Key4":"Value4",}
print(my_function()['Key1']) # Prints Value1
Note that a function ends when its execution hits a return statement (even if it's in the middle of a loop); you can't execute one return now, one return later, keep going, and return two things when you hit the end of the function. If you want to do that, keep a list of things you want to return and return the list when you're done.

You send information into and out of functions with arguments and return values, respectively. This function, for example:
def square(number):
"""Return the square of a number."""
return number * number
... recieves information through the number argument, and sends information back with the return ... statement. You can use it like this:
>>> x = square(7)
>>> print(x)
49
As you can see, we passed the value 7 to the function, and it returned the value 49 (which we stored in the variable x).
Now, lets say we have another function:
def halve(number):
"""Return half of a number."""
return number / 2.0
We can send information between two functions in a couple of different ways.
Use a temporary variable:
>>> tmp = square(6)
>>> halve(tmp)
18.0
use the first function directly as an argument to the second:
>>> halve(square(8))
32.0
Which of those you use will depend partly on personal taste, and partly on how complicated the thing you're trying to do is.
Even though they have the same name, the number variables inside square() and halve() are completely separate from each other, and they're invisible outside those functions:
>>> number
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'number' is not defined
So, it's actually impossible to "see" the variable my_dict in your example function. What you would normally do is something like this:
def my_function(my_dict):
# do something with my_dict
return my_dict
... and define my_dict outside the function.
(It's actually a little bit more complicated than that - dict objects are mutable (which just means they can change), so often you don't actually need to return them. However, for the time being it's probably best to get used to returning everything, just to be safe).

programmatically determine if an evaluation of a function is being assigned?

I would like to do this:
def foo():
if <a magical condition>:
return x
else:
poof()
# or...
def foo():
x = <a magical object>
return x
def poof():
print 'poof!'
bar = foo() # bar points to <a magical object> but poof() is not called
foo() # prints 'poof!'
I guess it comes down to what the circumstanses are when the returned object's __del__ method is called. But maybe there is a better way. Like if the function itself knew it's returned value was being assigned. I guess I'm worried about relying on the timing of the garbage collection. Also I don't like that global at_end_of_program flag.
My solution:
class Magic:
def __del__(s):
poof()
def foo():
x = Magic()
return x
def poof():
if not at_end_of_program:
print 'poof!'
bar = foo() # No poof.
foo() # prints 'poof!'

I'm pretty confused by your question, but I think what you are trying to do is run a function when a value is reassigned.
Instead of doing tricky things with a __del__() method function, I suggest you just put your value into a class instance, and then overload __setattr__(). You could also overload __delattr__() to make sure you catch del(object.x) for your value x.
The very purpose of __setattr__() is to give you a hook to catch when something assigns to a member of your class. And you won't need any strange end_of_program flag. At the end of your program, just get rid of your overloaded function for __delattr__() so it doesn't get called for end-of-program cleanup.

A function can't tell what its return value is used for. Your solution will print poof if you re-assign to bar for example.
What's the real problem you are trying to solve?