Develop Reference

What does it mean that "actual search of names is done at runtime" in python?

The docs states that:
It is important to realize that scopes are determined textually: the global scope of a function defined in a module is that module’s namespace, no matter from where or by what alias the function is called. On the other hand, the actual search for names is done dynamically, at run time
I understand the first part: scopes are determined textually. But what does it mean that the actual search for names is done dynamically at run time? As opposed to what?
Let's try to compare this to what happens in C for instance, as I understand that this is the opposite of what happens in Python.
In C, consider the following code:
int a = 5
printf("The value of a is: %d\n", a);
So in C, the actual search for names is done at compile time - that means that the compiled machine code for the printf function will contain reference to the memory address of a whereas in Python
a = 5
print(a)
The compiled code of the print(a) will contain instructions for going looking in the namespace dictionary for what is pointed to by a and then access it.
Is that correct?

It means that a name can suddenly start resolving to something else, because it was redefined during the execution of the program. The alternative would be to resolve names when the program is read and parsed, and stick to this interpretation. (Which would be somewhat faster allow considerable additional optimization, e.g. by "knowing" things about the default behavior of Python built-in functions; but it is not how the language was designed.)
Here's an example that suddenly changes behavior:
for n in range(3):
print(max([10, 20, 30]))
max = min
This loop will print what you expect on the first iteration, but from then on the identifier max will refer to the local variable max, and will resolve to the builtin min(). Silly, but realistic use cases are a different question...

As opposed to being done statically at compile-time.
For instance in a language like C or Rust, by default symbols are looked up at compile-time, and at runtime the code just goes to whatever was resolved during compilation.
In Python however, every time you call a function the interpreter will look for that name in the relevant scope(s), then will use whatever's bound to that name at that point in time. Semantically if not necessarily technically.
So if e.g. you swap the object assigned to that name, then the code will call the remplacement instead of the original. Even if the replacement is not callable at all.

Tensorflow VariableScope: original_name_scope vs name

In TensorFlow, the VariableScope class has both a original_name_scope and name attribute. What are their differences and when should I use one over the other? I can't seem to find much documentation on them.
Use case:
I'm using the tf.get_collection(key, scope) method. Its second argument expects a string, but my variable my_scope has type VariableScope. I'm trying both
tf.get_collection(key, my_scope.name)
and
tf.get_collection(key, my_scope.original_scope_name)
. Both seem to work, but I'm not sure which is "right" and won't give me problems later down the road.

foo.name returns the name (String) of the scope. On the other hand, foo.original_name_scope returns the same string as foo.name, except when the scope is recreated. In that case, all sub-scopes are appended with a _# as needed to make all calls to foo.original_name_scope return something unique for each instance of a scope.
For example, in this code:
with tf.variable_scope('a') as a:
print(a.name)
print(a.original_name_scope)
print(a.original_name_scope)
with tf.variable_scope('a') as b:
print(b.name)
print(b.original_name_scope)
Returns
a
a/
a/
a
a_1/
Note that the calls to original_name_scope corresponding to different scope instances a return different values.
Presumably, this lets you distinguish between different scope instances with the same name.

How to bind a name with multiple objects or values in python

I saw in a book about language description that says
On the other hand, a name can be bound to no object (a dangling pointer),
one object (the usual case), or several objects (a parameter name in a
recursive function).
How can we bind a name to several objects? Isnt that what we call an array for example where all elements have the same name but with index? For a recursive function like the example here:
x = 0
def f(y):
global x
x += 1
if x < 4 :
y +=100
f(y)
else: return
f(100)
Is the name y binded with multiple values that are created recursively since the nametable has already the y name binded to an initial value which is being reproduced with recursion?
EDITED Just press here Visualizer and see what it generates. :)

No.
A name is bound to one single object . When we are talking about Python - it is either bound to a single object in a given context, or do not exist at all.
What happens, is that the inner workings may have the name defined in several "layers" - but your code will only see one of those.
If a name is a variable in a recursive function, you will only see whatver is bound to it in the current running context - each time there is a function call in Python, the execution frame, which is an object which holds several attributes of the running code, including a reference to the local variables, is frozen. On the called function, a new execuciton frame is created, and there, the variable names are bound again to whatever new values they have in the called context. Your code just "see" this instance.
Then, there is the issue of global variables and builtin objects in Python: if a name is not a local variable in the function execution context, it is searched in the globals variables for the module (again, just one of those will be visible).ANd if the name is not defiend in the globals, them, Python looks for it in globals().__builtins__ that is your last call.

If I understand you correctly, you're asking about what rules Python has for creating variables in different scopes. Python uses lexical scoping on the function level.
It's hard to tell exactly what you're getting at with the code you've written, but, while there may be a different value associated with y in different scopes (with a value of y defined at each level of recursion), your code will only ever be able to see one at a time (the value defined at the scope in which you're operating).
To really understand scoping rules in Python, I would have a look at PEP 227. Also, have a look at this Stack Overflow question.
Finally, to be able to speak intelligently about what a "name" is in Python, I suggest you read about how Python is a "Call-By-Object" language.
At this point, we are capable of understanding that, instead of a "nametable", python uses a dictionary to hold what is accessible in a given scope. See this answer for a little more detail. The implication of this is that you can never have two of the same name in a single scope (for the same reason you can't have two of the same key in a python dictionary). So, while y may exist in a dictionary for a different scope, you have no way of accessing it, since you can only access the variables in the current scope's dictionary.

The key is:
several objects (a parameter name in a recursive function).
The passage is almost certainly not referring to arrays, but simply to the fact that in a recursive function (or any function, but a recursive function is likely to have multiple activations at one time), a parameter may be bound to a different value in each recursive call.
This does not mean that you can access each such object in every stack frame; indeed the point of the technique is to ensure that only one such value is accessible in each stack frame.

Firstly, you should mention in the question that the sentence from the book is not related explicitly to Python (as jsbueno wrote, one name is bound to exactly one object in Python).
Anyway, name bound to no object is a bit inaccurate. Generally, names are related to variables, and name related to a dangling pointer is the name of that pointer variable.
When speaking about the variable scope (i.e. the part of code where the variable is used), one variable name can be used only for a single value at a time. However, there may be other parts of code, independent on the one where we think about that variable. In the other part of code, the same name can be used; however, the two variables with the same name are totally isolated. This is the case of local variables also in the case of function bodies. If the language allows recursion, it must be capable to create another isolated space of local variable even for another call of the same function.
In Python, each function can also access outer variables, but it is more usual to use the inner, local variables. Whenever you assign a name some value, it is created in the local space.

Where is nonlocals()?

How do I obtain the non-local variables for the current scope? The functions vars, locals, and globals exist, but is there a function to get the nonlocals?
Why aren't the nonlocals listed when calling vars?
Update
My issue is that there's no way to enumerate the variables available in the current scope, as neither vars or globals includes the non-locals AFAICT.
I frequently use vars in code such as the following:
'{meh[0]}/{meh[3]} {a}{b}{c}'.format(**vars())
Which fails if any of these variables are in the scope of a containing function.

From within running code, you can easily get the names of the nonlocal variables - but retriving their content in a way a call to locals gets you a dictionary is a bit trickier.
The used nonlocal variable names are stored in the current running code object, in the co_freevars attribute.
So, getting the nonlocal names is a matter of:
names = inspect.currentframe().f_code.co_freevars
The contents for these variables, however, are stored in the __closure__ attribute (func_closure, in Python 2), of the function object instead. (Not the code object). The problem is that, without "aid from outside", there is no easy way for a running code to get to the function object it is running on. You can get to the frame object, which links to the code object, but there are no links back to the function object. (For a top level defined function one could explicitly use the function known name, as used in the def statement` but for an enclosed function, that is returned to a caller, there is no way of knowing its name either).
So, one has to resort to a trick - getting all the objects that link to the current code object, by using the gc module (garbage collector) - there is a gc.get_referrers call - it will return all the function objects that link to the code object one holds.
So, inside a function with non_local variables one could do:
import inspect, gc
from types import FunctionType
def a(b):
b1 = 2
def c():
nonlocal b1
print (b)
code = inspect.currentframe().f_code
names = code.co_freevars
function = [func for func in gc.get_referrers(code) if isinstance(func, FunctionType)][0]
nonlocals = dict (zip(names, (x.cell_contents for x in function.__closure__ )))
print(nonlocals)
return inspect.currentframe()
return c
c = a(5)
f = c()
And therefore retrieve the names and values of the nonlocals. But this won't work if you have more than one instance of that function around (that is, if the function of interested was created more than once with more than one call to the functin that generates it) - becasue all of those instances will link to the same code object. The example above, assumes there is only one function running with the current code - and would work correctly in this case. Another call to the factrory function would create another function, with potentially other values for the nonlocal variables, but with the same code object - the function = list genrator above would retrieve all of those, and arbitrarily pick the first of those.
The "correct" function is the one on which the current code is executing - I am trying to think of a way of retrieving this information, but can't get to it. If I can, I will complete this answer, but for now, this can't help you to retrieve the nonlocals values values.
(just found out that trying to use "eval" with a nonlocal variable name won't work as well)
It looks like that the only thing linking the current running frame to the function object where the nonlocal variables values are held is created at run time inside the native side of the Python interpreter. I can't think of a way of getting to it short of using the ctypes module to look at interpreters data structures at runtime, which would, of course, be unsuitable for any actual production code.
The bottom line: you can reliably retrieve the nonlocal variable names. But it looks like you can't get their value given their name as a string (nor rebind then).
You could try opening a feature-request for a "nonlocals" call on Python's bug tracker or on Python-ideas mailing list.

Correctness about variable scope

I'm currently developing some things in Python and I have a question about variables scope.
This is the code:
a = None
anything = False
if anything:
a = 1
else:
a = 2
print a # prints 2
If I remove the first line (a = None) the code still works as before. However in this case I'd be declaring the variable inside an "if" block, and regarding other languages like Java, that variable would only be visible inside the "if".
How exactly variable scoping works in Python and what's the good way to program in cases like this?
Thanks!

As a rule of thumb, scopes are created in three places:
File-scope - otherwise known as module scope
Class-scope - created inside class blocks
Function-scope - created inside def blocks
(There are a few exceptions to these.)
Assigning to a name reserves it in the scope namespace, marked as unbound until reaching the first assignment. So for a mental model, you are assigning values to names in a scope.

I believe that Python uses function scope for local variables. That is, in any given function, if you assign a value to a local variable, it will be available from that moment onwards within that function until it returns. Therefore, since both branches of your code are guaranteed to assign to a, there is no need to assign None to a initially.
Note that when you can also access variables declared in outer functions -- in other words, Python has closures.
def adder(first):
def add(second):
return first + second
return add
This defines a function called adder. When called with an argument first, it will return a function that adds whatever argument it receives to first and return that value. For instance:
add_two = adder(2)
add_three = adder(3)
add_two(4) # = 6
add_three(4) # = 7
However, although you can read the value from the outer function, you can't change it (unlike in many other languages). For instance, imagine trying to implement an accumulator. You might write code like so:
def accumulator():
total = 0
def add(number):
total += number
return total
return add
Unfortunately, trying to use this code results in an error message:
UnboundLocalError: local variable 'total' referenced before assignment
This is because the line total += number tries to change the value of total, which cannot be done in this way in Python.

There is no problem assigning the variable in the if block.
In this case it is being assigned on both branches, so you can see it will definitely be defined when you come to print it.
If one of the branches did not assign to a then a NameError exception would be raise when you try to print it after that branch

Python doesn't need variables to be declared initially, so you can declare and define at arbitrary points. And yes, the scope is function scope, so it will be visible outside the if.

i'm quite a beginner programmer, but for what i know, in python private variables don't exist. see private variables in the python documentation for a detailed discussion.
useful informations can also be found in the section "scopes and namespaces" on the same page.
personally, i write code like the one you posted pretty much every day, especially when the condition relies in getting input from the user, for example
if len(sys.argv)==2:
f = open(sys.argv[1], 'r')
else:
print ('provide input file')
i do declare variables before using them for structured types, for example i declare an empty list before appending its items within a loop.
hope it helps.