Convert Variable Name to String?

Question

I would like to convert a python variable name into the string equivalent as shown. Any ideas how?

var = {}
print ???  # Would like to see \'var\'
something_         


        

Answer 1

It's not very Pythonesque but I was curious and found this solution. You need to duplicate the globals dictionary since its size will change as soon as you define a new variable.

def var_to_name(var):
    # noinspection PyTypeChecker
    dict_vars = dict(globals().items())

    var_string = None

    for name in dict_vars.keys():
        if dict_vars[name] is var:
            var_string = name
            break

    return var_string


if __name__ == "__main__":
    test = 3
    print(f"test = {test}")
    print(f"variable name: {var_to_name(test)}")

which returns:

test = 3
variable name: test

Answer 2

I think this is a cool solution and I suppose the best you can get. But do you see any way to handle the ambigious results, your function may return? As "is" operator behaves unexpectedly with integers shows, low integers and strings of the same value get cached by python so that your variablename-function might priovide ambigous results with a high probability. In my case, I would like to create a decorator, that adds a new variable to a class by the varialbename i pass it:

def inject(klass, dependency):
klass.__dict__["__"+variablename(dependency)]=dependency

But if your method returns ambigous results, how can I know the name of the variable I added?

var any_var="myvarcontent"
var myvar="myvarcontent"
@inject(myvar)
class myclasss():
    def myclass_method(self):
        print self.__myvar    #I can not be sure, that this variable will be set...

Maybe if I will also check the local list I could at least remove the "dependency"-Variable from the list, but this will not be a reliable result.

Answer 3

This is not possible.

In Python, there really isn't any such thing as a "variable". What Python really has are "names" which can have objects bound to them. It makes no difference to the object what names, if any, it might be bound to. It might be bound to dozens of different names, or none.

Consider this example:

foo = 1
bar = 1
baz = 1

Now, suppose you have the integer object with value 1, and you want to work backwards and find its name. What would you print? Three different names have that object bound to them, and all are equally valid.

In Python, a name is a way to access an object, so there is no way to work with names directly. There might be some clever way to hack the Python bytecodes or something to get the value of the name, but that is at best a parlor trick.

If you know you want print foo to print "foo", you might as well just execute print "foo" in the first place.

EDIT: I have changed the wording slightly to make this more clear. Also, here is an even better example:

foo = 1
bar = foo
baz = foo

In practice, Python reuses the same object for integers with common values like 0 or 1, so the first example should bind the same object to all three names. But this example is crystal clear: the same object is bound to foo, bar, and baz.

Answer 4

What are you trying to achieve? There is absolutely no reason to ever do what you describe, and there is likely a much better solution to the problem you're trying to solve..

The most obvious alternative to what you request is a dictionary. For example:

>>> my_data = {'var': 'something'}
>>> my_data['something_else'] = 'something'
>>> print my_data.keys()
['var', 'something_else']
>>> print my_data['var']
something

Mostly as a.. challenge, I implemented your desired output. Do not use this code, please!

#!/usr/bin/env python2.6
class NewLocals:
    """Please don't ever use this code.."""
    def __init__(self, initial_locals):
        self.prev_locals = list(initial_locals.keys())

    def show_new(self, new_locals):
        output = ", ".join(list(set(new_locals) - set(self.prev_locals)))
        self.prev_locals = list(new_locals.keys())
        return output
# Set up
eww = None
eww = NewLocals(locals())

# "Working" requested code

var = {}

print eww.show_new(locals())  # Outputs: var

something_else = 3
print eww.show_new(locals()) # Outputs: something_else

# Further testing

another_variable = 4
and_a_final_one = 5

print eww.show_new(locals()) # Outputs: another_variable, and_a_final_one