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_else = 3
print ???  # Would print 'something_else'
Asked By: Brandon Pelfrey

||

Answers:

print "var"
print "something_else"

Or did you mean something_else?

Answered By: Jonathan Feinberg

You somehow have to refer to the variable you want to print the name of. So it would look like:

print varname(something_else)

There is no such function, but if there were it would be kind of pointless. You have to type out something_else, so you can as well just type quotes to the left and right of it to print the name as a string:

print "something_else"
Answered By: sth

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.

Answered By: steveha

Technically the information is available to you, but as others have asked, how would you make use of it in a sensible way?

>>> x = 52
>>> globals()
{'__builtins__': <module '__builtin__' (built-in)>, '__name__': '__main__', 
'x': 52, '__doc__': None, '__package__': None}

This shows that the variable name is present as a string in the globals() dictionary.

>>> globals().keys()[2]
'x'

In this case it happens to be the third key, but there’s no reliable way to know where a given variable name will end up

>>> for k in globals().keys():
...   if not k.startswith("_"):
...     print k
...
x
>>>

You could filter out system variables like this, but you’re still going to get all of your own items. Just running that code above created another variable “k” that changed the position of “x” in the dict.

But maybe this is a useful start for you. If you tell us what you want this capability for, more helpful information could possibly be given.

Answered By: Todd

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
Answered By: dbr

TL;DR: Not possible. See ‘conclusion’ at the end.


There is an usage scenario where you might need this. I’m not implying there are not better ways or achieving the same functionality.

This would be useful in order to ‘dump’ an arbitrary list of dictionaries in case of error, in debug modes and other similar situations.

What would be needed, is the reverse of the eval() function:

get_indentifier_name_missing_function()

which would take an identifier name (‘variable’,’dictionary’,etc) as an argument, and return a
string containing the identifier’s name.


Consider the following current state of affairs:

random_function(argument_data)

If one is passing an identifier name (‘function’,’variable’,’dictionary’,etc) argument_data to a random_function() (another identifier name), one actually passes an identifier (e.g.: <argument_data object at 0xb1ce10>) to another identifier (e.g.: <function random_function at 0xafff78>):

<function random_function at 0xafff78>(<argument_data object at 0xb1ce10>)

From my understanding, only the memory address is passed to the function:

<function at 0xafff78>(<object at 0xb1ce10>)

Therefore, one would need to pass a string as an argument to random_function() in order for that function to have the argument’s identifier name:

random_function('argument_data')

Inside the random_function()

def random_function(first_argument):

, one would use the already supplied string 'argument_data' to:

  1. serve as an ‘identifier name’ (to display, log, string split/concat, whatever)

  2. feed the eval() function in order to get a reference to the actual identifier, and therefore, a reference to the real data:

    print("Currently working on", first_argument)
    some_internal_var = eval(first_argument)
    print("here comes the data: " + str(some_internal_var))
    

Unfortunately, this doesn’t work in all cases. It only works if the random_function() can resolve the 'argument_data' string to an actual identifier. I.e. If argument_data identifier name is available in the random_function()‘s namespace.

This isn’t always the case:

# main1.py
import some_module1

argument_data = 'my data'

some_module1.random_function('argument_data')


# some_module1.py
def random_function(first_argument):
    print("Currently working on", first_argument)
    some_internal_var = eval(first_argument)
    print("here comes the data: " + str(some_internal_var))
######

Expected results would be:

Currently working on: argument_data
here comes the data: my data

Because argument_data identifier name is not available in the random_function()‘s namespace, this would yield instead:

Currently working on argument_data
Traceback (most recent call last):
  File "~/main1.py", line 6, in <module>
    some_module1.random_function('argument_data')
  File "~/some_module1.py", line 4, in random_function
    some_internal_var = eval(first_argument)
  File "<string>", line 1, in <module>
NameError: name 'argument_data' is not defined

Now, consider the hypotetical usage of a get_indentifier_name_missing_function() which would behave as described above.

Here’s a dummy Python 3.0 code: .

# main2.py
import some_module2
some_dictionary_1       = { 'definition_1':'text_1',
                            'definition_2':'text_2',
                            'etc':'etc.' }
some_other_dictionary_2 = { 'key_3':'value_3',
                            'key_4':'value_4', 
                            'etc':'etc.' }
#
# more such stuff
#
some_other_dictionary_n = { 'random_n':'random_n',
                            'etc':'etc.' }

for each_one_of_my_dictionaries in ( some_dictionary_1,
                                     some_other_dictionary_2,
                                     ...,
                                     some_other_dictionary_n ):
    some_module2.some_function(each_one_of_my_dictionaries)


# some_module2.py
def some_function(a_dictionary_object):
    for _key, _value in a_dictionary_object.items():
        print( get_indentifier_name_missing_function(a_dictionary_object)    +
               "    " +
               str(_key) +
               "  =  " +
               str(_value) )
######

Expected results would be:

some_dictionary_1    definition_1  =  text_1
some_dictionary_1    definition_2  =  text_2
some_dictionary_1    etc  =  etc.
some_other_dictionary_2    key_3  =  value_3
some_other_dictionary_2    key_4  =  value_4
some_other_dictionary_2    etc  =  etc.
......
......
......
some_other_dictionary_n    random_n  =  random_n
some_other_dictionary_n    etc  =  etc.

Unfortunately, get_indentifier_name_missing_function() would not see the ‘original’ identifier names (some_dictionary_,some_other_dictionary_2,some_other_dictionary_n). It would only see the a_dictionary_object identifier name.

Therefore the real result would rather be:

a_dictionary_object    definition_1  =  text_1
a_dictionary_object    definition_2  =  text_2
a_dictionary_object    etc  =  etc.
a_dictionary_object    key_3  =  value_3
a_dictionary_object    key_4  =  value_4
a_dictionary_object    etc  =  etc.
......
......
......
a_dictionary_object    random_n  =  random_n
a_dictionary_object    etc  =  etc.

So, the reverse of the eval() function won’t be that useful in this case.


Currently, one would need to do this:

# main2.py same as above, except:

    for each_one_of_my_dictionaries_names in ( 'some_dictionary_1',
                                               'some_other_dictionary_2',
                                               '...',
                                               'some_other_dictionary_n' ):
        some_module2.some_function( { each_one_of_my_dictionaries_names :
                                     eval(each_one_of_my_dictionaries_names) } )
    
    
    # some_module2.py
    def some_function(a_dictionary_name_object_container):
        for _dictionary_name, _dictionary_object in a_dictionary_name_object_container.items():
            for _key, _value in _dictionary_object.items():
                print( str(_dictionary_name) +
                       "    " +
                       str(_key) +
                       "  =  " +
                       str(_value) )
    ######

In conclusion:

  • Python passes only memory addresses as arguments to functions.
  • Strings representing the name of an identifier, can only be referenced back to the actual identifier by the eval() function if the name identifier is available in the current namespace.
  • A hypothetical reverse of the eval() function, would not be useful in cases where the identifier name is not ‘seen’ directly by the calling code. E.g. inside any called function.
  • Currently one needs to pass to a function:
    1. the string representing the identifier name
    2. the actual identifier (memory address)

This can be achieved by passing both the 'string' and eval('string') to the called function at the same time. I think this is the most ‘general’ way of solving this egg-chicken problem across arbitrary functions, modules, namespaces, without using corner-case solutions. The only downside is the use of the eval() function which may easily lead to unsecured code. Care must be taken to not feed the eval() function with just about anything, especially unfiltered external-input data.

Answered By: Petru Zaharia

Does Django not do this when generating field names?

http://docs.djangoproject.com/en/dev//topics/db/models/#verbose-field-names

Seems reasonable to me.

Answered By: Joe

This will work for simnple data types (str, int, float, list etc.)

>>> def my_print(var_str) : 
      print var_str+':', globals()[var_str]
>>> a = 5
>>> b = ['hello', ',world!']
>>> my_print('a')
a: 5
>>> my_print('b')
b: ['hello', ',world!']
Answered By: Hadsbjerg

I searched for this question because I wanted a Python program to print assignment statements for some of the variables in the program. For example, it might print “foo = 3, bar = 21, baz = 432”. The print function would need the variable names in string form. I could have provided my code with the strings “foo”,”bar”, and “baz”, but that felt like repeating myself. After reading the previous answers, I developed the solution below.

The globals() function behaves like a dict with variable names (in the form of strings) as keys. I wanted to retrieve from globals() the key corresponding to the value of each variable. The method globals().items() returns a list of tuples; in each tuple the first item is the variable name (as a string) and the second is the variable value. My variablename() function searches through that list to find the variable name(s) that corresponds to the value of the variable whose name I need in string form.

The function itertools.ifilter() does the search by testing each tuple in the globals().items() list with the function lambda x: var is globals()[x[0]]. In that function x is the tuple being tested; x[0] is the variable name (as a string) and x[1] is the value. The lambda function tests whether the value of the tested variable is the same as the value of the variable passed to variablename(). In fact, by using the is operator, the lambda function tests whether the name of the tested variable is bound to the exact same object as the variable passed to variablename(). If so, the tuple passes the test and is returned by ifilter().

The itertools.ifilter() function actually returns an iterator which doesn’t return any results until it is called properly. To get it called properly, I put it inside a list comprehension [tpl[0] for tpl ... globals().items())]. The list comprehension saves only the variable name tpl[0], ignoring the variable value. The list that is created contains one or more names (as strings) that are bound to the value of the variable passed to variablename().

In the uses of variablename() shown below, the desired string is returned as an element in a list. In many cases, it will be the only item in the list. If another variable name is assigned the same value, however, the list will be longer.

>>> def variablename(var):
...     import itertools
...     return [tpl[0] for tpl in 
...     itertools.ifilter(lambda x: var is x[1], globals().items())]
... 
>>> var = {}
>>> variablename(var)
['var']
>>> something_else = 3
>>> variablename(something_else)
['something_else']
>>> yet_another = 3
>>> variablename(something_else)
['yet_another', 'something_else']
Answered By: Ken Manly

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.

Answered By: fonzy

as long as it’s a variable and not a second class, this here works for me:

def print_var_name(variable):
 for name in globals():
     if eval(name) == variable:
        print name
foo = 123
print_var_name(foo)
>>>foo

this happens for class members:

class xyz:
     def __init__(self):
         pass
member = xyz()
print_var_name(member)
>>>member

ans this for classes (as example):

abc = xyz
print_var_name(abc)
>>>abc
>>>xyz

So for classes it gives you the name AND the properteries

Answered By: Alice

Here is a succinct variation that lets you specify any directory.
The issue with using directories to find anything is that multiple variables can have the same value. So this code returns a list of possible variables.

def varname( var, dir=locals()):
  return [ key for key, val in dir.items() if id( val) == id( var)]
Answered By: ZevGriner

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
Answered By: Frenmerican

By using the the unpacking operator:

>>> def tostr(**kwargs):
    return kwargs

>>> var = {}
>>> something_else = 3
>>> tostr(var = var,something_else=something_else)
{'var' = {},'something_else'=3}
Answered By: restrepo

I don’t know it’s right or not, but it worked for me

def varname(variable):
    for name in list(globals().keys()):
        expression = f'id({name})'
        if id(variable) == eval(expression):
            return name
Answered By: hamnghi

To get the variable name of var as a string:

var = 1000
var_name = [k for k,v in locals().items() if v == var][0] 
print(var_name) # ---> outputs 'var'
Answered By: tamtam

I’d like to point out a use case for this that is not an anti-pattern, and there is no better way to do it.

This seems to be a missing feature in python.

There are a number of functions, like patch.object, that take the name of a method or property to be patched or accessed.

Consider this:

patch.object(obj, "method_name", new_reg)

This can potentially start "false succeeding" when you change the name of a method. IE: you can ship a bug, you thought you were testing…. simply because of a bad method name refactor.

Now consider: varname. This could be an efficient, built-in function. But for now it can work by iterating an object or the caller’s frame:

Now your call can be:

patch.member(obj, obj.method_name, new_reg)

And the patch function can call:

varname(var, obj=obj)

This would: assert that the var is bound to the obj and return the name of the member. Or if the obj is not specified, use the callers stack frame to derive it, etc.

Could be made an efficient built in at some point, but here’s a definition that works. I deliberately didn’t support builtins, easy to add tho:

Feel free to stick this in a package called varname.py, and use it in your patch.object calls:

patch.object(obj, varname(obj, obj.method_name), new_reg)

Note: this was written for python 3.

import inspect

def _varname_dict(var, dct):
    key_name = None
    for key, val in dct.items():
        if val is var:
            if key_name is not None:
                raise NotImplementedError("Duplicate names not supported %s, %s" % (key_name, key))
            key_name = key
    return key_name

def _varname_obj(var, obj):
    key_name = None
    for key in dir(obj):
        val = getattr(obj, key)
        equal = val is var
        if equal:
            if key_name is not None:
                raise NotImplementedError("Duplicate names not supported %s, %s" % (key_name, key))
            key_name = key
    return key_name

def varname(var, obj=None):
    if obj is None:
        if hasattr(var, "__self__"):
            return var.__name__
        caller_frame = inspect.currentframe().f_back
        try:
            ret = _varname_dict(var, caller_frame.f_locals)
        except NameError:
            ret = _varname_dict(var, caller_frame.f_globals)
    else:
        ret = _varname_obj(var, obj)
    if ret is None:
        raise NameError("Name not found. (Note: builtins not supported)")
    return ret
Answered By: Erik Aronesty

Totally possible with the python-varname package (python3):

from varname import nameof

s = 'Hey!'

print (nameof(s))

Output:

s

Install:

pip3 install varname

Or get the package here:

https://github.com/pwwang/python-varname

Answered By: Synthase

This module works for converting variables names to a string:
https://pypi.org/project/varname/

Use it like this:

from varname import nameof

variable=0

name=nameof(variable)

print(name)

//output: variable

Install it by:

pip install varname

Thanks @restrepo, this was exactly what I needed to create a standard save_df_to_file() function. For this, I made some small changes to your tostr() function. Hope this will help someone else:

def variabletostr(**df):
        variablename = list(df.keys())[0]
        return variablename
    
    variabletostr(df=0)
Answered By: Arend

it is possible to a limited extent. the answer is similar to the solution by @tamtam .
The given example assumes the following assumptions –

  • You are searching for a variable by its value
  • The variable has a distinct value
  • The value is in the global namespace

Example:

testVar         = "unique value"
varNameAsString = [k for k,v in globals().items() if v == "unique value"]
#
# the variable "varNameAsString" will contain all the variable name that matches
# the value "unique value"
# for this example, it will be a list of a single entry "testVar"
#
print(varNameAsString)

Output : [‘testVar’]

You can extend this example for any other variable/data type

Answered By: Nomesis

The original question is pretty old, but I found an almost solution with Python 3. (I say almost because I think you can get close to a solution but I do not believe there is a solution concrete enough to satisfy the exact request).

First, you might want to consider the following:

  • objects are a core concept in Python, and they may be assigned a variable, but the variable itself is a bound name (think pointer or reference) not the object itself
  • var is just a variable name bound to an object and that object could have more than one reference (in your example it does not seem to)
  • in this case, var appears to be in the global namespace so you can use the global builtin conveniently named global
  • different name references to the same object will all share the same id which can be checked by running the id builtin id like so: id(var)

This function grabs the global variables and filters out the ones matching the content of your variable.

def get_bound_names(target_variable):
    '''Returns a list of bound object names.'''
    return [k for k, v in globals().items() if v is target_variable]

The real challenge here is that you are not guaranteed to get back the variable name by itself. It will be a list, but that list will contain the variable name you are looking for. If your target variable (bound to an object) is really the only bound name, you could access it this way:

bound_names = get_variable_names(target_variable)
var_string = bound_names[0]
Answered By: Jesse H.

Possible for Python >= 3.8 (with f'{var=}’ string )

Not sure if this could be used in production code, but in Python 3.8(and up) you can use f’ string debugging specifier. Add = at the end of an expression, and it will print both the expression and its value:

my_salary_variable = 5000
print(f'{my_salary_variable  =  }')

Output:
my_salary_variable  =  5000

To uncover this magic here is another example:

param_list = f'{my_salary_variable=}'.split('=')
print(param_list)

Output:
['my_salary_variable', '5000']

Explanation: when you put ‘=’ after your var in f’string, it returns a string with variable name, ‘=’ and its value. Split it with .split(‘=’) and get a List of 2 strings, [0] – your_variable_name, and [1] – actual object of variable.
Pick up [0] element of the list if you need variable name only.

my_salary_variable = 5000
param_list = f'{my_salary_variable=}'.split('=')
print(param_list[0])
Output:
my_salary_variable

or, in one line

my_salary_variable = 5000
print(f'{my_salary_variable=}'.split('=')[0])
Output:
my_salary_variable

Works with functions too:

def my_super_calc_foo(number):
    return number**3

print(f'{my_super_calc_foo(5)  =  }')
print(f'{my_super_calc_foo(5)=}'.split('='))

Output:

my_super_calc_foo(5)  =  125
['my_super_calc_foo(5)', '125']

Process finished with exit code 0
Answered By: Andrey Znamenskiy
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