Getting the name of a variable as a string

Question

This thread discusses how to get the name of a function as a string in Python: How to get a function name as a string?

How can I do the same for a variable? As opposed to functions, Python variables do not have the __name__ attribute.

In other words, if I have a variable such as:

foo = dict()
foo['bar'] = 2

I am looking for a function/attribute, e.g. retrieve_name() in order to create a DataFrame in Pandas from this list, where the column names are given by the names of the actual dictionaries:

# List of dictionaries for my DataFrame
list_of_dicts = [n_jobs, users, queues, priorities]
columns = [retrieve_name(d) for d in list_of_dicts] 

Answer 1

TL;DR

Use the Wrapper helper from python-varname:

from varname.helpers import Wrapper

foo = Wrapper(dict())

# foo.name == 'foo'
# foo.value == {}
foo.value['bar'] = 2

For list comprehension part, you can do:

n_jobs = Wrapper(<original_value>) 
users = Wrapper(<original_value>) 
queues = Wrapper(<original_value>) 
priorities = Wrapper(<original_value>) 

list_of_dicts = [n_jobs, users, queues, priorities]
columns = [d.name for d in list_of_dicts]
# ['n_jobs', 'users', 'queues', 'priorities']
# REMEMBER that you have to access the <original_value> by d.value

I am the author of the python-varname package. Please let me know if you have any questions or you can submit issues on Github.

The long answer

Is it even possible?

Yes and No.

We are retrieving the variable names at runtime, so we need a function to be called to enable us to access the previous frames to retrieve the variable names. That's why we need a Wrapper there. In that function, at runtime, we are parsing the source code/AST nodes in the previous frames to get the exact variable name.

However, the source code/AST nodes in the previous frames are not always available, or they could be modified by other environments (e.g: pytest's assert statement). One simple example is that the codes run via exec(). Even though we are still able to retrieve some information from the bytecode, it needs too much effort and it is also error-prone.

How to do it?

First of all, we need to identify which frame the variable is given. It's not always simply the direct previous frame. For example, we may have another wrapper for the function:

from varname import varname

def func():
  return varname()

def wrapped():
  return func()

x = wrapped()

In the above example, we have to skip the frame inside wrapped to get to the right frame x = wrapped() so that we are able to locate x. The arguments frame and ignore of varname allow us to skip some of these intermediate frames. See more details in the README file and the API docs of the package.

Then we need to parse the AST node to locate where the variable is assigned value (function call) to. It's not always just a simple assignment. Sometimes there could be complex AST nodes, for example, x = [wrapped()]. We need to identify the correct assignment by traversing the AST tree.

How reliable is it?

Once we identify the assignment node, it is reliable.

varname is all depending on executing package to look for the node. The node executing detects is ensured to be the correct one (see also this).

It partially works with environments where other AST magics apply, including pytest, ipython, macropy, birdseye, reticulate with R, etc. Neither executing nor varname is 100% working with those environments.

Do we need a package to do it?

Well, yes and no, again.

If your scenario is simple, the code provided by @juan Isaza or @scohe001 probably is enough for you to work with the case where a variable is defined at the direct previous frame and the AST node is a simple assignment. You just need to go one frame back and retrieve the information there.

However, if the scenario becomes complicated, or we need to adopt different application scenarios, you probably need a package like python-varname, to handle them. These scenarios may include to:

1. present more friendly messages when the source code is not available or AST nodes are not accessible
2. skip intermediate frames (allows the function to be wrapped or called in other intermediate frames)
3. automatically ignores calls from built-in functions or libraries. For example: x = str(func())
4. retrieve multiple variable names on the left-hand side of the assignment
5. etc.

How about the f-string?

Like the answer provided by @Aivar Paalberg. It's definitely fast and reliable. However, it's not at runtime, meaning that you have to know it's foo before you print the name out. But with varname, you don't have to know that variable is coming:

from varname import varname

def func():
  return varname()

# In external uses
x = func() # 'x'
y = func() # 'y'

Finally

python-varname is not only able to detect the variable name from an assignment, but also:

• Retrieve variable names directly, using nameof
• Detect next immediate attribute name, using will
• Fetch argument names/sources passed to a function using argname

Read more from its documentation.

However, the final word I want to say is that, try to avoid using it whenever you can.

Because you can't make sure that the client code will run in an environment where the source node is available or AST node is accessible. And of course, it costs resources to parse the source code, identify the environment, retrieve the AST nodes and evaluate them when needed.

Answer 2

With Python 3.8 one can simply use f-string debugging feature:

>>> foo = dict()
>>> f'{foo=}'.split('=')[0]
'foo' 

One drawback of this method is that in order to get 'foo' printed you have to add f'{foo=}' yourself. In other words, you already have to know the name of the variable. In other words, the above code snippet is exactly the same as just

>>> 'foo'

Please see the other answers here that might be applicable to answer the question.

Answer 3

Even if variable values don't point back to the name, you have access to the list of every assigned variable and its value, so I'm astounded that only one person suggested looping through there to look for your var name.

Someone mentioned on that answer that you might have to walk the stack and check everyone's locals and globals to find foo, but if foo is assigned in the scope where you're calling this retrieve_name function, you can use inspect's current frame to get you all of those local variables.

My explanation might be a little bit too wordy (maybe I should've used a "foo" less words), but here's how it would look in code (Note that if there is more than one variable assigned to the same value, you will get both of those variable names):

import inspect

x, y, z = 1, 2, 3

def retrieve_name(var):
    callers_local_vars = inspect.currentframe().f_back.f_locals.items()
    return [var_name for var_name, var_val in callers_local_vars if var_val is var]

print(retrieve_name(y))

---

If you're calling this function from another function, something like:

def foo(bar):
    return retrieve_name(bar)

foo(baz)

And you want the baz instead of bar, you'll just need to go back a scope further. This can be done by adding an extra .f_back in the caller_local_vars initialization.

See an example here: ideone

Answer 4

The only objects in Python that have canonical names are modules, functions, and classes, and of course there is no guarantee that this canonical name has any meaning in any namespace after the function or class has been defined or the module imported. These names can also be modified after the objects are created so they may not always be particularly trustworthy.

What you want to do is not possible without recursively walking the tree of named objects; a name is a one-way reference to an object. A common or garden-variety Python object contains no references to its names. Imagine if every integer, every dict, every list, every Boolean needed to maintain a list of strings that represented names that referred to it! It would be an implementation nightmare, with little benefit to the programmer.

Answer 5

On python3, this function will get the outer most name in the stack:

import inspect


def retrieve_name(var):
        """
        Gets the name of var. Does it from the out most frame inner-wards.
        :param var: variable to get name from.
        :return: string
        """
        for fi in reversed(inspect.stack()):
            names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
            if len(names) > 0:
                return names[0]

It is useful anywhere on the code. Traverses the reversed stack looking for the first match.

Answer 6

I don't believe this is possible. Consider the following example:

>>> a = []
>>> b = a
>>> id(a)
140031712435664
>>> id(b)
140031712435664

The a and b point to the same object, but the object can't know what variables point to it.

Answer 7

>> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v == my_var][0]
>> my_var_name 
'my_var'

In case you get an error if myvar points to another variable, try this (suggested by @mherzog)-

 >> my_var = 5
>> my_var_name = [ k for k,v in locals().items() if v is my_var][0]
>> my_var_name 
'my_var'

locals() - Return a dictionary containing the current scope's local variables. by iterating through this dictionary we can check the key which has a value equal to the defined variable, just extracting the key will give us the text of variable in string format.

from (after a bit changes) https://www.tutorialspoint.com/How-to-get-a-variable-name-as-a-string-in-Python

Answer 8

def name(**variables):
    return [x for x in variables]

It's used like this:

name(variable=variable)

Answer 9

I wrote the package sorcery to do this kind of magic robustly. You can write:

from sorcery import dict_of

columns = dict_of(n_jobs, users, queues, priorities)

and pass that to the dataframe constructor. It's equivalent to:

columns = dict(n_jobs=n_jobs, users=users, queues=queues, priorities=priorities)

Answer 10

Here's one approach. I wouldn't recommend this for anything important, because it'll be quite brittle. But it can be done.

Create a function that uses the inspect module to find the source code that called it. Then you can parse the source code to identify the variable names that you want to retrieve. For example, here's a function called autodict that takes a list of variables and returns a dictionary mapping variable names to their values. E.g.:

x = 'foo'
y = 'bar'
d = autodict(x, y)
print d

Would give:

{'x': 'foo', 'y': 'bar'}

Inspecting the source code itself is better than searching through the locals() or globals() because the latter approach doesn't tell you which of the variables are the ones you want.

At any rate, here's the code:

def autodict(*args):
    get_rid_of = ['autodict(', ',', ')', '\n']
    calling_code = inspect.getouterframes(inspect.currentframe())[1][4][0]
    calling_code = calling_code[calling_code.index('autodict'):]
    for garbage in get_rid_of:
        calling_code = calling_code.replace(garbage, '')
    var_names, var_values = calling_code.split(), args
    dyn_dict = {var_name: var_value for var_name, var_value in
                zip(var_names, var_values)}
    return dyn_dict

The action happens in the line with inspect.getouterframes, which returns the string within the code that called autodict.

The obvious downside to this sort of magic is that it makes assumptions about how the source code is structured. And of course, it won't work at all if it's run inside the interpreter.

Answer 11

This function will print variable name with its value:

import inspect

def print_this(var):
    callers_local_vars = inspect.currentframe().f_back.f_locals.items()
    print(str([k for k, v in callers_local_vars if v is var][0])+': '+str(var))

***Input & Function call:***
my_var = 10

print_this(my_var)

***Output**:*
my_var: 10

Answer 12

>>> locals()['foo']
{}
>>> globals()['foo']
{}

If you wanted to write your own function, it could be done such that you could check for a variable defined in locals then check globals. If nothing is found you could compare on id() to see if the variable points to the same location in memory.

If your variable is in a class, you could use className.dict.keys() or vars(self) to see if your variable has been defined.

Answer 13

I have a method, and while not the most efficient...it works! (and it doesn't involve any fancy modules).

Basically it compares your Variable's ID to globals() Variables' IDs, then returns the match's name.

def getVariableName(variable, globalVariables=globals().copy()):
    """ Get Variable Name as String by comparing its ID to globals() Variables' IDs

        args:
            variable(var): Variable to find name for (Obviously this variable has to exist)

        kwargs:
            globalVariables(dict): Copy of the globals() dict (Adding to Kwargs allows this function to work properly when imported from another .py)
    """
    for globalVariable in globalVariables:
        if id(variable) == id(globalVariables[globalVariable]): # If our Variable's ID matches this Global Variable's ID...
            return globalVariable # Return its name from the Globals() dict

Answer 14

In Python, the def and class keywords will bind a specific name to the object they define (function or class). Similarly, modules are given a name by virtue of being called something specific in the filesystem. In all three cases, there's an obvious way to assign a "canonical" name to the object in question.

However, for other kinds of objects, such a canonical name may simply not exist. For example, consider the elements of a list. The elements in the list are not individually named, and it is entirely possible that the only way to refer to them in a program is by using list indices on the containing list. If such a list of objects was passed into your function, you could not possibly assign meaningful identifiers to the values.

Python doesn't save the name on the left hand side of an assignment into the assigned object because:

1. It would require figuring out which name was "canonical" among multiple conflicting objects,
2. It would make no sense for objects which are never assigned to an explicit variable name,
3. It would be extremely inefficient,
4. Literally no other language in existence does that.

So, for example, functions defined using lambda will always have the "name" <lambda>, rather than a specific function name.

The best approach would be simply to ask the caller to pass in an (optional) list of names. If typing the '...','...' is too cumbersome, you could accept e.g. a single string containing a comma-separated list of names (like namedtuple does).

Answer 15

I think it's so difficult to do this in Python because of the simple fact that you never will not know the name of the variable you're using. So, in his example, you could do:

Instead of:

list_of_dicts = [n_jobs, users, queues, priorities]

dict_of_dicts = {"n_jobs" : n_jobs, "users" : users, "queues" : queues, "priorities" : priorities}

Answer 16

Many of the answers return just one variable name. But that won't work well if more than one variable have the same value. Here's a variation of Amr Sharaki's answer which returns multiple results if more variables have the same value.

def getVariableNames(variable):
    results = []
    globalVariables=globals().copy()
    for globalVariable in globalVariables:
        if id(variable) == id(globalVariables[globalVariable]):
            results.append(globalVariable)
    return results

a = 1
b = 1
getVariableNames(a)
# ['a', 'b']

Answer 17

just another way to do this based on the content of input variable:

(it returns the name of the first variable that matches to the input variable, otherwise None. One can modify it to get all variable names which are having the same content as input variable)

def retrieve_name(x, Vars=vars()):
    for k in Vars:
        if isinstance(x, type(Vars[k])):
            if x is Vars[k]:
                return k
    return None

Answer 18

If the goal is to help you keep track of your variables, you can write a simple function that labels the variable and returns its value and type. For example, suppose i_f=3.01 and you round it to an integer called i_n to use in a code, and then need a string i_s that will go into a report.

def whatis(string, x):
    print(string+' value=',repr(x),type(x))
    return string+' value='+repr(x)+repr(type(x))
i_f=3.01
i_n=int(i_f)
i_s=str(i_n)
i_l=[i_f, i_n, i_s]
i_u=(i_f, i_n, i_s)

## make report that identifies all types
report='\n'+20*'#'+'\nThis is the report:\n'
report+= whatis('i_f ',i_f)+'\n'
report+=whatis('i_n ',i_n)+'\n'
report+=whatis('i_s ',i_s)+'\n'
report+=whatis('i_l ',i_l)+'\n'
report+=whatis('i_u ',i_u)+'\n'
print(report)

This prints to the window at each call for debugging purposes and also yields a string for the written report. The only downside is that you have to type the variable twice each time you call the function.

I am a Python newbie and found this very useful way to log my efforts as I program and try to cope with all the objects in Python. One flaw is that whatis() fails if it calls a function described outside the procedure where it is used. For example, int(i_f) was a valid function call only because the int function is known to Python. You could call whatis() using int(i_f**2), but if for some strange reason you choose to define a function called int_squared it must be declared inside the procedure where whatis() is used.

Answer 19

Maybe this could be useful:

def Retriever(bar):
    return (list(globals().keys()))[list(map(lambda x: id(x), list(globals().values()))).index(id(bar))]

The function goes through the list of IDs of values from the global scope (the namespace could be edited), finds the index of the wanted/required var or function based on its ID, and then returns the name from the list of global names based on the acquired index.

Answer 20

Whenever I have to do it, mostly while communicating json schema and constants with the frontend I define a class as follows

class Param:
    def __init__(self, name, value):
        self.name = name
        self.value = value

Then define the variable with name and value.

frame_folder_count = Param({'name':'frame_folder_count', 'value':10})

Now you can access the name and value using the object.

>>> frame_folder_count.name
'frame_folder_count'

Answer 21

>>> def varname(v, scope=None):
        d = globals() if not scope else vars(scope); return [k for k in d if d[k] == v]
...
>>> d1 = {'a': 'ape'}; d2 = {'b': 'bear'}; d3 = {'c': 'cat'}
>>> ld = [d1, d2, d3]
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3']]
>>> d5 = d3
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]
>>> def varname(v, scope=None):
        d = globals() if not scope else vars(scope); return [k for k in d if d[k] is v]
...
>>> [varname(d) for d in ld]
[['d1'], ['d2'], ['d3', 'd5']]

As you see and is noted here, there can be multiple variables with the same value or even address, so using a wrapper to keep the names with the data is best.

Answer 22

Following method will not return the name of variable but using this method you can create data frame easily if variable is available in global scope.

class CustomDict(dict):
    def __add__(self, other):
        return CustomDict({**self, **other})

class GlobalBase(type):
    def __getattr__(cls, key):
        return CustomDict({key: globals()[key]})

    def __getitem__(cls, keys):
        return CustomDict({key: globals()[key] for key in keys})

class G(metaclass=GlobalBase):
    pass

x, y, z = 0, 1, 2

print('method 1:', G['x', 'y', 'z']) # Outcome: method 1: {'x': 0, 'y': 1, 'z': 2}
print('method 2:', G.x + G.y + G.z) # Outcome: method 2: {'x': 0, 'y': 1, 'z': 2}

A = [0, 1]
B = [1, 2]
pd.DataFrame(G.A + G.B) # It will return a data frame with A and B columns

Answer 23

I try to get name from inspect locals, but it cann't process var likes a[1], b.val. After it, I got a new idea --- get var name from the code, and I try it succ! code like below:

#direct get from called function code
def retrieve_name_ex(var):
    stacks = inspect.stack()
    try:
        func = stacks[0].function
        code = stacks[1].code_context[0]
        s = code.index(func)
        s = code.index("(", s + len(func)) + 1
        e = code.index(")", s)
        return code[s:e].strip()
    except:
        return ""

Answer 24

You can try the following to retrieve the name of a function you defined (does not work for built-in functions though):

import re
def retrieve_name(func):
    return re.match("<function\s+(\w+)\s+at.*", str(func)).group(1)

def foo(x):
    return x**2

print(retrieve_name(foo))
# foo

Answer 25

When finding the name of a variable from its value,
you may have several variables equal to the same value,
for example var1 = 'hello' and var2 = 'hello'.

My solution:

def find_var_name(val):

    dict_list = []
    global_dict = dict(globals())

    for k, v in global_dict.items():
        dict_list.append([k, v])
   
    return [item[0] for item in dict_list if item[1] == val]

var1 = 'hello'
var2 = 'hello'
find_var_name('hello')

Outputs

['var1', 'var2']

Answer 26

Compressed version of iDilip's answer:

import inspect
def varname(x):
  return [k for k,v in inspect.currentframe().f_back.f_locals.items() if v is x][0]

hi = 123
print(varname(hi))

Answer 27

Some of the previous cases would fail if there are two variables with the same value. So it is convenient to alert it:

Defining function:

# Variable to string of variable name

def var_name(variable,i=0):

  results = []
  for name in globals():   
     if eval(name) == variable:
       results.append(name)

  if len(results) > 1:
    print('Warning:' )
    print('   var_name() has found',len(results), 'possible outcomes.')
    print('   Please choose the suitable parameter "i". Where "i" is the index')
    print('   that matches your choice from the list below.')
    print('  ',results) ; print('')

  return results[i]

Use:

var_1 = 10
var_name(var_1) # Output will be "var_1"

If you have 2 variables with the same value like var_1 = 8 and var_2 = 8, then a warning will appear.

var_1 = 8
var_2 = 8
var_name(var_2)  # Output will be "var_1" too but Warning will appear

Answer 28

It's totally possible to get the name of an instance variable, so long as it is the property of a class.

I got this from Effective Python by Brett Slatkin. Hope it helps someone:

The class must implement the get, set, and set_name dunder methods, which are part of the "Descriptor Protocol"

This worked when I ran it:

class FieldThatKnowsItsName():
    def __init__(self):
        self.name = None
        self._value= None
        self.owner = None
 
    def __set_name__(self, owner, name):
        self.name = name
        self.owner = owner
        self.owner.fields[self.name] = self

    def __get__(self, instance, instance_type):
        return self

    def __set__(self, instance, value):
        self = value

class SuperTable:
    fields = {}
    field_1=FieldThatKnowsItsName()
    field_2=FieldThatKnowsItsName()

table = SuperTable()
print(table.field_1.name)
print(table.field_2.name)

You can then add methods and or extend your datatype as you like.

As a bonus, the set_name(self, owner, name) dunder also passes the parent instance, so the Field class instance can register itself with the parent.

I got this from Effective Python by Brett Slatkin. It took a while to figure out how to implement.

Answer 29

You can get your variable as kwargs and return it as string:

var=2
def getVarName(**kwargs):
    return list(kwargs.keys())[0]

print (getVarName(var = var))

Note: variable name must be equal to itself.