python - 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

Using the python-varname package, you can easily retrieve the name of the variables

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

As of v0.6.0, in your case, you can do:

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

Update: 07/08/2021

To get the name of a data frame, you can write your own wrapper using argname2 (as of v0.7.1):

>>> # argname is superseded by argname2
>>> from varname import argname2 
>>> from pandas import DataFrame
>>> 
>>> df1 = DataFrame(dict(a=[1]))
>>> df2 = DataFrame(dict(b=[1]))
>>> 
>>> def handle_dfs(*dfs):
...     names = argname2('*dfs')
...     for name, df in zip(names, dfs):
...         df.__dfname__ = name
...     
...     for df in dfs:
...         print(f'name: {df.__dfname__}, value: {df}')
... 
>>> handle_dfs(df1, df2)
name: df1, value:    a
0  1
name: df2, value:    b
0  1

---

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

Answer 2

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

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

Answer 3

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 4

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 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

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

It's used like this:

name(variable=variable)

Answer 8

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 9

>> 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 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

>>> 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 12

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 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

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 16

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 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

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 20

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 21

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 22

>>> 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 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']