How to get a variable name as a string in Python?

Question

I would like to be able to get the name of a variable as a string but I don't know if Python has that much introspection capabilities. Something like:

>>> print(my_var.__name__)
'my_var'

I want to do that because I have a bunch of vars I'd like to turn into a dictionary like :

bar=True
foo=False
>>> my_dict=dict(bar=bar, foo=foo)
>>> print mydict
>>> print my_dict 
{'foo': False, 'bar': True}

But I'd like something more automatic than that.

Python have locals() and vars(), so I guess there is a way.

Answer 1

This is not possible in Python, which really doesn't have "variables". Python has names, and there can be more than one name for the same object.

Answer 2

Most objects don't have a __name__ attribute. (Classes, functions, and modules do; any more builtin types that have one?)

What else would you expect for print(my_var.__name__) other than print("my_var")? Can you simply use the string directly?

You could "slice" a dict:

def dict_slice(D, keys, default=None):
  return dict((k, D.get(k, default)) for k in keys)

print dict_slice(locals(), ["foo", "bar"])
# or use set literal syntax if you have a recent enough version:
print dict_slice(locals(), {"foo", "bar"})

Alternatively:

throw = object()  # sentinel
def dict_slice(D, keys, default=throw):
  def get(k):
    v = D.get(k, throw)
    if v is not throw:
      return v
    if default is throw:
      raise KeyError(k)
    return default
  return dict((k, get(k)) for k in keys)

Answer 3

As unwind said, this isn't really something you do in Python - variables are actually name mappings to objects.

However, here's one way to try and do it:

 >>> a = 1
 >>> for k, v in list(locals().iteritems()):
         if id(v) == id(a):
             a_as_str = k
 >>> a_as_str
 a
 >>> type(a_as_str)
 'str'

Answer 4

This is a hack. It will not work on all Python implementations distributions (in particular, those that do not have traceback.extract_stack.)

import traceback

def make_dict(*expr):
    (filename,line_number,function_name,text)=traceback.extract_stack()[-2]
    begin=text.find('make_dict(')+len('make_dict(')
    end=text.find(')',begin)
    text=[name.strip() for name in text[begin:end].split(',')]
    return dict(zip(text,expr))

bar=True
foo=False
print(make_dict(bar,foo))
# {'foo': False, 'bar': True}

Note that this hack is fragile:

make_dict(bar,
          foo)

(calling make_dict on 2 lines) will not work.

Instead of trying to generate the dict out of the values foo and bar, it would be much more Pythonic to generate the dict out of the string variable names 'foo' and 'bar':

dict([(name,locals()[name]) for name in ('foo','bar')])

Answer 5

Are you trying to do this?

dict( (name,eval(name)) for name in ['some','list','of','vars'] )

Example

>>> some= 1
>>> list= 2
>>> of= 3
>>> vars= 4
>>> dict( (name,eval(name)) for name in ['some','list','of','vars'] )
{'list': 2, 'some': 1, 'vars': 4, 'of': 3}