Simpler way to create dictionary of separate varia

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 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

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 v is a:
             a_as_str = k
 >>> a_as_str
 a
 >>> type(a_as_str)
 'str'

Answer 2

I've wanted to do this quite a lot. This hack is very similar to rlotun's suggestion, but it's a one-liner, which is important to me:

blah = 1
blah_name = [ k for k,v in locals().iteritems() if v is blah][0]

Python 3+

blah = 1
blah_name = [ k for k,v in locals().items() if v is blah][0]

Answer 3

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 4

Well, I encountered the very same need a few days ago and had to get a variable's name which was pointing to the object itself.

And why was it so necessary?

In short I was building a plug-in for Maya. The core plug-in was built using C++ but the GUI is drawn through Python(as its not processor intensive). Since I, as yet, don't know how to return multiple values from the plug-in except the default MStatus, therefore to update a dictionary in Python I had to pass the the name of the variable, pointing to the object implementing the GUI and which contained the dictionary itself, to the plug-in and then use the MGlobal::executePythonCommand() to update the dictionary from the global scope of Maya.

To do that what I did was something like:

import time

class foo(bar):

    def __init__(self):
        super(foo, self).__init__()
        self.time = time.time() #almost guaranteed to be unique on a single computer

    def name(self):
        g = globals()
        for x in g:
            if isinstance(g[x], type(self)):
                if g[x].time == self.time:
                    return x
                    #or you could:
                    #return filter(None,[x if g[x].time == self.time else None for x in g if isinstance(g[x], type(self))])
                    #and return all keys pointing to object itself

I know that it is not the perfect solution in in the globals many keys could be pointing to the same object e.g.:

a = foo()
b = a
b.name()
>>>b
or
>>>a

and that the approach isn't thread-safe. Correct me if I am wrong.

At least this approach solved my problem by getting the name of any variable in the global scope which pointed to the object itself and pass it over to the plug-in, as argument, for it use internally.

I tried this on int (the primitive integer class) but the problem is that these primitive classes don't get bypassed (please correct the technical terminology used if its wrong). You could re-implement int and then do int = foo but a = 3 will never be an object of foo but of the primitive. To overcome that you have to a = foo(3) to get a.name() to work.

Answer 5

Python3. Use inspect to capture the calling local namespace then use ideas presented here. Can return more than one answer as has been pointed out.

def varname(var):
  import inspect
  frame = inspect.currentframe()
  var_id = id(var)
  for name in frame.f_back.f_locals.keys():
    try:
      if id(eval(name)) == var_id:
        return(name)
    except:
      pass

Answer 6

Maybe I'm overthinking this but..

str_l = next((k for k,v in locals().items() if id(l) == id(v)))


>>> bar = True
>>> foo = False
>>> my_dict=dict(bar=bar, foo=foo)
>>> next((k for k,v in locals().items() if id(bar) == id(v)))
'bar'
>>> next((k for k,v in locals().items() if id(foo) == id(v)))
'foo'
>>> next((k for k,v in locals().items() if id(my_dict) == id(v)))
'my_dict'