138

Say I have a variable named choice it is equal to 2. How would I access the name of the variable? Something equivalent to

In [53]: namestr(choice)
Out[53]: 'choice'

for use in making a dictionary. There's a good way to do this and I'm just missing it.

EDIT:

The reason to do this is thus. I am running some data analysis stuff where I call the program with multiple parameters that I would like to tweak, or not tweak, at runtime. I read in the parameters I used in the last run from a .config file formated as

filename
no_sig_resonance.dat

mass_peak
700

choice
1,2,3

When prompted for values, the previously used is displayed and an empty string input will use the previously used value.

My question comes about because when it comes to writing the dictionary that these values have been scanned into. If a parameter is needed I run get_param which accesses the file and finds the parameter.

I think I will avoid the problem all together by reading the .config file once and producing a dictionary from that. I avoided that originally for... reasons I no longer remember. Perfect situation to update my code!

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8
  • 1
    python.net/~goodger/projects/pycon/2007/idiomatic/… Commented Feb 26, 2009 at 22:33
  • The link provides some pictures that in 2 minutes will teach you what are names in Python. Commented Feb 26, 2009 at 22:34
  • 1
    also related: stackoverflow.com/questions/932818/… Commented Dec 17, 2013 at 0:11
  • 22
    There is little worse on Stack Overflow than people questioning the existence of a question itself. Commented Jan 17, 2020 at 22:28
  • 2
    Finally, a perfectly valid solution with Python 3.8 f-string = syntax : stackoverflow.com/a/57225950/4080671 Commented Oct 17, 2020 at 19:32

8 Answers 8

Reset to default
156

If you insist, here is some horrible inspect-based solution.

import inspect, re

def varname(p):
  for line in inspect.getframeinfo(inspect.currentframe().f_back)[3]:
    m = re.search(r'\bvarname\s*\(\s*([A-Za-z_][A-Za-z0-9_]*)\s*\)', line)
    if m:
      return m.group(1)

if __name__ == '__main__':
  spam = 42
  print varname(spam)

I hope it will inspire you to reevaluate the problem you have and look for another approach.

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6 Comments

There is related use case for inspect.currentframe().f_back.f_locals
Sometimes a night of debugging makes us do things we're not proud of.
upvoted for literal LOL
I like how Jake's comment comes 5 years after the question and yet it still has 70+ up votes of people that, like me, reluctantly searched how to do this shameless thing in a moment of desperation
Yes it does. When it is passed to the function varname the variable name is indeed x.
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121

To answer your original question:

def namestr(obj, namespace):
    return [name for name in namespace if namespace[name] is obj]

Example:

>>> a = 'some var'
>>> namestr(a, globals())
['a']

We can get caller's namespace using inspect module:

import inspect

def names_in_caller(obj, depth=2) -> list[str]:
    """Return *obj*'s names in the caller namespace.

    Default *depth* is 2 meaning that caller's caller names 
    are returned
    """ 
    f = inspect.currentframe()
    for _ in range(depth):
        f = f.f_back
    return namestr(obj, f.f_locals)

Example:

>>> def f(n):
...     n_name = name_in_caller(n)  # find name in the caller
...     return f"{n_name}={n}"  # like f"{n=}" but with callers' name
... 
>>> a = 1
>>> f(a)
"['a']=1"
>>> b = 1
>>> f(a)
"['a', 'b']=1"

As @rbright already pointed out whatever you do there are probably better ways to do it.

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9 Comments

This fails if there is more than one variable name referring to the same value.
@Greg Hewgill: You might have noticed that namestr returns list. It is a hint that there could be more than one name.
I'd say most integers don't share their representations.
I like the flow of your answer code, but I may be working with integers that may be repeated several times in the code... sigh
@Piotr Dobrogost: "share their representions" refers to CPython optimization that caches small integers therefore one of the sets is not countable it is finite.
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18

If you are trying to do this, it means you are doing something wrong. Consider using a dict instead.

def show_val(vals, name):
    print "Name:", name, "val:", vals[name]

vals = {'a': 1, 'b': 2}
show_val(vals, 'b')

Output:

Name: b val: 2
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3 Comments

+1: The original question is not sensible. It's a simple dictionary. Or -- perhaps -- use a different language.
Lets say I have choice = 'He chose 3!'. Can I add this to a dict without typing out "choice" when specifying the key?
key = "choice"; print vals[key]
11

Rather than ask for details to a specific solution, I recommend describing the problem you face; I think you'll get better answers. I say this since there's almost certainly a better way to do whatever it is you're trying to do. Accessing variable names in this way is not commonly needed to solve problems in any language.

That said, all of your variable names are already in dictionaries which are accessible through the built-in functions locals and globals. Use the correct one for the scope you are inspecting.

One of the few common idioms for inspecting these dictionaries is for easy string interpolation:

>>> first = 'John'
>>> last = 'Doe'
>>> print '%(first)s %(last)s' % globals()
John Doe

This sort of thing tends to be a bit more readable than the alternatives even though it requires inspecting variables by name.

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2 Comments

I want to try to and print 'first' and 'last', the values that I call the references, not what they are equal to.
I know, I was just letting you know one reason people commonly look at those dictionaries at all. Usually you shouldn't need to. Unless you want to give us more detail, just dig around in them for what you're looking for. E.g., globals().keys()
10

You can't, as there are no variables in Python but only names.

For example:

> a = [1,2,3]
> b = a
> a is b
True

Which of those two is now the correct variable? There's no difference between a and b.

There's been a similar question before.

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7 Comments

The difference between a and b is the name.
gs is right. a and b refer to the same object, so a function receiving this object has no way to know which name was used to refer to the object in the function call.
He's kinda right, in that there is no way for you to write a function in Python to differentiate. Still, there's no reason there couldn't be a built-in function that accomplishes this.
dis.dis() returns some names (not always).
"There are no variables in Python" Silly misinformation. You mean, "variables in Python work differently than in C."
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7

Will something like this work for you?

>>> def namestr(**kwargs):
...     for k,v in kwargs.items():
...       print "%s = %s" % (k, repr(v))
...
>>> namestr(a=1, b=2)
a = 1
b = 2

And in your example:

>>> choice = {'key': 24; 'data': None}
>>> namestr(choice=choice)
choice = {'data': None, 'key': 24}
>>> printvars(**globals())
__builtins__ = <module '__builtin__' (built-in)>
__name__ = '__main__'
__doc__ = None
namestr = <function namestr at 0xb7d8ec34>
choice = {'data': None, 'key': 24}
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Comments

5

With eager evaluation, variables essentially turn into their values any time you look at them (to paraphrase). That said, Python does have built-in namespaces. For example, locals() will return a dictionary mapping a function's variables' names to their values, and globals() does the same for a module. Thus:

for name, value in globals().items():
    if value is unknown_variable:
        ... do something with name

Note that you don't need to import anything to be able to access locals() and globals().

Also, if there are multiple aliases for a value, iterating through a namespace only finds the first one.

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2 Comments

This fails if there is more than one variable name with the same value.
Not much you can do about that, although I'll add a warning. ;)
3

For the revised question of how to read in configuration parameters, I'd strongly recommend saving yourself some time and effort and use ConfigParser or (my preferred tool) ConfigObj.

They can do everything you need, they're easy to use, and someone else has already worried about how to get them to work properly!

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