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'
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]
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}
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')])
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