[Python-ideas] Re: Extract variable name from itself

2023-09-14 Thread Dom Grigonis 
Good ideas, however not robust:
a = 1
b = 1
print(id(a))# 4536318072
print(id(b))# 4536318072

> On 14 Sep 2023, at 16:35, Jonathan Fine  wrote:
> 
> POSTSCRIPT
> 
> We can also use locals() to 'inverse search' to get the name, much as in the 
> original post.
> 
> >>> locals()['x']
> (0, 1, 2, 3)
> >>> id(x)
> 139910226553296
> >>> id(locals()['x'])
> 139910226553296
> -- 
> Jonathan
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[Python-ideas] Re: Extract variable name from itself

2023-09-14 Thread Stephen J. Turnbull 
Jonathan Fine writes:

 > We can also use locals() to 'inverse search' to get the name, much
 > as in the original post.

As has already been explained, locals() (and any namespace for that
matter) is a many-one mapping, and therefore the inverse is not
well-defined.

At least for the 'print(f"count is {count}")' example, you really need
the compiler's help to get it right, unless you are willing to do it
the other way around:

def debug_name(name: str) -> None:
print(f"{name} is {eval(name)}")

but in general that's fraught with all the problems of using eval().

Steve



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[Python-ideas] Re: Extract variable name from itself

2023-09-14 Thread Jonathan Fine 
POSTSCRIPT

We can also use locals() to 'inverse search' to get the name, much as in
the original post.

>>> locals()['x']
(0, 1, 2, 3)
>>> id(x)
139910226553296
>>> id(locals()['x'])
139910226553296
-- 
Jonathan
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[Python-ideas] Re: Extract variable name from itself

2023-09-14 Thread Jonathan Fine 
Perhaps use the id of an object instead of its value. Here's how it might
work

>>> store = {}

>>> def wibble(fn):
...   val = fn()
...   name = fn.__name__
...   store[id(val)] = name
...   return val

>>> @wibble
... def ddd():
... return tuple(range(4))

>>> ddd
(0, 1, 2, 3)
>>> store[id(ddd)]
'ddd'
>>>

>>> x = ddd
>>> ddd = None
>>> store[id(x)]
'ddd'

I hope this helps.

-- 
Jonathan
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