[Python-ideas] Re: Extract variable name from itself
On 12/09/2023 11:54, Dom Grigonis wrote:
Yes, Thank you!
So 2 solutions now. They both solve what I have encountered. Beyond that, they
differ by:
a) f-string print(f’{=a.method}’) # ‘a.method’
No new builtin needed.
Simply reprints expression representation.
I don't understand your semantics either. What would be the difference
between your proposed
print(f’{=a.method}’)
and simply writing
print('a.method')
?
Would it be just that the syntax inside the curly braces is checked for
legality,
so that
print(f'{=!?}')
would not be allowed (presumably it would cause a SyntaxError)
?
Or do you want to check that the expression can be evaluated at run time?
You could achieve that by simply writing
a.method
As for the example in your first post:
var = 710
variable_name = [k fork, v inlocals().items()ifv == 710][0]
print("Your variable name is "+ variable_name)
it does "work", but it doesn't make much sense with Python's semantics.
You could have two identifiers bound to the same object; which one you
got hold of would be essentially random.
Puzzled.
Rob Cliffe___
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[Python-ideas] Re: Extract variable name from itself
I agree.
What I meant is that it is unreliable with `v is 710` and even more unreliable
with `v == 710` as “==“ is less strict than “is”.
DG
> On 12 Sep 2023, at 21:03, Rob Cliffe wrote:
>
>
>>> As for the example in your first post:
>>> var = 710
>>> variable_name = [k for k, v in locals().items() if v == 710][0]
>>> print("Your variable name is " + variable_name)
>>>
>>> it does "work", but it doesn't make much sense with Python's semantics.
>>> You could have two identifiers bound to the same object; which one you got
>>> hold of would be essentially random.
>> Yes, if `==` was replaced by `is`. Currently it is even more random as it
>> would return the first one which evaluates __eq__() positively.
>>
> I'm not sure what you're saying. In:
> var1 = 710
> var2 = 710
> variable_names = [k for k, v in locals().items() if v is 710]
> Depending on the Python implementation, variable_names may be ['var1',
> 'var2'] or it may be empty (depending on whether 710 is interned). It could
> also in theory contain one of 'var1', 'var2' but not the other, though I
> would be surprised if that happened in practice. The behaviour is not
> guaranteed and should not be relied on.
> Rob Cliffe
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[Python-ideas] Re: Extract variable name from itself
As for the example in your first post:
var = 710
variable_name = [k fork, v inlocals().items()ifv == 710][0]
print("Your variable name is "+ variable_name)
it does "work", but it doesn't make much sense with Python's
semantics. You could have two identifiers bound to the same object;
which one you got hold of would be essentially random.
Yes, if `==` was replaced by `is`. Currently it is even more random as
it would return the first one which evaluates __eq__() positively.
I'm not sure what you're saying. In:
var1 = 710
var2 = 710
variable_names = [k for k, v in locals().items() if v is 710]
Depending on the Python implementation, variable_names may be ['var1',
'var2'] or it may be empty (depending on whether 710 is interned). It
could also in theory contain one of 'var1', 'var2' but not the other,
though I would be surprised if that happened in practice. The behaviour
is not guaranteed and should not be relied on.
Rob Cliffe___
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[Python-ideas] Re: Extract variable name from itself
Good points, thanks for reply.
> Or do you want to check that the expression can be evaluated at run time?
> You could achieve that by simply writing
> a.method
Yes, the point is that it would be checked for legality of expression itself.
Whether to check if it actually evaluates without an error or not would both
work. I see pros and cons in both cases. Maybe not evaluating is better as
evaluation can always be checked separately if needed and eventually will be
checked on actual usage anyway.
> As for the example in your first post:
> var = 710
> variable_name = [k for k, v in locals().items() if v == 710][0]
> print("Your variable name is " + variable_name)
>
> it does "work", but it doesn't make much sense with Python's semantics. You
> could have two identifiers bound to the same object; which one you got hold
> of would be essentially random.
Yes, if `==` was replaced by `is`. Currently it is even more random as it would
return the first one which evaluates __eq__() positively.
That's exactly why I started this. This was to illustrate that I could not find
any robust way to do it.
DG.
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[Python-ideas] Re: Extract variable name from itself
Yes, Thank you!
So 2 solutions now. They both solve what I have encountered. Beyond that, they
differ by:
a) f-string print(f’{=a.method}’) # ‘a.method’
No new builtin needed.
Simply reprints expression representation.
b) print(nameof(a.method)) # ‘method’
New builtin.
Has its own logic.
Its extra functionality on top of what I have suggested can already
achieved in python via a.method.__name__
I think f-string approach if optimal. It adds new feature, without
functionality overlap and doesn’t crowd the namespace. Also, implementation is
simpler as all what is required seems to be in the place already.
I have little preference regarding syntax, but
f’{=expression}’ does seem to make sense.
Regards,
DG
> On 12 Sep 2023, at 13:14, Valentin Berlier wrote:
>
> Do you mean something like C# nameof()?
> https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/operators/nameof
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[Python-ideas] Re: Extract variable name from itself
An addition to f-strings:
a = 1
b = 1
print(f'{a+b}') # '2'
print(f'{a+b=}') # 'a+b=2'
# I suggest adding this too
print(f'{=a+b}')# 'a+b'
It is different from putting it in quotes in a way how editors interpret it.
E.g. changing the variable name in PyCharm would unambiguously change f-string
as well, while it would not change the string literal.
Also, one could argue that it can just be extracted by
`f’{a+b=}’.split(‘=‘)[0]`, but it is of course sub-optimal given the
possibility that evaluation is expensive and not required.
DG.
> On 12 Sep 2023, at 13:06, Chris Angelico wrote:
>
> On Tue, 12 Sept 2023 at 20:05, Dom Grigonis wrote:
>>
>> Please read the next part of my e-mail too. It actually answer your question.
>>
>
> I did read it, and it didn't answer the question. Your desired
> semantics are not clear.
>
> ChrisA
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[Python-ideas] Re: Extract variable name from itself
Do you mean something like C# nameof()? https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/operators/nameof ___ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-le...@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/USPQDMEA7ACWH2T56ZYTWBGG7SMT3RIF/ Code of Conduct: http://python.org/psf/codeofconduct/
[Python-ideas] Re: Extract variable name from itself
On Tue, 12 Sept 2023 at 20:05, Dom Grigonis wrote: > > Please read the next part of my e-mail too. It actually answer your question. > I did read it, and it didn't answer the question. Your desired semantics are not clear. ChrisA ___ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-le...@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/P5RHZPHTVREEYOG2VSBZYLA6UU33T2X3/ Code of Conduct: http://python.org/psf/codeofconduct/
[Python-ideas] Re: Extract variable name from itself
Please read the next part of my e-mail too. It actually answer your question. > On 12 Sep 2023, at 13:00, Chris Angelico wrote: > > On Tue, 12 Sept 2023 at 19:51, Dom Grigonis wrote: >> >> It wouldn’t. I know this is weird and not in line of how things work. This >> is more about simply getting reference variable name in locals() from the >> reference itself. But how would one access the variable name, when once >> called it returns the object it points to, not itself. So this would >> obviously require an exception in parser itself. >> >> a = object() >> b = a >> print(a.__varname__) # ‘a' >> print(b.__varname__) # 'b' >> print((a + b).__varname__) # Undefined??? ‘a + b’? >> > > I don't understand your desired semantics. Do you just want to take > whatever is given and put it in quotes? > > ChrisA > ___ > Python-ideas mailing list -- python-ideas@python.org > To unsubscribe send an email to python-ideas-le...@python.org > https://mail.python.org/mailman3/lists/python-ideas.python.org/ > Message archived at > https://mail.python.org/archives/list/python-ideas@python.org/message/ZWO3FPC2G2AWA5TNINBX4ZHDMKFBWQZN/ > Code of Conduct: http://python.org/psf/codeofconduct/ ___ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-le...@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/U3R6MHLYKC6R7RX2LG7EACVLOPB6TH5N/ Code of Conduct: http://python.org/psf/codeofconduct/
[Python-ideas] Re: Extract variable name from itself
On Tue, 12 Sept 2023 at 19:51, Dom Grigonis wrote: > > It wouldn’t. I know this is weird and not in line of how things work. This is > more about simply getting reference variable name in locals() from the > reference itself. But how would one access the variable name, when once > called it returns the object it points to, not itself. So this would > obviously require an exception in parser itself. > > a = object() > b = a > print(a.__varname__) # ‘a' > print(b.__varname__) # 'b' > print((a + b).__varname__) # Undefined??? ‘a + b’? > I don't understand your desired semantics. Do you just want to take whatever is given and put it in quotes? ChrisA ___ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-le...@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/ZWO3FPC2G2AWA5TNINBX4ZHDMKFBWQZN/ Code of Conduct: http://python.org/psf/codeofconduct/
[Python-ideas] Re: Extract variable name from itself
It wouldn’t. I know this is weird and not in line of how things work. This is
more about simply getting reference variable name in locals() from the
reference itself. But how would one access the variable name, when once called
it returns the object it points to, not itself. So this would obviously require
an exception in parser itself.
a = object()
b = a
print(a.__varname__) # ‘a'
print(b.__varname__) # 'b'
print((a + b).__varname__) # Undefined??? ‘a + b’?
Actually, it could be easily achieved by what is happening in f-strings:
a = 1
b = 2
print(f'{a+b=}')# 'a+b=2'
# So I would like to be able to have something like
print(f'{=a+b}')# 'a+b'
So now it really looks just an addition to f-strings. They can print
1. value
2. `expression literal`=value
3. What I am asking is: `expression literal` (without value)
One can of course argue that this is the same as just typing it, but the
difference is that things within curly braces are interpreted as code by the
editor, which is the whole point.
DG.
> On 12 Sep 2023, at 12:24, Antoine Rozo wrote:
>
> Why would an object always be linked to a variable name, and why only one
> name?
>
> Le mar. 12 sept. 2023 à 10:23, Dom Grigonis > a écrit :
> As far as I understand, it is not possible. Is it?
>
> Something similar to:
>
> var = 710
> variable_name = [k for k, v in locals().items() if v == 710][0]
> print("Your variable name is " + variable_name)
>
> ,except robust.
>
> Possible ways to expose it:
> * builtin function `varname(obj: object) —> str`
> * object.__varname__ dunder (this one wouldn’t clutter general namespace)
>
> I am working on some code and just got to the point where it seems I could
> make use of it. The case is as follows:
>
> # What I do now:
> class A:
> defaults = dict()
>
> def set_default(self, a):
> self.defaults['a'] = a
>
> def make_something(self, a=None):
> a = a if a is not None else self.defaults.get('a')
> return Something(a)
>
> # What I would like to be able to do:
> class A:
> defaults = dict()
>
> def set_default(self, a):
> self.defaults[a.__varname__] = a
>
> def make_something(self, a=None):
> a = a if a is not None else self.defaults.get(a.__varname__)
> return Something(a)
>
> In this case I like second one, because it allows a tight coupling of
> variable names, which has the following benefits:
> 1. General simplification, where one doesn’t have to think variable name and
> its key value in dictionary. Code looks cleaner.
> 2. If one was to change a variable name of `set_default` it would immediately
> surface as an error.
> 3. For someone working with IDEs or making use of clever multiple selection
> tools, it would be easy to change all in one go.
>
> Any reasons why the above would not be a good practice?
> Any reasons why exposing it is not a good idea?
> Would this be difficult to achieve from python dev’s perspective?
>
>
> — This one can’t be solved with deferred eval :/ —
> Regards,
> DG
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[Python-ideas] Re: Have del return a value
Makes sense. On Tue, Sep 12, 2023 at 2:55 AM Rob Cliffe via Python-ideas < python-ideas@python.org> wrote: > > > On 08/09/2023 22:19, Christopher Barker wrote: > > On Fri, Sep 8, 2023 at 11:00 AM Barry Scott > wrote: > >> I see no need for del to return anything, you already have the reference >> in foo. >> The times that foo is dropped at module level are rare enough to not need >> special syntax. >> > > I agree - using del is very rare in my code. The use case of passing a > name into a function (or somewhere else?) and then immediately deleting it > is rare indeed -- not worth new syntax. > > -Chris > > > +1. The only time I can remember using del to delete a variable is after > creating a structure with a large memory footprint that is only needed > temporarily. > Best wishes > Rob Cliffe > ___ > Python-ideas mailing list -- python-ideas@python.org > To unsubscribe send an email to python-ideas-le...@python.org > https://mail.python.org/mailman3/lists/python-ideas.python.org/ > Message archived at > https://mail.python.org/archives/list/python-ideas@python.org/message/RMS3RID2R2NO4V5A3AOKJCUUOCMHXB4Z/ > Code of Conduct: http://python.org/psf/codeofconduct/ > ___ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-le...@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/LTPC4FNQFAWYNZEV5F63TOKU2H3ISUYH/ Code of Conduct: http://python.org/psf/codeofconduct/
[Python-ideas] Re: Extract variable name from itself
Why would an object always be linked to a variable name, and why only one
name?
Le mar. 12 sept. 2023 à 10:23, Dom Grigonis a
écrit :
> As far as I understand, it is not possible. Is it?
>
> Something similar to:
>
> var = 710
> variable_name = [k for k, v in locals().items() if v == 710][0]
> print("Your variable name is " + variable_name)
>
> ,except robust.
>
> Possible ways to expose it:
> * builtin function `varname(obj: object) —> str`
> * object.__varname__ dunder (this one wouldn’t clutter general namespace)
>
> I am working on some code and just got to the point where it seems I could
> make use of it. The case is as follows:
>
> # What I do now:class A:
> defaults = dict()
>
> def set_default(self, a):
> self.defaults['a'] = a
>
> def make_something(self, a=None):
> a = a if a is not None else self.defaults.get('a')
> return Something(a)
> # What I would like to be able to do:class A:
> defaults = dict()
>
> def set_default(self, a):
> self.defaults[a.__varname__] = a
>
> def make_something(self, a=None):
> a = a if a is not None else self.defaults.get(a.__varname__)
> return Something(a)
>
>
> In this case I like second one, because it allows a tight coupling of
> variable names, which has the following benefits:
> 1. General simplification, where one doesn’t have to think variable name
> and its key value in dictionary. Code looks cleaner.
> 2. If one was to change a variable name of `set_default` it would
> immediately surface as an error.
> 3. For someone working with IDEs or making use of clever multiple
> selection tools, it would be easy to change all in one go.
>
> Any reasons why the above would not be a good practice?
> Any reasons why exposing it is not a good idea?
> Would this be difficult to achieve from python dev’s perspective?
>
>
> — This one can’t be solved with deferred eval :/ —
> Regards,
> DG
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--
Antoine Rozo
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[Python-ideas] Extract variable name from itself
As far as I understand, it is not possible. Is it?
Something similar to:
var = 710
variable_name = [k for k, v in locals().items() if v == 710][0]
print("Your variable name is " + variable_name)
,except robust.
Possible ways to expose it:
* builtin function `varname(obj: object) —> str`
* object.__varname__ dunder (this one wouldn’t clutter general namespace)
I am working on some code and just got to the point where it seems I could make
use of it. The case is as follows:
# What I do now:
class A:
defaults = dict()
def set_default(self, a):
self.defaults['a'] = a
def make_something(self, a=None):
a = a if a is not None else self.defaults.get('a')
return Something(a)
# What I would like to be able to do:
class A:
defaults = dict()
def set_default(self, a):
self.defaults[a.__varname__] = a
def make_something(self, a=None):
a = a if a is not None else self.defaults.get(a.__varname__)
return Something(a)
In this case I like second one, because it allows a tight coupling of variable
names, which has the following benefits:
1. General simplification, where one doesn’t have to think variable name and
its key value in dictionary. Code looks cleaner.
2. If one was to change a variable name of `set_default` it would immediately
surface as an error.
3. For someone working with IDEs or making use of clever multiple selection
tools, it would be easy to change all in one go.
Any reasons why the above would not be a good practice?
Any reasons why exposing it is not a good idea?
Would this be difficult to achieve from python dev’s perspective?
— This one can’t be solved with deferred eval :/ —
Regards,
DG___
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[Python-ideas] Re: Have del return a value
On 08/09/2023 22:19, Christopher Barker wrote: On Fri, Sep 8, 2023 at 11:00 AM Barry Scott wrote: I see no need for del to return anything, you already have the reference in foo. The times that foo is dropped at module level are rare enough to not need special syntax. I agree - using del is very rare in my code. The use case of passing a name into a function (or somewhere else?) and then immediately deleting it is rare indeed -- not worth new syntax. -Chris +1. The only time I can remember using del to delete a variable is after creating a structure with a large memory footprint that is only needed temporarily. Best wishes Rob Cliffe___ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-le...@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/RMS3RID2R2NO4V5A3AOKJCUUOCMHXB4Z/ Code of Conduct: http://python.org/psf/codeofconduct/
