Robert Madole wrote:

So I just wrote my 1-gazillionth "if hasattr('method', my_object):" and wanted badly to write this.

if 'method' on my_object:

I'm not the most experienced Python user, but as far as I can tell hasattr is the preferred way of checking if a method exists.

Please correct me if there is a better way

Most Python code should just assume the method is present and let the AttributeError fly if it isn't.

If you genuinely need to check for presence of a method, then either catching the AttributeError or using hasattr is fine.

This is a relatively rare requirement though, so the chance of getting dedicated syntax (and a new keyword, no less!) to support it is basically nil.

Cheers, Nick.

I think this is a great idea, but I would not want to introduce a new keyword. How about

foo is in bar = hasattr(foo,bar) foo is not in bar = not hasattr(foo,bar)

But there are other common checks I do all the time that I would really like to have shortcuts for. Here are ways to do them without new keywords:

foo is class bar = isinstance(foo,bar) foo is from bar = issubclass(foo,bar) foo is with bar = type(foo) == type(bar)

Can we make an exception to the moratorium, just for today?

--- Bruce http://www.vroospeak.com

On Thu, Apr 1, 2010 at 8:15 AM, Robert Madole robmadole@gmail.com wrote:

So I just wrote my 1-gazillionth "if hasattr('method', my_object):" and wanted badly to write this.

   if 'method' on my_object:

I'm not the most experienced Python user, but as far as I can tell hasattr is the preferred way of checking if a method exists.

Please correct me if there is a better way, otherwise I'd like to see what everyone thinks about an "on" statement being added to the Python language.

Thanks,

Rob Madole _______________________________________________ Python-ideas mailing list Python-ideas@python.org http://mail.python.org/mailman/listinfo/python-ideas

Bruce Leban wrote:

I think this is a great idea, but I would not want to introduce a new keyword. How about

foo is in bar     = hasattr(foo,bar)
foo is not in bar = not hasattr(foo,bar)

But there are other common checks I do all the time that I would really like to have shortcuts for. Here are ways to do them without new keywords:

foo is class bar = isinstance(foo,bar)

To me, this is asking whether 'foo' is a class. Better would be:

foo in class bar

foo is from bar  = issubclass(foo,bar)

Perhaps:

foo from class bar

(although I'm not happy with it.)

foo is with bar  = type(foo) == type(bar)

Can we make an exception to the moratorium, just for today?

Only for foolish ideas. :-)

--- Bruce http://www.vroospeak.com

On Thu, Apr 1, 2010 at 8:15 AM, Robert Madole <robmadole@gmail.com mailto:robmadole@gmail.com> wrote:

So I just wrote my 1-gazillionth "if hasattr('method', my_object):"
and wanted badly to write this.

       if 'method' on my_object:

I'm not the most experienced Python user, but as far as I can tell
hasattr is the preferred way of checking if a method exists.

Please correct me if there is a better way, otherwise I'd like to
see what everyone thinks about an "on" statement being added to the
Python language.

Thanks,

Rob Madole
_______________________________________________
Python-ideas mailing list
Python-ideas@python.org <mailto:Python-ideas@python.org>
http://mail.python.org/mailman/listinfo/python-ideas

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Python-ideas mailing list Python-ideas@python.org http://mail.python.org/mailman/listinfo/python-ideas

On 2010-04-01 14:04 PM, Oleg Broytman wrote:

On Thu, Apr 01, 2010 at 11:54:53AM -0700, Bruce Leban wrote:

...

 foo is in bar     = hasattr(foo,bar)

foo is in bar

is currently a valid and widely used expression. It means bar.__contains__(foo).

No, it's not.

...

...

1 is in range(5)

File "<stdin>", line 1 1 is in range(5) ^ SyntaxError: invalid syntax

On Thu, Apr 1, 2010 at 8:54 PM, Bruce Leban bruce@leapyear.org wrote:

I think this is a great idea, but I would not want to introduce a new keyword. How about     foo is in bar     = hasattr(foo,bar)     foo is not in bar = not hasattr(foo,bar) But there are other common checks I do all the time that I would really like to have shortcuts for. Here are ways to do them without new keywords:     foo is class bar = isinstance(foo,bar)     foo is from bar  = issubclass(foo,bar)     foo is with bar  = type(foo) == type(bar) Can we make an exception to the moratorium, just for today?

I hope so! I'm tossing in a few more suggestions that would make some idioms more intuitive:

* iterable except for item := (x for x in iterable if x!=item)

* pass from iterable := for i in iterable: pass

* global return value := sys.exit(value)

* import obj in module as name

import module setattr(module, name, obj) del module

* not raise: BLOCK

try: BLOCK except: pass

* with x as is: BLOCK

_x = deepcopy(x) try: BLOCK finally: x = _x; del _x

* try for i in iterable: BLOCK1 else: BLOCK2

for i in iterable: try: BLOCK1; break except: pass else: BLOCK2

George

Am 01.04.2010 22:54, schrieb George Sakkis:

I hope so! I'm tossing in a few more suggestions that would make some idioms more intuitive:

• import obj in module as name

  import module setattr(module, name, obj) del module

My favorite!

Georg

On Sun, Apr 4, 2010 at 05:35, Greg Ewing greg.ewing@canterbury.ac.nzwrote:

+1. Also we must absolutely have

foo is totally bar

for those *$&@*&(*@& moments we all experience while coding.

Indeed. On that note, I think I'm going to start implementing Google's recently-announced Google Annotations Gallery as Python decorators. I actually did once implement their @noop...

http://google-opensource.blogspot.com/2010/03/google-annotations-gallery.htm...

On 2010-04-04 14:17 , Stefan Behnel wrote:

George Sakkis, 01.04.2010 22:54:

...

• with x as is:

BLOCK

_x = deepcopy(x) try: BLOCK finally: x = _x; del _x

I think this use case is important enough to implement it immediately. However, since the moratorium is currently in place for Python, would it be ok to fork Python 3 to integrate it into the language? I propose Mython as the name of the fork. I also propose to drop the entire Subversion history, so that we can start using Mercurial instantaneously. That will greatly improve the freedom of Mython as a language.

The name Mython is taken:

http://mython.org/

However, you can use the existing Mython to implement this syntax pretty easily.