[Python-ideas] [Python-Dev] Dict access with double-dot (syntactic sugar)

[Josiah Carlson]

On Fri, Mar 25, 2011 at 9:00 AM, Jameson Quinn <jameson.quinn at gmail.com> wrote:
> I realized that python already has a way to access the string-based members
> of a dict without using quotes:
> def expect_a_chair(chair, **kw):
>   print "Thanks. That chair is %s." % chair
>   if kw:
>     for key, val in kw.iteritems():
>       print "I wasn't expecting the (%s) %s!" % (val, key)
> d = json.loads('{"chair":"comfy","inquisition":"Spanish"}')
> expect_a_chair(**d)
> try:
>   expect_a_chair({})
> except TypeError:
>   print "No chair."
> The ** operator does this. Notice that nowhere in that python code (not
> counting the json) do I have to put "chair" in quotes.
> Honestly, this solves my current use case. I can use functions like
> expect_a_chair for everything I need right now.
> But perhaps, if there were a quote-free way to access string-based dict
> items, it should be based on this. The problem is, this ** operator is a
> unary operator, and the non-unary ** is already taken.
> So, I don't have a perfect name for my proposed quoteless synonym for
> '["..."]'. My best option is '*.'. Note that this could also be used for
> unpacking, and with defaults:
> d*.(x,y,z) #=== (d["x"], d["y"], d['z'])
> e=d*.(e=None) #like 'e=d.get("e", None)'.
> Does this sound worth-it to anyone?

Gut reaction:
Dear gods, how and why does * have a . operator, and why is there a
function being called on it?!?
Huh; x,y,z aren't defined in that scope, that's gotta return a
NameError for sure.
Or wait, is that some new way to specify a vector for multiplication?
When did Python get a vector type and literal?

If punctuation is the answer, Perl is the question. Since this is
Python, and punctuation is decidedly not the answer, I'm going to go a
little further than before: -sys.maxint

Regards,
 - Josiah