2011/3/25 Josiah Carlson <josiah.carlson@gmail.com>

On Fri, Mar 25, 2011 at 9:00 AM, Jameson Quinn <jameson.quinn@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

OK. As I said, I think that the existing functionality as I showed in expect_a_chair solves my dislike for using quotes for something which logically is an attribute name. (It would be even better if json had a way to distinguish object-like data, with well-defined parameters, from more free-form data, but that is out of scope here.) Jameson ps. A true masochist could do something like my unpacking '*.' by a horrendous abuse of the lambda syntax: (lambda x,y,z,e=None, **kw:oh_please_god_no(locals()))(**d) I leave the definition of oh_please_god_no as an exercise for the reader. :)

2011/3/25 Josiah Carlson <josiah.carlson@gmail.com>

On Fri, Mar 25, 2011 at 9:00 AM, Jameson Quinn <jameson.quinn@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

OK. As I said, I think that the existing functionality as I showed in expect_a_chair solves my dislike for using quotes for something which logically is an attribute name. (It would be even better if json had a way to distinguish object-like data, with well-defined parameters, from more free-form data, but that is out of scope here.) Jameson ps. A true masochist could do something like my unpacking '*.' by a horrendous abuse of the lambda syntax: (lambda x,y,z,e=None, **kw:oh_please_god_no(locals()))(**d) I leave the definition of oh_please_god_no as an exercise for the reader. :)