Subject: Re: iterating over DictRow
To: Karsten Hilbert <Karsten.Hilbert@gmx.net>, psycopg@lists.postgresql.org
References: <20200923215444.GA25406@hermes.hilbert.loc>
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From: Adrian Klaver <adrian.klaver@aklaver.com>
Message-ID: <65a4f678-4c7f-d88b-3709-e0686a561461@aklaver.com>
Date: Fri, 25 Sep 2020 15:19:12 -0700
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On 9/25/20 2:16 PM, Karsten Hilbert wrote:
> On Fri, Sep 25, 2020 at 09:06:43AM -0700, Adrian Klaver wrote:
> 
>>> In py2 one *had* to do DictRow.keys() to iterate over the
>>> keys. In py3
>>>
>>> 	for key in DictRow:
>>>
>>> is the suggested idiom for that which, however, iterates over
>>> DictRow as a list (as it always did).
>>>
>>> DictRow.keys() still exists on dicts in py3 (and is not
>>> deprec(i?)ated to my knowledge) but now returns a memoryview
>>> (dict_keys, that is) rather than a list, which brings with it
>>> its own set of issues (dict and keys "list" are not
>>> independant objects anymore).
>>>
>>> So, neither using py2's
>>>
>>> 	for key in DictRow.keys():
>>>
>>> under py3 nor changing to py3's
>>>
>>> 	for key in DictRow:			# beep: variable wrongly named
>>>
>>> leads to fully equivalent code. So this is a py2/py3 Gotcha
>>> in psycopg2.
>>
>> Well you can do, borrowing from previous example:
>>
>> for ky in r0._index:
>>      print(ky)
>>
>> for ky in r0._index:
>>      print(r0[ky])
>>
>> Where _index is a substitute for *.keys().
> 
> Sure, there's a number of solutions to my immediate problem,
> the fitting of which is
> 
> 	for key in dict(DictRow):
> 
> That's the best fit because my
> 
> 	def _escape_dict(the_dict, ...):
> 
> was inaptly named. It should have been (and now is)
> 
> 	def _escape_dict_like(dict_like, ...):
> 
> within which
> 
> 	dict(dict_like)
> 
> is quite the thing to do despite having to make something a
> duck which already nearly quacks like one is somehwat
> unfortunate.

I'm pretty sure DictRow has had the same behavior for some time so:

Are you migrating from Python 2?

Or what changed that made this show up?


> 
> Karsten
> --
> GPG  40BE 5B0E C98E 1713 AFA6  5BC0 3BEA AC80 7D4F C89B
> 
> 


-- 
Adrian Klaver
adrian.klaver@aklaver.com