I concur about the weakness in Django, when it is, as you say, a
relatively simple SQL statement to pull in that data.
I'd go for the list comp idea if you've got a small data set.
Alternatively you could select all the answers and then use
itertools.groupby to group by question. That's also efficient as it
won't use a lot of memory to do the transformation.
Euan
On Jun 22, 2:01 am, JeffH <holtzma...@gmail.com> wrote:
> Doing a nested list comp has to be less efficient than an outer join
> done at the db level. For my issue, it's a relatively small data set,
> so I'd rather be more pythonic than eg running raw sql. Nonetheless, I
> view this as a weakness in Django's ORM, and would plead with TPTB to
> provide a solution, not that it would affect my current problem.
>
> On Jun 21, 11:46 am, "euan.godd...@googlemail.com"
>
> <euan.godd...@gmail.com> wrote:
> > AFAIK there is no direct way to do this sort of thing in a single
> > query. select_related will only get foreign key relations into the
> > query so the ORM will always do an SQL query for each row in your many
> > to many. In these types of situations I tend to select everything I
> > need and then run some sort of pre-processing to put the two sets
> > together, thus ensuring only 2 queries rather than n+1 (where n is the
> > number of items in your original queryset).
>
> > Scott Gould's answer is about as good as you'll get I reckon.
>
> > On Jun 21, 3:37 pm, JeffH <holtzma...@gmail.com> wrote:
>
> > > That looks reasonable... but I wonder if the ORM can do it directly
> > > somehow. Anyone?
>
> > > On Jun 21, 10:14 am, Scott Gould <zinck...@gmail.com> wrote:
>
> > > > There may well be a better way to do this, especially since it's been
> > > > a good year since I was struggling with this myself. (Very similar
> > > > case to yours, different subject matter of course.)
>
> > > > The way I ended up doing it was to use a template tag and some list
> > > > comprehensions to whittle things down. E.g.:
>
> > > > questions = Questions.objects.all()
> > > > answers = Answers.objects.filter(candidate=my_candidate)
>
> > > > questions_and_answers = [(q, [a for a in answers if a.question = q])
> > > > for q in questions]
>
> > > > ...which should give you a list of (question, <list of answers>)
> > > > tuples.
>
> > > > On Jun 21, 10:00 am, JeffH <holtzma...@gmail.com> wrote:
>
> > > > > To clarify: Each race has a set of questions. The candidate may have
> > > > > responded to none, some, or all. The answers are linked to the
> > > > > candidate (and to the question). For each candidate, I want to display
> > > > > all the questions, with or without answer. The way it works currently,
> > > > > only the questions with answers get displayed.
>
> > > > > On Jun 21, 8:41 am, Daniel Roseman <dan...@roseman.org.uk> wrote:
>
> > > > > > On Jun 21, 12:51 pm, JeffH <holtzma...@gmail.com> wrote:
>
> > > > > > > I have some models that (simplified) look like the following.
>
> > > > > > > class Answer(models.Model):
> > > > > > > id = models.CharField(max_length=32, primary_key=True)
> > > > > > > text = models.TextField(blank=False)
> > > > > > > question = models.ForeignKey(Question)
> > > > > > > candidate = models.ForeignKey(Candidate)
>
> > > > > > > class Question(models.Model):
> > > > > > > id = models.CharField(max_length=32, primary_key=True)
> > > > > > > text = models.TextField(blank=False)
>
> > > > > > > class Candidate(models.Model):
> > > > > > > id = models.CharField(max_length=32, primary_key=True)
> > > > > > > name = models.CharField(max_length=32, blank=False)
>
> > > > > > > class Race(models.Model):
> > > > > > > id = models.CharField(max_length=32, primary_key=True)
> > > > > > > name = models.CharField(max_length=128, blank=False)
> > > > > > > questions = models.ManyToManyField(Question)
> > > > > > > candidates = models.ManyToManyField(Candidate)
>
> > > > > > > So, a Race has Candidates and Questions, and a Candidate has
> > > > > > > Answers.
> > > > > > > Each answer is associated with a Question and a Candidate.
> > > > > > > Displaying
> > > > > > > the question associated with an answer is easy:
>
> > > > > > > # context variable in view
> > > > > > > answers = Answer.objects.filter(candidate=candidate)
>
> > > > > > > # template code
> > > > > > > <table>
> > > > > > > {% for answer in answers %}
> > > > > > > <tr>
> > > > > > > <td>{{answer.question.text}}</td>
> > > > > > > <td>{{answer.text}}</td>
> > > > > > > </tr>
> > > > > > > {% endfor %}
> > > > > > > </table>
>
> > > > > > > From the point of view of the Candidate, I need to display all the
> > > > > > > questions, including the ones without Answers. I know how to to do
> > > > > > > this using raw sql and an outer join. How to do it in the orm?
>
> > > > > > > Thanks in advance for any ideas.
>
> > > > > > > --Jeff
>
> > > > > > Not quite enough information here to answer. What are you wanting to
> > > > > > join? If you just want to display all the questions, why do you
> > > > > > need a
> > > > > > join at all?
> > > > > > --
> > > > > > DR.
--
You received this message because you are subscribed to the Google Groups
"Django users" group.
To post to this group, send email to django-us...@googlegroups.com.
To unsubscribe from this group, send email to
django-users+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/django-users?hl=en.
