So, jek, if you're listening, or anyone else -- is there an already
existing, working implementation of a Dict of Lists or Dict of Sets
collection class?


On Mar 10, 1:55 pm, Eric Ongerth <ericonge...@gmail.com> wrote:
> Ach, I did it again... proceeded as if column_mapped_collection and
> attribute_mapped_collection provide collection classes that are dicts
> of lists.  No, they're just dicts of scalar values!
>
> Again and again I need dicts of lists.  They seem to really fit the
> data that I tend to work with; A's related to C's indexed by B's (the
> B typically indicating the *type* of relationship between the A and
> the C).
>
> Eventually I need to work my own implementation of dict-of-lists and
> dict-of-dicts collection classes up to full strength so I can share
> them as a recipe or extension or something.
>
> On Mar 10, 1:02 pm, Eric Ongerth <ericonge...@gmail.com> wrote:
>
> > I make occasional use of alternate collection classes for my
> > sqlalchemy relationships.  For example, column_mapped_collection.  The
> > existing collection classes in sqlalchemy.orm.collections work really
> > well for me; I have a lot of relational data that very naturally
> > belongs in dicts rather than lists because it makes sense as one thing
> > indexed or mapped by another.
>
> > I find that I often need to add an item to one of these dict-mapped
> > collections and would rather not have to check whether the key already
> > exists in the dict.  So I'm doing, essentially:
> > collection.setdefault(key, []).append(value).
>
> > As a matter of syntactic sugar (which alternative collection classes
> > already are, anyway) I would like to have collection classes backed by
> > dicts with the capability of python's defaultdict, so that I could
> > write more simply, and with equivalent results:
> > collection[key].append(value)
>
> > I understand how to create my own collection class which will behave
> > as desired.  My question is, has anyone already got code for this that
> > I could borrow rather than reinvent the wheel?  Thanks in advance.  --
> > Eric

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