Re: django template loop collections.defaultdict(lambda: collections.defaultdict(list))

Thanks Tom for replying! 

The code used to create the dict is:

my_dict = collections.defaultdict(lambda: collections.defaultdict(list))

for obj in queryset:

    my_dict[int(obj.position.split('-')[0])][int(obj.position.split('-')[2])].append(obj)

Basically I have one field in the queryset called position (the format is N-N-N-N) and I need to create the dict as in my question.

Sincerely, I don't know how I can create the dict without using defaultdict.

D

On Saturday, August 24, 2013 2:51:56 PM UTC+2, tom wrote:

On 23 Aug 2013, at 16:58, Daviddd <davide...@gmail.com> wrote:

In my template I tried:

{% for key, groups in queryset.iteritems %}      groups = {{ groups }} <br>      {% for group_key, cols in groups.iteritems %}        cols = {{ group_key }} <br>        {% for objs in cols %}            {# rest of the code #}  

But only the first loop is evaluated

groups = (1, defaultdict(<type 'list'>, {1: [<Obj: Obj 1 by daviddd>, <Obj: Obj 2 by daviddd>, <Obj: Obj3 by daviddd>], 2: [<Obj: Obj 4 by daviddd>], 3: [<Obj: Obj 5 by daviddd>, <Obj: Obj 6 by daviddd>, <Obj: Obj 7 by daviddd>]}))    groups = (2, defaultdict(<type 'list'>, {1: [<Obj: Obj 7.7 by daviddd>]}))    groups = (3, defaultdict(<type 'list'>, {1: [<Obj: Obj 7.8 by daviddd>]}))  

Maybe I'm misreading, but aren't your inner "groupses" still defaultdicts? I think they need to be coerced to normal dicts in order for Django's template system to be able to properly iterate over them.

Tom

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