How to create this nested python dictionary?

Question

This is the what-I-want ( warning: Python noob here, this might be an incorrect way of representing, but I hope you get what I want )

forms = { 
            { 'id' : '1',
              'form' : form_something,
            },
            { 'id'  : '4',
              'form' : form_something2,
            }
        }

Now my question is how do I create this dictionary in my django view which goes like this, so far->

links = Links.objects.all()
forms = {}

for link in links:
    form = LinkForm(instance = link )
    forms.update({ 'id' : link.id, 'form' : form})
    # which is obviously the wrong way to do it

your forms is not a valid python dictionary. It is a set of two dictionaries, and you can't do it, because dictionaries are not hashable. — Ruggero Turra
– Ruggero Turra, Commented Jul 23, 2011 at 9:45

agf · Accepted Answer · 2011-07-23 08:57:23Z

2

This will create a list of dictionaries:

links = Links.objects.all()
forms = []
for link in links:
    form = LinkForm(instance = link)
    forms.append({'id': link.id, 'form': form})

If you want to create a dictionary of dictionaries, you have to have keys:

links = Links.objects.all()
forms = {}
for link in links:
    form = LinkForm(instance = link)
    # you need something to use as a key
    forms[key] = {'id': link.id, 'form': form}

Notice I changed where you had spaces in your code to match the standard Python way, but it doesn't really matter.

The form of the nested dictionary would be:

forms = { 
       'key1': { 'id' : '1',
          'form' : form_something,
        },
       'key2': { 'id'  : '4',
          'form' : form_something2
        }
    }

I added keys and removed the comma from after form_something2.

edited Jul 23, 2011 at 8:57

answered Jul 23, 2011 at 8:51

agf

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Comments

K Mehta · Accepted Answer · 2011-07-23 09:46:16Z

1

I've never worked with django, but in pure python, it'd go something like (you want a list/tuple of forms, not a dictionary of forms):

links = Links.objects.all()
forms = []
for link in links:
  form = LinkForm(instance = link)
  # the comma tells python to treat the contents inside the ( ) as a list
  forms.append({ 'id' : link.id, 'form' : form })

EDIT:

Based on the discussion in the comments, lists are a better option in comparison to tuples in terms of performance. Edited the code above to reflect this change.

edited Jul 23, 2011 at 9:46

answered Jul 23, 2011 at 8:54

K Mehta

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6 Comments

Sai Krishna Over a year ago

The thing is I've got to use this dictionary in django templates, not sure if django template renders lists

agf Over a year ago

This code doesn't work at all... a tuple is immutable, so you can't add to it, and the syntax for an empty tuple is tuple() not (). Also, only use += it when you have multiple items, it's the equivalent of extend. If you have one item, use append.

K Mehta Over a year ago

I've tested the code, and it definitely runs on python 2.6. not sure if its any different in django

K Mehta Over a year ago

but I'll keep your pointers in mind the next time I code in python :)

agf Over a year ago

If you do id(forms) after each time, you'll see you're not adding to it, you're making a copy with one new item every time. With a list you can just add to the same one. I was wrong about () -- I totally forgot that worked.

|

Hassek · Accepted Answer · 2011-07-23 09:10:12Z

0

i think what you want to do is something like this:

forms = { '1': form_something,
      '4':form_something2,
     }

This way you will look for your forms with their id (which is a common way to find things)

you can do this this way:

forms = {}

for link in Links.objects.all():
    form = LinkForm(instance = link )
    forms[form.id] = form`

Now I am curious why you want to save forms in a dictionary? seems odd to me.

Hope this helps

answered Jul 23, 2011 at 9:10

Hassek

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1 Comment

agf Over a year ago

Django templates take a dictionary of objects to render.

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