How to sort OrderedDict of OrderedDict - Python

Question

I'm trying to sort OrderedDict in OrderedDict by 'depth' key. Is there any solution to sort that Dictionary ?

OrderedDict([
  (2, OrderedDict([
    ('depth', 0),  
    ('height', 51), 
    ('width', 51),   
    ('id', 100)
  ])), 
  (1, OrderedDict([
    ('depth', 2),  
    ('height', 51), 
    ('width', 51),  
    ('id', 55)
  ])), 
  (0, OrderedDict([
    ('depth', 1),  
    ('height', 51), 
    ('width', 51),  
    ('id', 48)
  ])),
]) 

Sorted dict should look like this:

OrderedDict([
  (2, OrderedDict([
    ('depth', 0),  
    ('height', 51), 
    ('width', 51),   
    ('id', 100)
  ])), 
  (0, OrderedDict([
    ('depth', 1),  
    ('height', 51), 
    ('width', 51),  
    ('id', 48)
  ])),
  (1, OrderedDict([
    ('depth', 2),  
    ('height', 51), 
    ('width', 51),  
    ('id', 55)
  ])), 
]) 

any idea how to get it?

Answer 1

You'll have to create a new one since OrderedDict is sorted by insertion order.

In your case the code would look like this:

foo = OrderedDict(sorted(foo.iteritems(), key=lambda x: x[1]['depth']))

See http://docs.python.org/dev/library/collections.html#ordereddict-examples-and-recipes for more examples.

Note for Python 3 you will need to use .items() instead of .iteritems().

Answer 2

>>> OrderedDict(sorted(od.items(), key=lambda item: item[1]['depth']))

Answer 3

Sometimes you might want to keep the initial dictionary and not create a new one.

In that case you could do the following:

temp = sorted(list(foo.items()), key=lambda x: x[1]['depth'])
foo.clear()
foo.update(temp)