How to sort OrderedDict of OrderedDict?

问题

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?

回答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().

回答2:

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

回答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)

来源：https://stackoverflow.com/questions/8031418/how-to-sort-ordereddict-of-ordereddict

标签

python

sorting

nested

ordereddictionary