How to sort OrderedDict of OrderedDict?
PythonSortingNestedOrdereddictionaryPython Problem Overview
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?
Python Solutions
Solution 1 - Python
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().
Solution 2 - Python
>>> OrderedDict(sorted(od.items(), key=lambda item: item[1]['depth']))
Solution 3 - Python
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)