How to sort OrderedDict using a sorted list of keys?

Question

Suppose I have a collections.OrderedDict object and a re-arranged list of its keys:

ordereddict = collections.OrderedDict((
    ('key_78', 'value'),
    ('key_40', 'value'),
    ('key_96', 'value'),
    ('key_53', 'value'),
    ('key_04', 'value'),
    ('key_89', 'value'),
    ('key_52', 'value'),
    ('key_86', 'value'),
    ('key_16', 'value'),
    ('key_63', 'value'),
))

# Example only; actual list will **not** == sorted(ordereddict)
key_list = ['key_04', 'key_16', 'key_40', 'key_52', 'key_53', 'key_63', 'key_78', 'key_86', 'key_89', 'key_96']

How can I sort the OrderedDict so that it is ordered in the same way as the key_list?

Answer 1

Use the following:

def sort_by_list(dict_, list_):
    for key in list_:
        dict_.move_to_end(key)

sort_by_list(ordereddict, key_list)

This only works if the list_ contains all the keys in the dict, and on Python 3.2 or later.

Answer 2

Just create a new OrderedDict:

newdct = OrderedDict((key, olddct[key]) for key in sortedlist)

If you really need this to happen in place, you can clear the olddct and update it with the new one:

olddct.clear()
olddct.update(newdct)