How can i sort this dict per datetime?

How can i sort this dict per datetime? Is it possible?

{
1: {'l':{'created_on': datetime.datetime(2011, 9, 29, 17,开发者_StackOverflow社区 39, 26)}},
2: {'l':{'created_on': datetime.datetime(2011, 9, 23, 17, 39, 26)}},
3: {'l':{'created_on': datetime.datetime(2011, 9, 30, 17, 39, 26)}}
}

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You can call the sorted() function on the values, with the datetime as the comparison key:

sorted(d.values(), key=lambda item: item["l"]["created_on"])

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You can get a sorted list of keys for your dictionary d (dictionaries cannot be sorted by themselves):

>>> sorted(d.keys(), key=lambda k: d[k]['l']['created_on'])
[2, 1, 3]

PS: If you really need a dictionary with sorted keys, the now standard OrderedDict type is your friend (as described in the comments by rocksportrocker and agf after Frédéric's answer).

---

You can not sort a dict() itself. The dict has no implict order:

>>> print dict(a=3, b=2, c=1)
{'a': 3, 'c': 1, 'b': 2}

If you want to keep the full structure of your dict, you have to flatten your dict dd into a list li first. Then you can sort with the datetime value as a key:

import operator 

dd = { 1: ..... }

li = [] 

for (i, it) in dd.items():
    for (l, it1) in it.items():
        for (c, dt) in it.items():
            li.append((i, l, c, dt))
li.sort(key=operator.itemgetter(3))

print li