I am trying to sort a dictionary containing dictionaries. Here is that dictionary:
mydict = {
'b': {'play': 2, 'scratch': 5, 'face': 8},
'c': {'do': 4, 'chew': 6},
'a': {'wash': 1, 'drink': 10, 'give': 20, 'eat': 30}
}
I want the following result after sorting:
{
'a': {'eat': 30, 'give': 20, 'drink': 10, 'wash': 1},
'b': {'face': 8, 'scratch': 5, 'play': 2},
'c': {'chew': 6, 'do': 4}
}
I will appreciate if you tell me how to solve this issue.
Let's start with your dictionary:
>>> mydict = {
... 'b': {'play': 2, 'scratch': 5, 'face': 8},
... 'c': {'do': 4, 'chew': 6},
... 'a': {'wash': 1, 'drink': 10, 'give': 20, 'eat': 30}
... }
Ordinary dictionaries are unordered. Ordered dictionaries, however are available from the collections module:
>>> from collections import OrderedDict
We can convert your dictionary to an ordered dictionary as follows:
>>> d = OrderedDict(sorted(mydict.items()))
>>> d
OrderedDict([('a', {'give': 20, 'drink': 10, 'eat': 30, 'wash': 1}), ('b', {'scratch': 5, 'play': 2, 'face': 8}), ('c', {'do': 4, 'chew': 6})])
As you can see above, d is ordered as we want. Alternatively, we can look at just the keys and verify they are in the order that we want:
>>> d.keys()
odict_keys(['a', 'b', 'c'])
In other ways, our ordered dictionary d behaves just like a regular dictionary:
>>> d['a']
{'give': 20, 'drink': 10, 'eat': 30, 'wash': 1}
If we want the dictionaries inside mydict to be sorted in descending order of value, we use an OrderedDict again:
>>> mydict['a']
{'give': 20, 'drink': 10, 'eat': 30, 'wash': 1}
>>> OrderedDict(sorted(mydict['a'].items(), key=lambda v: -v[-1]))
OrderedDict([('eat', 30), ('give', 20), ('drink', 10), ('wash', 1)])
If we want to apply this ordering to all entries of mydict:
>>> d = OrderedDict( sorted( (key1, OrderedDict(sorted(value.items(), key=lambda v: -v[-1]))) for (key1, value) in mydict.items()) )
>>> d
OrderedDict([('a', OrderedDict([('eat', 30), ('give', 20), ('drink', 10), ('wash', 1)])), ('b', OrderedDict([('face', 8), ('scratch', 5), ('play', 2)])), ('c', OrderedDict([('chew', 6), ('do', 4)]))])
