Sort dictionary of dictionaries by value Python ( 3 Level Dict )

Question

I have a dict like this:

{
 'INT-ABC1': 
    {
        'acc1': {'val': -22313.7381693064, 'Qua': -241.0}, 
        'acc2': {'val': -1312.940854148, 'Qua': -13.0}
    }, 
 'INT-ABC2': 
    {
        'acc1': {'val': -131.2510359399, 'Qua' : -23.0}, 
        'acc3': {'val': -131.40521409002, 'Qua' : -13.0},
        'acc5': {'val': -12312.7688190937, 'Qua' : -1313.0}
    }
}

I need to sort it by 'val' and get a similar dict out of it.

Can you show us what have you tried and where did you get stuck ? — Nir Alfasi, Apr 24 '15 at 16:45
What have you tried so far? What do you want the results to be? Do you want to print them in sorted order, do you want some other object(s) to contain some information about the sort order? Something else? — Eric Renouf, Apr 24 '15 at 16:57
possible duplicate of [Sort a Python dictionary by value](http://stackoverflow.com/questions/613183/sort-a-python-dictionary-by-value) — viki.omega9, Apr 24 '15 at 17:02
Python `dict`s are unsorted. If you want a sorted dictionary you should use `collections.OrderedDict` — kylieCatt, Apr 24 '15 at 20:03

viki.omega9 · Answer 1 · 2015-04-27T06:17:42.607

Consider using the sorted function. Use this as a reference. For the case that you have, you have to use a key function that tells python how to compare two entities of the dictionary you have in your question. The sorted function then returns a list that you would iterate.

For example,

>>>sorted(student_tuples, key=lambda student: student[2])   # sort by age
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]

For your case you need to get the items from the dict.

EDIT: First like one of the comments said, you can't directly sort a dict in python. I don't think an ordered dict is what you need. You can only sort a dict that has been converted to a list tuples. For example {'a': 'b', 'c': 'd'}, has to be first converted to [('a', 'b'), ('c', 'd')] and then you can choose to sort as you wish. The problem here is that once you do this, you can't store it back into a normal dict. The reason is that's not how normal dicts in work. You can use an ordereddict in the python collections that "remembers" the order of insert and therefore is a "sorted" dict. But without doing some work you cannot easily create a "sorted" dict.

For your input, the 'b' in my example is a dict. So you have to do some work to get the key that you want. I am assuming you want to sort between the 'acc' values for each 'INT-ABC'.

>>> k_dict
{'INT-ABC1': {'acc1': {'Qua': -241.0, 'val': -22313.7381693064}, 'acc2': {'Qua': -13.0, 'val': -1312.940854148}}}
>>> k_dict.items()
[('INT-ABC1', {'acc1': {'Qua': -241.0, 'val': -22313.7381693064}, 'acc2': {'Qua': -13.0, 'val': -1312.940854148}})]
>>> k_dict.items()[0][1].items()
[('acc1', {'Qua': -241.0, 'val': -22313.7381693064}), ('acc2', {'Qua': -13.0, 'val': -1312.940854148})]
>>> sorted(k_dict.items()[0][1].items(), key=lambda  x: x[1]['val'])
[('acc1', {'Qua': -241.0, 'val': -22313.7381693064}), ('acc2', {'Qua': -13.0, 'val': -1312.940854148})]

Given this,

>>> from collections import OrderedDict
>>> t = OrderedDict()
>>> k_dict.items()[0][0]
'INT-ABC1'
>>> t[k_dict.items()[0][0]] = sorted(k_dict.items()[0][1].items(), key=lambda  x: x[1]['val'])
>>> t
OrderedDict([('INT-ABC1', [('acc1', {'Qua': -241.0, 'val': -22313.7381693064}), ('acc2', {'Qua': -13.0, 'val': -1312.940854148})])])

my dict look like this { 'INT-ABC1': { 'acc1': {'val': -22313.7381693064, 'Qua': -241.0}, 'acc2': {'val': -1312.940854148, 'Qua': -13.0} }, 'INT-ABC2': { 'acc1': {'val': -131.2510359399, 'Qua' : -23.0}, 'acc3': {'val': -131.40521409002, 'Qua' : -13.0}, 'acc5': {'val': -12312.7688190937, 'Qua' : -1313.0} } } I need to sort and get a similar dict out of it. I need to sort it by 'val'. Kindly advise. thanks a lot — Marshall Wilson, Apr 24 '15 at 18:46
@MarshallWilson Check my EDIT, I hope you can follow now and finish up with a for loop. — viki.omega9, Apr 27 '15 at 06:19
@MarshallWilson If you still need help send me a message. But this answer should be accepted as it's doing what you want to do. — viki.omega9, Apr 27 '15 at 06:21

Pallav Agarwal · Answer 2 · 2015-04-24T17:26:10.247

1

Something like this would work, but I think If you are starting with python, you should learn about the sorted function in python.

>>> sorted(k.items(), key = lambda x: x[1][0].itervalues().next()['val'])

Where k is your dictionary.

Link provided by viki.omega9 in his answer should be a good starting point.

Note: The command will return a list of tuples (or pairs) in sorted order of the form:

[(key1, val1), (key2, val2)]

because dictionaries by nature are unordered.

edited Apr 24 '15 at 17:26

answered Apr 24 '15 at 17:20

Pallav Agarwal

108
10

Thanks but sorry my dict has different struture – Marshall Wilson Apr 24 '15 at 18:44
{ 'INT-ABC1': { 'acc1': {'val': -22313.7381693064, 'Qua': -241.0}, 'acc2': {'val': -1312.940854148, 'Qua': -13.0} }, 'INT-ABC2': { 'acc1': {'val': -131.2510359399, 'Qua' : -23.0}, 'acc3': {'val': -131.40521409002, 'Qua' : -13.0}, 'acc5': {'val': -12312.7688190937, 'Qua' : -1313.0} } } – Marshall Wilson Apr 24 '15 at 18:44
What exactly do you want the output to be?? I don't think I understand what you are trying to do.. – Pallav Agarwal Apr 25 '15 at 16:18

Sort dictionary of dictionaries by value Python ( 3 Level Dict )

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