Sort nested dictionary by value, and remainder by another value, in Python

Question

Consider this dictionary format.

{'KEY1':{'name':'google','date':20100701,'downloads':0},
 'KEY2':{'name':'chrome','date':20071010,'downloads':0},
 'KEY3':{'name':'python','date':20100710,'downloads':100}}

I'd like the dictionary sorted by downloads first, and then all items with no downloads sorted by date. Obviously a dictionary cannot be sorted, I just need a sorted listed of keys I can iterate over.

['KEY3','KEY1','KEY2']

I can already sort the list by either value using sorted, but how do I sort by second value too?

Amber · Accepted Answer · 2010-11-05T22:48:16.453

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Use the key argument for sorted(). It lets you specify a function that, given the actual item being sorted, returns a value that should be sorted by. If this value is a tuple, then it sorts like tuples sort - by the first value, and then by the second value.

sorted(your_list, key=lambda x: (your_dict[x]['downloads'], your_dict[x]['date']))

edited Nov 05 '10 at 22:48

answered Nov 05 '10 at 22:43

Amber

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Very nice line! Here is how you can reverse the order of the sorting: just add `reverse=True` as a parameter to the `sorted` function, at the end. Thanks – otmezger Apr 29 '13 at 17:27
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Change to the above one. your_list = sorted(your_dict, key=lambda x: (your_dict[x]['downloads'], your_dict[x]['date'])) – Yogesh Kamat Jun 24 '16 at 15:22
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Hi, I am getting an error `TypeError: unhashable type: 'dict'` when calling like: `sorted(d.items(), key=lambda x: (d[x]['downloads'], d[x]['date']))`, where d is equal to the dictionary in the OP. Please if anyone knows the problem, let me know. Thanks – ChickenFeet May 09 '17 at 05:33
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@ChickenFeet use `d.keys()` not `d.items()` – Amber May 12 '17 at 04:34
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This just returns keys, is there a way to get keys and the items you're sorting by? For example, is there a way to do `sorted((d.keys(), lambda y: d[y]['downloads']), key=lambda x: (d[x]['downloads'])` that's actually correct? That code as is is throwing me the same TypeError @ChickenFeet got – mgrogger Jul 16 '19 at 15:04
sorted out. Thx. – kta Sep 13 '19 at 03:29

Eli Courtwright · Answer 2 · 2010-11-05T22:49:05.497

You can pass a key function to sorted which returns a tuple containing the two things you wish to sort on. Assuming that your big dictionary is called d:

def keyfunc(tup):
    key, d = tup
    return d["downloads"], d["date"]

items = sorted(d.items(), key = keyfunc)

You can do this with a lambda if you prefer, but this is probably more clear. Here's the equivalent lambda-based code:

items = sorted(d.items(), key = lambda tup: (tup[1]["downloads"], tup[1]["date"]))

Incidentally, since you mentioned that you wanted to sort by "downloads" first, the above two examples sort according to download counts in ascending order. However, from context it sounds like you might want to sort in decreasing order of downloads, in which case you'd say

return -d["downloads"], d["date"]

in your keyfunc. If you wanted something like sorting in ascending order for non-zero download numbers, then having all zero-download records after that, you could say something like

return (-d["downloads"] or sys.maxint), d["date"]

I tried both, but instead of a list of keys I got a list of tuples. I can use this too but it's unnecessary complexity. `[('KEY3',{'name':'python','date':20100710,'downloads':100})]` — , Nov 05 '10 at 23:32
@user479870 You can get the keys from the tuples with `[x[0] for x in items]`. — questionto42, Sep 12 '21 at 12:44

score 2 · Answer 3 · edited May 23 '17 at 12:18

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My other answer was wrong (as are most of the answers here)

sorted_keys = sorted((key for key in outer_dict if outer_dict[key]['downloads']),
                     key=lambda x: (outer_dict[key]['downloads'],
                                    outer_dict[key]['downloads'])
                     reverse=True)

sorted_keys += sorted((key for key in outer_dict if not outer_dict[key]['downloads']),
                      key=lambda x: outer_dict[key]['date'])

This will create a list with the items that have been downloaded sorted in descending order at the front of it and the rest of the items that have not been downloaded sorted by date after those that have.

But actually, the last part of Eli Courtwrights answer is the best.

edited May 23 '17 at 12:18

Community

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answered Nov 05 '10 at 22:42

aaronasterling

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Er, why are you `import`-ing `operator` and then never using it? – Amber Nov 05 '10 at 22:47
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@Amber because I thought I was going to do something else and then it started raining. I was in a rush. Inside now. – aaronasterling Nov 05 '10 at 22:49
Without having tested it. Why a tuple of two `outer_dict[key]['downloads']` after the first `lambda x:`? – questionto42 Sep 12 '21 at 13:26

score 2 · Answer 4 · edited May 22 '21 at 03:40

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your_dict = dict(sorted(your_dict.items(), key = lambda x: (x[1]["downloads"], x[1]["date"])))

edited May 22 '21 at 03:40

Josef

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answered May 21 '21 at 03:15

Phoenix

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While this answer may answer the question it would be good if you explained the code a bit, e.g. why it is built that way, how it works etc. – Dominik May 22 '21 at 12:41

score 1 · Answer 5 · answered Nov 05 '10 at 22:42

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a = {'KEY1':{'name':'google','date':20100701,'downloads':0},
 'KEY2':{'name':'chrome','date':20071010,'downloads':0},
 'KEY3':{'name':'python','date':20100710,'downloads':100}}


z = a.items()

z.sort(key=lambda x: (x[1]['downloads'], x[1]['date']))

answered Nov 05 '10 at 22:42

mouad

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`z.sort(key=lambda x: (x[1]['downloads'], x[1]['date']))` returns None. Any idea why? – ChickenFeet May 09 '17 at 05:35

Sort nested dictionary by value, and remainder by another value, in Python

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