Using .sort with PyMongo

Question

With PyMongo, when I try to retrieve objects sorted by their 'number' and 'date' fields like this:

db.test.find({"number": {"$gt": 1}}).sort({"number": 1, "date": -1})

I get this error:

TypeError: if no direction is specified, key_or_list must be an instance of list

What's wrong with my sort query?

georg · Accepted Answer · 2019-05-06T11:00:50.253

235

sort should be a list of key-direction pairs, that is

db.test.find({"number": {"$gt": 1}}).sort([("number", 1), ("date", -1)])

The reason why this has to be a list is that the ordering of the arguments matters and dicts are not ordered in Python < 3.6

edited May 06 '19 at 11:00

answered Apr 20 '12 at 07:54

georg

33

The reason that this is a list in Python is that the ordering of the arguments to `sort()` matters and dicts are not ordered in Python. – André Laszlo Nov 16 '12 at 12:11
@AndréLaszlo can an OrderedDict() be used? – zakdances Mar 12 '13 at 23:16
@yourfriendzak, I'm afraid not. It's a bit unpythonic, I guess. [Here's the reason why](http://api.mongodb.org/python/1.0/pymongo-pysrc.html#L70), at line 70. An OrderedDict is not an instance of types.ListType. – André Laszlo Mar 14 '13 at 12:02
9

This saved me a ton of research. The Little MongoDB Book misleads in sort examples. – Dogukan Tufekci Apr 20 '13 at 22:22
6

If it is only one field, it can be .sort("_id", 1) – Cyber Square Professional Mar 15 '18 at 11:51
Kinda a bummer that mongodb's documentation examples use dictionaries instead of list of tuples, but it makes sense why pymongo needs them. JSONs are ordered dictionaries (as strings) -- pymong's error message could be clearer on what to change to solve the problem. – Marc Maxmeister May 08 '18 at 19:46
2

in python3.6+ dicts are ordered so if anyone is up for it, might be worth a pull request to pymongo to bring it in line with generic mongodb syntax. Of course this wouldn't work when running pymongo on older python versions.. – thiezn May 08 '18 at 21:41
if you want to sort a single field, this is another way .sort('{}'.format('created_on'), 1 if sort_type == 'asc' else -1) – Md. Tanvir Raihan Apr 29 '20 at 05:17

1 Answers1