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Home/ Questions/Q 9241931
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T08:28:20+00:00 2026-06-18T08:28:20+00:00

I am running a mongodb find query with an $in operator: collection.find({name: {$in: [name1,

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I am running a mongodb find query with an $in operator:

collection.find({name: {$in: [name1, name2, ...]}})

I would like the results to be sorted in the same order as my name array: [name1, name2, ...]. How do I achieve this?

Note: I am accessing MongoDb through pymongo, but I don’t think that’s of any importance.

EDIT: as it’s impossible to achieve this natively in MongoDb, I ended up using a typical Python solution:

names = [name1, name2, ...]
results = list(collection.find({"name": {"$in": names}}))
results.sort(key=lambda x: names.index(x["name"]))
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  1. Editorial Team
    Editorial Team
    2026-06-18T08:28:21+00:00Added an answer on June 18, 2026 at 8:28 am

    Impossible. $in operator checks the presence. The list is treated as set.

    Options:

    • Split for several queries for name1 … nameN or filter the result the same way.
      More names – more queries.
    • Use itertools groupby/ifilter. In that case – add the “sorting precedence” flag to every document and match name1 to PREC1, name2 to PREC2, …., then isort by PREC then group by PREC.

    If your collection has the index on “name” field – option 1 is better.
    If doest not have the index or you cannot create it due to high write/read ratio – option 2 is for you.

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