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Home/ Questions/Q 8411313
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T00:23:52+00:00 2026-06-10T00:23:52+00:00

Let’s assume that we have 2 sets of objects set1 = [{‘id’:’1′, ‘x’:’1′, ‘y’:’2′},

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Let’s assume that we have 2 sets of objects

set1 = [{'id':'1', 'x':'1', 'y':'2'}, {'id':'2', 'x':'2', 'y':'2'}]
set2 = [{'id':'1', 'z':'1'}, {'id':'2', 'z':'2'}]

We want:

set3 = set1.join(set2).on('id'); 

>> set3 
[{'id':'1', 'x':'1', 'y':'2', 'z':'1'},{'id':'2', 'x':'2', 'y':'2', 'z':'2'}]

What’s right tools for achieving this functionality?
May underscore help here?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T00:23:53+00:00Added an answer on June 10, 2026 at 12:23 am

    OPTION 1, plain js

    I would suggest that you transform each of the lists into a set by id, e.g.

    {1: {x: 1, y: 1}, 2: {x: 2, y: 2}}
    

    Then run a for over one (or both) of the sets and create a new dictionary with attributes from these two — this latter bit depends on whether you’re looking for the inner or the outer join. This should result in a roughly linear runtime, the javascript implementation of dictionaries is pretty efficient.

    OPTION 2, underscore, for dense id sets, using _.zip()

    If the id‘s are relatively dense and you’d like the outer join or know in advance that the sets of ids are exactly the same, another option is to stuff the data into three arrays — one for each attribute and then use the underscore’s zip() method.

    OPTION 3, underscore, using _.groupBy()

    A yet another possibility in to run _.groupBy() on the lists you have with a custom comparison method, that will also allow joining on multiple keys. Some simple postprocessing will be required, though, since the direct result will be a dictionary of the form

    {1: [{'id':'1', 'x':'1', 'y':'2'}, {'id':'1', 'z':'1'}],
     2: [{'id':'2', 'x':'2', 'y':'2'}, {'id':'2', 'z':'2'}]}
    

    An inner join behaviour in the latter case can be achieved by filtering out those items in the resulting dictionary that don’t have the maximum number of items in the list (2, in the example).

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