Merge two arrays of objects in Ruby

Question

I have an array of objects with unique IDs:

[{id: 1, score: 33}, {id: 23, score: 50}, {id:512, score: 27}, ...]

I also have an array of User records with matching IDs. The user records have "name" but not "score":

[{id: 1, name: "Jon"}, {id: 23, name: "Tom"}, {id: 512, name: "Joey"}, ...]

How can I create a single array with each id, name, and score?

[{id: 1, name: "Jon", score: 33}, {id: 23, name: "Tom", score: 50}, {id: 512, name: "Joey", score: 27}, ...]

I tried merge, combine, filter, etc but haven't found the Ruby function to accomplish this.

Paweł Dawczak · Accepted Answer · 2015-06-04 10:23:37Z

3

Assuming that in users there is always record with corresponding :id from scores:

scores = [{id: 1, score: 33}, {id: 23, score: 50}, {id:512, score: 27}]
users  = [{id: 1, name: "Jon"}, {id: 23, name: "Tom"}, {id: 512, name: "Joey"}]

scores = scores.map { |score| score.merge(users.find { |user| user[:id] == score[:id] }) }
# => [{:id=>1, :score=>33, :name=>"Jon"}, {:id=>23, :score=>50, :name=>"Tom"}, {:id=>512, :score=>27, :name=>"Joey"}]

Hope that puts you in proper direction!

answered Jun 4, 2015 at 10:23

Paweł Dawczak

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2 Comments

Don P Over a year ago

Interesting - thanks Pawel! Do you also know a function that lets me find just the matching object from scores given an ID?

Paweł Dawczak Over a year ago

Actually, this is what find (from Enumerable module, which is included in Array) does. You should go with: scores.find { |element| element[:id] == 123 }. Check documentation for more examples. Hope that helps!

Santhosh · Accepted Answer · 2015-06-04 10:49:00Z

1

You can use an intermediate Hash.

hsh = Hash[ a1.map {|h| [h[:id], h[:score]]} ]
# => {1=>33, 23=>50, 512=>27}
a2.map {|h| h[:score] = hsh[h[:id]]; h}
# => [{:id=>1, :name=>"Jon", :score=>33}, {:id=>23, :name=>"Tom", :score=>50}, {:id=>512, :name=>"Joey", :score=>27}]

answered Jun 4, 2015 at 10:49

Santhosh

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1 Comment

Cary Swoveland Over a year ago

I like your solution, but would prefer a2.each_with_object({}) { |g,h| h.update(g[id] => g) }, as it doesn't mutate a2 and IMO it reads better.

Cary Swoveland · Accepted Answer · 2015-06-05 18:33:49Z

1

If, as in the example, scores[i][:id] = users[i][:id] for all i, and you are using v1.9+ (where key insertion order is maintained), you could write:

scores.zip(users).each_with_object({}) do |(sh,uh),h|
   h.update(sh).update(uh)
end

Would I use this? Would you?

answered Jun 5, 2015 at 18:33

Cary Swoveland

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