How to merge array of hash based on the same keys in ruby?
example :
a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]
How to get result like this?
a = [{:a=>[1, 10]},{:b=>8},{:c=>[7, 2]}]
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1Me guess that keeping polymorphic values (sometimes 7, sometimes [ 7, 7 ]) is not such a good practice. Why not just keep this in a single hash of arrays rather than an array of hashes whose values are sometimes arrays?Boris Stitnicky– Boris Stitnicky2014-03-11 09:11:15 +00:00Commented Mar 11, 2014 at 9:11
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4 Answers
Try
a.flat_map(&:entries)
.group_by(&:first)
.map{|k,v| Hash[k, v.map(&:last)]}
Comments
Another alternative:
a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]
p a.each_with_object({}) { |h, o| h.each { |k,v| (o[k] ||= []) << v } }
# => {:a=>[1, 10], :b=>[8], :c=>[7, 2]}
It also works when the Hashes have multiple key/value combinations, e.g:
b = [{:a=>1, :b=>5, :x=>10},{:a=>10, :y=>2},{:b=>8},{:c=>7},{:c=>2}]
p b.each_with_object({}) { |h, o| h.each { |k,v| (o[k] ||= []) << v } }
# => {:a=>[1, 10], :b=>[5, 8], :x=>[10], :y=>[2], :c=>[7, 2]}
1 Comment
Arup Rakshit
If you don't mind... Can I use this
(hsh[k] ||= []) << v in my code? I was looking for this. But couldn't remember anyway.Minor addition to answer by Arie Shaw to match required answer:
a.flat_map(&:entries)
.group_by(&:first)
.map{|k,v| Hash[k, v.size.eql?(1) ? v.last.last : v.map(&:last) ]}
#=> [{:a=>[1, 10]}, {:b=>8}, {:c=>[7, 2]}]
Comments
I'd do :
a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]
merged_hash = a.each_with_object({}) do |item,hsh|
k,v = item.shift
hsh[k] = hsh.has_key?(k) ? [ *Array( v ), hsh[k] ] : v
end
merged_hash.map { |k,v| { k => v } }
# => [{:a=>[10, 1]}, {:b=>8}, {:c=>[2, 7]}]
update
A better taste :
a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]
merged_hash = a.each_with_object({}) do |item,hsh|
k,v = item.shift
(hsh[k] ||= []) << v
end
merged_hash.map { |k,v| { k => v } }
# => [{:a=>[10, 1]}, {:b=>8}, {:c=>[2, 7]}]
3 Comments
Boris Stitnicky
Arup, your solution is much cleaner than the other two (I don't like flat_map), but you made one mistake. If you try
a = [{a: 1}, {a: 2}, {a: 3}], you will get a nested array. I'm correcting it.Arup Rakshit
@BorisStitnicky Can I write now
hsh[k] = hsh.has_key?(k) ? [*hsh[k]] << v : v ?Arup Rakshit
@BorisStitnicky I didn't check that although. :(