Trying to add values of same key but with iterating it. here is my array
arr = [ {a: 10, b: 5, c: 2}, {a: 5, c: 3}, { b: 15, c: 4}, {a: 2}, {} ]
Want to converted it like
{a: 17, b: 20, c: 9}
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3 Answers
Here is one way to do this by making use of Enumerable#reduce and Hash#merge:
arr.reduce {|acc, h| acc.merge(h) {|_,v1,v2| v1 + v2 }}
#=> {:a=>17, :b=>20, :c=>9}
5 Comments
falsetru
(Hash.new {|h, k| 0}) can be omitted, but merge! should be replaced with merge (otherwise, arr[0] will be modified)
Wand Maker
@falsetru Thanks for the tip - will update the answer
Stefan
You can remove
({}) if you replace merge! with mergeWand Maker
@Stefan Thanks for that nice tip.
Stefan
I think that is what @falsetru tried to suggest earlier. Note the parentheses in the comment ;-)
each_with_object to the rescue:
result = arr.each_with_object(Hash.new(0)) do |hash, result|
hash.each { |key, value| result[key] += value}
end
p result
1 Comment
Pascal
Thanks. I prefer each_with_object over reduce/inject (because it has a more logical order of the block parameters) and i think it is more readable than using merge. That being said: the merge solution is pretty sleek.
#1 Use a "counting hash"
Edit: I just noticed that @Pascal posted the same solution earlier. I'll leave mine for the explanation.
arr.each_with_object(Hash.new(0)) { |h,g| h.each { |k,v| g[k] += v } }
#=> {:a=>17, :b=>20, :c=>9}
h = Hash.new(0) is a counting hash with a default value of zero. That means that if h does not have a key k, h[k] returns zero (but the hash is not altered). Ruby expands h[k] += 4 to h[k] = h[k] + 4, so if h does not have a key k, h[k] on the right side of the equality equals zero.
#2 Use Hash#update
Specifically, use the form of this method (aka merge!) that employs a block to determine the values of keys that are present in both hashes being merged.
arr.each_with_object({}) { |h,g| g.update(h) { |_,o,n| o+n } }
#=> {:a=>17, :b=>20, :c=>9}
3 Comments
Anuj
Thanks @Stefan : P
Stefan
@CarySwoveland you don't need a default value in #2, the inner block is only executed if the key exists in both hashes.
Cary Swoveland
Thanks, @Stefan. It's late and I'm bushed from building stairs all day, so the little grey cells seem to have nodded off.