Sum of particular values in a Ruby hash

Question

I have two hashes:

first = { 1 => [15, 15, 15, 8], 4 => [11, 12, 7, 7], 5 => [14, 17, 13, 13],
          6 => [19, 19, 15, 15], 7 => [5, 12, 12, 12], 8 => [10, 14, 14, 14], 
          9 => [8, 7, 8, 8] } 

second = { 1 => [0, 1, 2], 4 => [2, 3], 5 => [2, 3], 6 => [0, 1, 2, 3], 
           7 => [1, 2, 3], 8 => [1, 2, 3], 9 => [2, 3] }

As you see they both use same keys, but values are different. second's values contains ranges of indexes of the first's values and I need to sum only those first's values that are in those ranges.

The expected output:

result = { 1 => [45], 4 => [14], 5 => [26], etc }

Explain this result = { 1 => [45], 4 => [14], 5 => [26], etc } properly — gates, Apr 04 '17 at 16:37
What is the code you are having trouble with? [so] is not a Write-My-Code-For-Me-Service. Those *do* exist, they are called "programmers" and you can hire them for a fee. — Jörg W Mittag, Apr 04 '17 at 17:42

mu is too short · Accepted Answer · 2019-03-13T17:17:05.603

8

A straightforward way is to say exactly what you mean:

first.map { |k, v| [ k, [ v.values_at(*second[k]).sum ] ] }.to_h

If your version of Ruby doesn't have Array#sum then use inject(:+):

first.map { |k, v| [ k, [ v.values_at(*second[k]).inject(:+) ] ] }.to_h

You could also skip the map/to_h business by using each_with_object instead:

first.each_with_object({}) { |(k, v), h| h[k] = [ v.values_at(*second[k]).inject(:+) ] }

A little time with the Array#values_at documentation might be fruitful as would the Enumerable documentation.

edited Mar 13 '19 at 17:17

answered Apr 04 '17 at 16:48

mu is too short

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, why enclose each sum in an array? I see @llya did the same, so I'm probably missing something. – Cary Swoveland Apr 04 '17 at 18:07
@CarySwoveland Because that's the format the question asked for. I agree that it looks a bit weird but... – mu is too short Apr 04 '17 at 18:42
Egad, how did I miss that? – Cary Swoveland Apr 04 '17 at 18:58

Ilya · Answer 2 · 2017-04-06T06:37:32.077

3

More natural way is to use Hash#merge, I think:

first.merge(second) { |_, f, s| [f.values_at(*s).sum] }
#=> {1=>[45], 4=>[14], 5=>[26], 6=>[68], 7=>[36], 8=>[42], 9=>[16]}

Of course, you can use inject(:+) instead of sum if your Ruby version < 2.4

edited Apr 06 '17 at 06:37

answered Apr 04 '17 at 17:15

Ilya

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score 2 · Answer 3 · answered Apr 04 '17 at 17:59

Since you say the elements of second are ranges, you might write them as such:

second = { 1=>0..2, 4=>2..3, 5=>2..3, 6=>0..3, 7=>1..3, 8=>1..3, 9=>2..3 }

Then

second.merge(second) { |k,o,_| first[k].values_at(*o).sum }
  #=> {1=>45, 4=>14, 5=>26, 6=>68, 7=>36, 8=>42, 9=>16}

This uses the form of Hash#merge that employs a block to determine the values of keys that are present in both hashes being merged, which here is all keys. See the doc for details, particularly the values of the three block variables k, o, and n. (Here I've replaced n with _ to signify that it is not used in the block calculation.)

score 1 · Answer 4 · answered Apr 04 '17 at 16:54

1

Try this:

def sum(list, indices)
    sum = 0
    indices.each do |index|
        sum += list[index]
    end
   return sum
end

And then

second.map { |k, v| [k, sum(first[k], v)] }.to_h

answered Apr 04 '17 at 16:54

Samantha Klonaris

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5

Sum of particular values in a Ruby hash

4 Answers4