Reduce Hash Values

Question

I am having trouble with the syntax for reduce. I have a hash of the following format:

H = {"Key1" => 1, "Key2" => 2}

I would like to use reduce to find the sum of the values in this function.

Something Like

H.reduce(0) {|memo, elem| memo+=elem}

I know this is wrong. I dont understand how I can make elem the value of the hash.

Try: H.values.reduce(:+) – Diego Basch Dec 14 '12 at 16:57 — Diego Basch, Dec 14 '12 at 16:57

score 86 · Answer 1 · edited Feb 01 '22 at 08:47

86

You can make elem contain the value by splitting it up in 2 variables:

H.reduce(0) {|memo, (key, val)| memo + val}

edited Feb 01 '22 at 08:47

adamors

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answered Dec 14 '12 at 17:17

steenslag

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I would have written `memo + val` to avoid letting people think that updating `memo` is helpful. – Ludovic Kuty Nov 15 '16 at 13:29
Strongly agreeing with @lkuty, the += mutation of memo is unnecessary and misleading. – Pierre May 25 '17 at 04:50

maerics · Accepted Answer · 2017-01-24T19:29:26.337

Use Enumerable#reduce, if you're ok with getting nil if the hash happens to be empty:

H.values.reduce(:+) # => 3
Hash.new.values.reduce(:+) # => nil

To safely get 0 when the hash is empty, use:

H.values.reduce(0) { |sum,x| sum + x } # or...
H.reduce(0) { |sum,(key,val)| sum + val } # ...if you need to inspect the key

Here's a quick benchmark, for kicks. Note that it appears to be slightly faster to reduce just the values rather than values from the key/value pairs:

                               user     system      total        real
H.values.reduce(:+)        4.510000   0.080000   4.590000 (  4.595229)
H.values.reduce(0) {...}   4.660000   0.080000   4.740000 (  4.739708)
H.reduce(0) {...}          5.160000   0.070000   5.230000 (  5.241916)

require 'benchmark'

size = 1_000
hash = Hash[* Array.new(size*2) { rand } ]

N=10_000
Benchmark.bm(24) do |x|
  x.report('H.values.reduce(:+)')      { N.times { hash.dup.values.reduce(:+) } }
  x.report('H.values.reduce(0) {...}') { N.times { hash.dup.values.reduce(0) { |sum,x| sum + x } } }
  x.report('H.reduce(0) {...}')        { N.times { hash.dup.reduce(0) { |sum,(_,v)| sum + v } } }
end

Note that this doesn't return 0 if a hash is empty. You'd need to use the memo version below — Josh Bodah, Jul 29 '15 at 17:44

score 6 · Answer 3 · answered Dec 14 '12 at 16:58

6

Try this:

H.reduce(0) { |memo, elem| memo += elem[1] }

or

H.reduce(0) { |memo, (key, value)| memo += value }

answered Dec 14 '12 at 16:58

yasu

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I would have written `memo + value` to avoid letting people think that updating `memo` is helpful. – Ludovic Kuty Nov 15 '16 at 13:29

astorije · Answer 4 · 2017-05-16T21:54:16.933

2

I know I'm excavating this one, but if you happen to use Rails, the .sum method can help:

H = {"Key1" => 1, "Key2" => 2}
=> {"Key1"=>1, "Key2"=>2}
> H.values.sum
=> 3

Advantage is that it returns 0 on empty hashes:

> {}.values.sum
=> 0
> {}.values.reduce(:+)
=> nil

I noticed it was Rails-specific only after typing this answer. I know the OP didn't add the Rails tag, but I figured it might be useful for people stopping by.

Note that as of Ruby 2.4.0, .sum is now available.

edited May 16 '17 at 21:54

answered Apr 29 '16 at 21:28

astorije

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`.sum` is now implemented in Ruby 2.4 [Enumerable#sum](https://ruby-doc.org/core-2.4.0/Enumerable.html#method-i-sum). – user3162553 May 16 '17 at 19:25

score 1 · Answer 5 · answered Dec 14 '12 at 16:56

1

h = {"Key1" => 1, "Key2" => 2}

h.values.inject(0){|f,v| f += v.to_i }
# => 3

or

h.values.inject(:+)
# => 3

answered Dec 14 '12 at 16:56

score 0 · Answer 6 · answered Jul 02 '20 at 18:47

0

In case of a complex hash it might be easier to map it first to an array of values, then reduce:

values = H.map do |k, v|
    # some complex logic here
end
values.reduce(:+)

Or values.reduce(0, :+) if the array might be empty.

answered Jul 02 '20 at 18:47

x-yuri

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Reduce Hash Values

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