Reduce Hash Values

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

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

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

H.values.reduce(:+) # => 3Hash.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        realH.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_000hash = Hash[* Array.new(size*2) { rand } ]N=10_000Benchmark.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

Try this:

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

or

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