Convert Array to Hash removing duplicate keys and adding values at the same time

Question

I have an array that looks like so:

f = [["Wed, 12-31", 120.0],["Thu, 01-01", 120.0], ["Thu, 01-01", 120.0]]

I can convert it to a hash and remove the duplicate keys:

h = Hash[ *f.collect { |v| [v] }.flatten ]
# => {"Wed, 12-31"=>120.0, "Thu, 01-01"=>120.0}

which is almost there, but I'd like to sum the value for elements with the identical date strings, the desired result from the above array would be this:

{"Wed, 12-31"=>120.0, "Thu, 01-01"=>240.0}

How can I accomplish this?

Note, you can convert your array to a hash discarding duplicate keys without the .collect.flatten nonsense. Just use Hash[array]. f.collect { |v| [v] } is just mapping the array to itself. It doesn't do anything. .flatten on an already flat array is equally useless. — user229044
– user229044 ♦, Commented Jan 1, 2015 at 22:50

Kenny Meyer · Accepted Answer · 2015-01-01 22:59:54Z

2

This works:

result = Hash.new(0)
f = [["Wed, 12-31", 120.0],["Thu, 01-01", 120.0], ["Thu, 01-01", 120.0]]
f.each { |subarray| result[subarray[0]] += subarray[1] }
puts result

If you would like to be fancy you could use .inject()

answered Jan 1, 2015 at 22:59

Kenny Meyer

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the Tin Man Over a year ago

If you'd like to be fancier, and lazier, use each_with_object instead of inject. It works more cleanly.

the Tin Man · Accepted Answer · 2015-01-01 23:35:15Z

2

I'd use each_with_object:

ary = [["Wed, 12-31", 120.0], ["Thu, 01-01", 120.0], ["Thu, 01-01", 120.0]]

hash = ary.each_with_object(Hash.new(0)){ |(k,v), h| h[k] += v }
=> {"Wed, 12-31"=>120.0, "Thu, 01-01"=>240.0}

answered Jan 1, 2015 at 23:35

the Tin Man

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2 Comments

Sonny Parlin Over a year ago

I would just like to add that while I accepted @kennymeyer's answer, I later switched to this solution.

the Tin Man Over a year ago

Just to reiterate something you'll see occasionally on Stack Overflow: Don't pick the first answer that comes along and answers your question. There are many ways to do something, and people aren't poised at their computers waiting to pounce on questions. Answers sometimes take a while to formulate; I often wait to answer hoping someone will respond with what I consider to be the right way to do something, and then later will add mine if nobody has done it or answered as thoroughly as I think the answer should be. So, ask, then wait a day and see what shows up.

Cary Swoveland · Accepted Answer · 2015-01-02 06:08:18Z

2

The other standard approach to this type of problem is to use Enumerable#group_by:

Hash[ary.group_by(&:first).map { |d,a| [d, a.reduce(0) { |t,(_,n)| t+n }] }]
  #=> {"Wed, 12-31"=>120.0, "Thu, 01-01"=>240.0}

We have:

a = ary.group_by(&:first)
  #=> {"Wed, 12-31"=>[["Wed, 12-31", 120.0]],
  #    "Thu, 01-01"=>[["Thu, 01-01", 120.0], ["Thu, 01-01", 120.0]]} 
b = a.map { |d,a| [d, a.reduce(0) { |t,(_,n)| t+n }] }
  #=> [["Wed, 12-31", 120.0], ["Thu, 01-01", 240.0]] 
Hash[b] 
  #=> {"Wed, 12-31"=>120.0, "Thu, 01-01"=>240.0}

or with Ruby 2.0+:

b.to_h
  #=> {"Wed, 12-31"=>120.0, "Thu, 01-01"=>240.0}

A variant of this is:

ary.group_by(&:first).map { |d,a| [d, a.transpose.last.reduce(:+) ] }.to_h

edited Jan 2, 2015 at 6:08

answered Jan 1, 2015 at 23:54

Cary Swoveland

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1 Comment

Kenny Meyer Over a year ago

Thanks for Ruby's 2.0+ to_h approach! That's rad.

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