Ruby sum from nested hash - ruby-on-rails

Question

How can I return the total scores, strokes and rounds from the following array?
players = [{"Angel Cabrera"=>{"score"=>2, "strokes"=>146, "rounds"=>3}},
{"Jason Day"=>{"score"=>1, "strokes"=>145, "rounds"=>3}},
{"Bryson DeChambeau"=>{"score"=>0, "strokes"=>144, "rounds"=>3}},
{"Sergio Garcia"=>{"score"=>0, "strokes"=>144, "rounds"=>3}},
{"Ian Poulter"=>{"score"=>5, "strokes"=>162, "rounds"=>3}},
{"Vijay Singh"=>nil},
{"Jordan Spieth"=>{"score"=>-4, "strokes"=>140, "rounds"=>3}}]
I can get the strokes by doing the following but I know that isn't the best way to do it.
players.each do |x|
x.values()[0]["strokes"]
end
How can I return the sum of the strokes given the array above?

Answer 1

Here are three ways of doing that.
Use the form of Hash#update that employs a block to determine the values of keys that are present in both hashes being merged
players.map { |g| g.first.last }.
compact.
each_with_object({}) { |g,h| h.update(g) { |_,o,v| o+v } }
#=> {"score"=>4, "strokes"=>881, "rounds"=>18}
The steps:
a = players.map { |g| g.first.last }
#=> [{"score"=> 2, "strokes"=>146, "rounds"=>3},
# {"score"=> 1, "strokes"=>145, "rounds"=>3},
# {"score"=> 0, "strokes"=>144, "rounds"=>3},
# {"score"=> 0, "strokes"=>144, "rounds"=>3},
# {"score"=> 5, "strokes"=>162, "rounds"=>3},
# nil,
# {"score"=>-4, "strokes"=>140, "rounds"=>3}]
b = a.compact
#=> [{"score"=> 2, "strokes"=>146, "rounds"=>3},
# {"score"=> 1, "strokes"=>145, "rounds"=>3},
# {"score"=> 0, "strokes"=>144, "rounds"=>3},
# {"score"=> 0, "strokes"=>144, "rounds"=>3},
# {"score"=> 5, "strokes"=>162, "rounds"=>3},
# {"score"=>-4, "strokes"=>140, "rounds"=>3}]
b.each_with_object({}) { |g,h| h.update(g) { |_,o,v| o+v } }
#=> {"score"=>4, "strokes"=>881, "rounds"=>18}
Here, Hash#update (aka merge!) uses the block { |_,o,v| o+v }) to determine the values of keys that are present in both hashes. The first block variable (which is not used, and therefore can be represented by the local variable _) is the key, the second (o, for "old") is the value of the key in h and the third (n, for "new") is the value of the key in g.
Use a counting hash
players.map { |g| g.first.last }.
compact.
each_with_object(Hash.new(0)) { |g,h| g.keys.each { |k| h[k] += g[k] } }
Hash.new(0) creates an empty hash with a default value of zero, represented by the block variable g. This means that if a hash h does not have a key k, h[k] returns the default value (but does not alter the hash). h[k] += g[k] above expands to:
h[k] = h[k] + g[k]
If h does not have a key k, h[k] on the right side is therefore replaced by 0.
Sum values and then convert to a hash
If you are using Ruby v1.9+ and the keys are guaranteed to have the same order in each hash, a third way it could be done is as follows:
["scores", "strokes", "rounds"].zip(
players.map { |g| g.first.last }.
compact.
map(&:values).
transpose.
map { |arr| arr.reduce(:+) }
).to_h
#=> {"scores"=>4, "strokes"=>881, "rounds"=>18}
The steps (starting from b above) are:
c = b.map(&:values)
#=> [[ 2, 146, 3],
# [ 1, 145, 3],
# [ 0, 144, 3],
# [ 0, 144, 3],
# [ 5, 162, 3],
# [-4, 140, 3]]
d = c.transpose
#=> [[ 2, 1, 0, 0, 5, -4],
# [146, 145, 144, 144, 162, 140],
# [ 3, 3, 3, 3, 3, 3]]
totals = d.map { |arr| arr.reduce(:+) }
#=> [4, 881, 18]
e = ["scores", "strokes", "rounds"].zip(totals)
#=> [["scores", 4], ["strokes", 881], ["rounds", 18]]
e.to_h
#=> {"scores"=>4, "strokes"=>881, "rounds"=>18}

Answer 2

Use this code:
#total= 0
players.each do |x|
a= x.values[0]
if a.class == Hash
#total += a["strokes"]
end
end
puts #total