如何基于红宝石中的相同键合并哈希数组？

Question

如何基于红宝石中的相同键合并哈希数组？

示例：

a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]

如何获得这样的结果？

a = [{:a=>[1, 10]},{:b=>8},{:c=>[7, 2]}]

Answer 1

尝试

a.flat_map(&:entries)
  .group_by(&:first)
  .map{|k,v| Hash[k, v.map(&:last)]}

Answer 2

另一种选择：

a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]

p a.each_with_object({}) { |h, o| h.each { |k,v| (o[k] ||= []) << v } }
# => {:a=>[1, 10], :b=>[8], :c=>[7, 2]}

当散列具有多个键/值组合时也可以使用，例如：

b = [{:a=>1, :b=>5, :x=>10},{:a=>10, :y=>2},{:b=>8},{:c=>7},{:c=>2}]
p b.each_with_object({}) { |h, o| h.each { |k,v| (o[k] ||= []) << v } }
# => {:a=>[1, 10], :b=>[5, 8], :x=>[10], :y=>[2], :c=>[7, 2]}

Answer 3

对Arie Shaw的回答进行了少量补充，以匹配所需的答案：

a.flat_map(&:entries)
  .group_by(&:first)
  .map{|k,v| Hash[k, v.size.eql?(1) ? v.last.last : v.map(&:last) ]}
#=> [{:a=>[1, 10]}, {:b=>8}, {:c=>[7, 2]}]

Answer 4

我愿意：

a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]
merged_hash = a.each_with_object({})  do |item,hsh|
  k,v = item.shift
  hsh[k] = hsh.has_key?(k) ? [ *Array( v ), hsh[k] ] : v
end

merged_hash.map { |k,v| { k => v } }
# => [{:a=>[10, 1]}, {:b=>8}, {:c=>[2, 7]}]

更新

更好的口味：

a = [{:a=>1},{:a=>10},{:b=>8},{:c=>7},{:c=>2}]
merged_hash = a.each_with_object({})  do |item,hsh|
  k,v = item.shift
 (hsh[k] ||= []) << v
end

merged_hash.map { |k,v| { k => v } }
# => [{:a=>[10, 1]}, {:b=>8}, {:c=>[2, 7]}]