如何基于相同的哈希键值合并两个哈希数组？

Question

所以我有两个哈希数组：

a = [{"b"=>123,"c"=>456}, {"b"=>456,"c"=>555}]
b = [{"c"=>456,"d"=>789},  {"b"=>222,"c"=>444}]

如果键 c 的值在 a 和 b 中相等，我将如何连接它们？这意味着我希望能够与a['c'] == b['c'] 的条件连接

这是我想要得到的结果：

final_array = [{"b"=>123,"c"=>456,"d"=>789}, {"b"=>456,"c"=>555}, {"b"=>222,"c"=>444}]

Answer 1

a = [{"b"=>123,"c"=>456}, {"b"=>456,"c"=>555}]
b = [{"c"=>456,"d"=>789}, {"b"=>222,"c"=>444}]

p a.zip(b).map{|h1,h2| h1["c"] == h2["c"] ? h1.merge(h2) : [h1 ,h2]}.flatten
# => [{"b"=>123, "c"=>456, "d"=>789}, {"b"=>456, "c"=>555}, {"b"=>222, "c"=>444}]

Answer 2

a = [{"b"=>123,"c"=>456}, {"b"=>456,"c"=>555}]
b = [{"c"=>456,"d"=>789},  {"b"=>222,"c"=>444}]

def merge_hashes_with_equal_values(array_of_hashes, key)
  array_of_hashes.sort { |a,b| a[key] <=> b[key] }.
    chunk { |h| h[key] }.
    each_with_object([]) { |h, result|  result << h.last.inject(&:merge) }
end

p merge_hashes_with_equal_values(a + b, 'c')

# => [{"b"=>222, "c"=>444}, {"c"=>456, "d"=>789, "b"=>123}, {"b"=>456, "c"=>555}]

首先连接数组，然后将其传递给带有哈希键的方法以进行组合。对该数组进行排序，然后将要合并的散列放在另一个数组中，这使得合并更容易编程。在这里，我选择了#chunk 来检测连续运行的具有相等键的哈希进行合并，并选择#each_with_object 来编译最终数组。

由于此方法需要一个数组来处理，因此起始数组的长度不需要相等，并且这些数组的顺序无关紧要。缺点是要操作的键必须包含可排序的值（例如，没有 nil）。

这是解决问题的另一种方法，这个方法使用哈希来构建结果：

def merge_hashes_with_equal_values(array_of_hashes, key)
  result = Hash.new { |h,k| h[k] = {} }
  remainder = []
  array_of_hashes.each_with_object(result) do |h, answer|
    if h.has_key?(key)
      answer[h.fetch(key)].merge!(h)
    else
      remainder << h
    end
  end.values + remainder
end

Answer 3

Enumerable#flat_map 和 Hash#update 是实现此目的的完美方法：

a = [{"b"=>123,"c"=>456}, {"b"=>456,"c"=>555}]    
b = [{"c"=>456,"d"=>789}, {"b"=>222,"c"=>444}]
p a.zip(b).flat_map{|k,v| next k.update(v) if k["c"] == v["c"];[k,v]}
# >> [{"b"=>123, "c"=>456, "d"=>789}, {"b"=>456, "c"=>555}, {"b"=>222, "c"=>444}]