带有哈希的数组，如何合并相同的键并添加它的值答案

【问题标题】：Array with hash, how to merge same keys and add its value带有哈希的数组，如何合并相同的键并添加它的值
【发布时间】：2016-05-03 11:50:33
【问题描述】：

我有一个包含哈希的数组。如果它们具有相同的键，我只想添加它的值。

@receivers << result

@receivers
=> [{:email=>"user_02@yorlook.com", :amount=>10.00}]
result
=> {:email=>"user_02@yorlook.com", :amount=>7.00}

我希望上面的结果看起来像这样

[{:email=>"user_02@yorlook.com", :amount=>17.00}]

有人知道怎么做吗？

这是整个方法

  def receivers
    @receivers = []
    orders.each do |order|
      product_email = order.product.user.paypal_email
      outfit_email  = order.outfit_user.paypal_email
      if order.user_owns_outfit?
        result = { email: product_email, amount: amount(order.total_price) }
      else
        result = { email: product_email, amount: amount(order.total_price, 0.9),
                   email: outfit_email,  amount: amount(order.total_price, 0.1) }
      end
      @receivers << result
    end
  end

【问题讨论】：

“相同的键”是指相同的:email 值？
@Stefan，这是个好问题..

标签： ruby

【解决方案1】：

使用Enumerable#group_by

@receivers.group_by {|h| h[:email]}.map do |k, v|
  {email: k, amount: v.inject(0){|s,h| s + h[:amount] } }
end
# => [{:email=>"user_02@yorlook.com", :amount=>17.0}]

使用Enumerable#each_with_object

@receivers.each_with_object(Hash.new(0)) {|h, nh| nh[h[:email]]+= h[:amount] }.map do |k, v|
 {email: k, amount: v}
end

【讨论】：

【解决方案2】：

# Output: [{ "em@il.one" => 29.0 }, { "em@il.two" => 39.0 }]
def receivers
  return @receivers if @receivers
  # Produces: { "em@il.one" => 29.0, "em@il.two" => 39.0 }
  partial_result = orders.reduce Hash.new(0.00) do |result, order|
    product_email = order.product.user.paypal_email
    outfit_email  = order.outfit_user.paypal_email

    if order.user_owns_outfit?
      result[product_email] += amount(order.total_price)
    else
      result[product_email] += amount(order.total_price, .9)
      result[outfit_email]  += amount(order.total_price, .1)
    end

    result
  end

  @receivers = partial_result.reduce [] do |result, (email, amount)|
    result << { email => amount }
  end
end

【讨论】：

【解决方案3】：

我会这样写代码：

def add(destination, source)
  if destination.nil?
    return nil
  end
  if source.class == Hash
    source = [source]
  end
  for item in source
    target = destination.find {|d| d[:email] == item[:email]}
    if target.nil?
      destination << item
    else
      target[:amount] += item[:amount]
    end
  end
  destination
end

用法：

@receivers = []
add(@receivers, {:email=>"user_02@yorlook.com", :amount=>10.00})
=> [{:email=>"user_02@yorlook.com", :amount=>10.0}]
add(@receivers, @receivers)
=> [{:email=>"user_02@yorlook.com", :amount=>20.0}]

【讨论】：

显式类型检查和for 循环不是很惯用。
好吧，它不在循环中，但想法是您可以提供不同的方法来调用该方法
纯粹是风格，但请考虑return nil if destination.nil?; source = [source] if source_class == Hash。（或if Hash === source）。您可以编写for item in [*source]，而不是将source 转换为数组。（这是我第一次写for... :-)）。

【解决方案4】：

a = [
  {:email=>"user_02@yorlook.com", :amount=>10.0},
  {:email=>"user_02@yorlook.com", :amount=>7.0}
]

a.group_by { |v| v.delete :email } # group by emails
 .map { |k, v| [k, v.inject(0) { |memo, a| memo + a[:amount] } ] } # sum amounts
 .map { |e| %i|email amount|.zip e } # zip to keys
 .map &:to_h # convert nested arrays to hashes

【讨论】：

【解决方案5】：

据我了解，您可以只使用.inject：

a = [{:email=>"user_02@yorlook.com", :amount=>10.00}]
b = {:email=>"user_02@yorlook.com", :amount=>7.00}
c = {email: 'user_03@yorlook.com', amount: 10}

[a, b, c].flatten.inject({}) do |a, e|
  a[e[:email]] ||= 0
  a[e[:email]] += e[:amount]
  a
end

=> {
  "user_02@yorlook.com" => 17.0,
  "user_03@yorlook.com" => 10
}

【讨论】：