Question

我有一个包含事务哈希的事务数组。我想循环遍历事务哈希数组，并且对于每个财政年度，将利息和主要值相加。然后需要将它们添加到各个阵列中，按财务年度分组，如下所示。

简而言之：

每个财政年度：

1.Sum the interest keys of the interest键

2.创建哈希

3.将哈希添加到相应的数组

4.重复主要值

我在第一步有点困惑：对于每个财政年度，做x - 任何帮助都将受到赞赏。

Ruby 2.3.1

鉴于

 @transactions = [
  {"interest" => "120.00", "principal" => "250", "financial_year" => "2017"},
  {"interest" => "120.00", "principal" => "250", "financial_year" => "2017"},
  {"interest" => "120.00", "principal" => "250", "financial_year" => "2017"},
  {"interest" => "120.00", "principal" => "250", "financial_year" => "2018"},
  {"interest" => "120.00", "principal" => "250", "financial_year" => "2018"},
  {"interest" => "120.00", "principal" => "250", "financial_year" => "2019"},
  {"interest" => "120.00", "principal" => "250", "financial_year" => "2019"}
 ]

我想计算

 @interest_totals = [
    {"financial_year" => "2017", "total" => "360"}
    {"financial_year" => "2018", "total" => "240"}
    {"financial_year" => "2019", "total" => "240"}
 ]

和

 @principal_totals = [
    {"financial_year" => "2017", "total" => "750"}
    {"financial_year" => "2018", "total" => "500"}
    {"financial_year" => "2019", "total" => "500"}
 ]

Answer 1

使用inject或reduce或each_with_object。我使用下面的each_with_object，以便我可以在一次迭代中计算两者的总和

interests, principals = @transactions.each_with_object([Hash.new(0), Hash.new(0)]) do |transaction, (interests, principals)|
  interests[transaction['financial_year']] += transaction['interest'].to_d
  principals[transaction['financial_year']] += transaction['principal'].to_d
end

p interests
# {"2017"=>#<BigDecimal:7fd1c6cc20f0,'0.36E3',9(18)>, "2018"=>#<BigDecimal:7fd1c6cc1d30,'0.24E3',9(18)>, "2019"=>#<BigDecimal:7fd1c6cc1970,'0.24E3',9(18)>}

p principals
# {"2017"=>#<BigDecimal:7fd1c6cc2050,'0.75E3',9(18)>, "2018"=>#<BigDecimal:7fd1c6cc1c90,'0.5E3',9(18)>, "2019"=>#<BigDecimal:7fd1c6cc18d0,'0.5E3',9(18)>}

这将为您提供一个哈希，其中年份是关键，值是总数。如果您确实需要一组哈希值，请使用zip

@interest_sum = interests.map do |interest|
  Hash[['financial_year', 'total'].zip(interest)]
end

Answer 2

您可以按年度对值进行分组，然后将原则和利息金额相加

@interest_totals =  @transactions.group_by{|a| a["financial_year"]}.map{|k, v| {"financial_year" => k, "total" => v.map{|e|e["interest"].to_i}.sum}}

#=> [{ "financial_year" => "2017", "total" => 360 },
#=>  { "financial_year" => "2018", "total" => 240 },
#=>  { "financial_year" => "2019", "total" => 240 }]

@principal_totals = @transactions.group_by{|a| a["financial_year"]}.map{|k, v| {"financial_year" => k, "total" => v.map{|e|e["principal"].to_i}.sum}}

#=> [{ "financial_year" => "2017", "total" => 750 },
#=>  { "financial_year" => "2018", "total" => 500 },
#=>  { "financial_year" => "2019", "total" => 500 }]

Answer 3

我建议您先计算以下哈希值。

h = @transactions.each_with_object({}) do |g,h|
  h.update(g["financial_year"]=>g.reject { |k,_| k=="financial_year" }) do |_,o,n|
    o.merge(n) { |_,oo,nn| (oo.to_i + nn.to_i).to_s }
  end
end
  #=> {"2017"=>{"interest"=>"360", "principal"=>"750"},
  #    "2018"=>{"interest"=>"240", "principal"=>"500"},
  #    "2019"=>{"interest"=>"240", "principal"=>"500"}}

这使用Hash#update（aka merge!）和Hash#merge的形式，它使用一个块来确定合并的两个哈希中存在的键的值。有关详细信息，请参阅文档，尤其是块变量的值，_，o，n，k，oo和nn。（在第一个块中，我使用公共密钥的下划线表示它未在块计算中使用。）

这种中间解决方案的一个优点是，如果添加了键（＆＃34; financial_year＆＃34;除外）（＆＃34;余额＆＃34;，比如说），删除或删除或不需要更改重命名。

您现在可以轻松构建其他对象。例如，

@interest_totals = h.map { |year,v| {"financial_year"=>year, "total"=> v["interest"] } }
  #=> [{"financial_year"=>"2017", "total"=>"360"},
  #    {"financial_year"=>"2018", "total"=>"240"},
  #    {"financial_year"=>"2019", "total"=>"240"}]

Ruby哈希数组 - 每个财政年度

3 个答案: