Question

我有以下带散列的数组：

[
    [
        {"created_at":"2013-11-16","position_f":1},
        {"created_at":"2013-11-17","position_f":3},
        {"created_at":"2013-11-18","position_f":0},
        {"created_at":"2013-11-19","position_f":1}
    ],
    [
        {"created_at":"2013-11-16","position_f":2},
        {"created_at":"2013-11-17","position_f":3},
        {"created_at":"2013-11-18","position_f":11},
        {"created_at":"2013-11-19","position_f":45}
    ]
]

预期结果：

 [
        {"created_at":"2013-11-16","position_f":3},
        {"created_at":"2013-11-17","position_f":6},
        {"created_at":"2013-11-18","position_f":11},
        {"created_at":"2013-11-19","position_f":46}
 ]

使用哈希对这些数组求和并在Expected result

下得到结果的简单方法是什么？

谢谢。

Answer 1

这是：

arr =[
        [
            {"created_at"=>"2013-11-16","position_f"=>10},
            {"created_at"=>"2013-11-17","position_f"=>0},
            {"created_at"=>"2013-11-18","position_f"=>0},
            {"created_at"=>"2013-11-19","position_f"=>1}
        ],
        [
            {"created_at"=>"2013-11-16","position_f"=>20},
            {"created_at"=>"2013-11-17","position_f"=>0},
            {"created_at"=>"2013-11-18","position_f"=>11},
            {"created_at"=>"2013-11-19","position_f"=>45}
        ]
    ]
p arr.inject(0){|sum,a| sum +=a[0]["position_f"]} # => 30

使用awesome_print

进行更新

require 'awesome_print'

arr = [
    [
        {"created_at"=>"2013-11-16","position_f"=>1},
        {"created_at"=>"2013-11-17","position_f"=>3},
        {"created_at"=>"2013-11-18","position_f"=>0},
        {"created_at"=>"2013-11-19","position_f"=>1}
    ],
    [
        {"created_at"=>"2013-11-16","position_f"=>2},
        {"created_at"=>"2013-11-17","position_f"=>3},
        {"created_at"=>"2013-11-18","position_f"=>11},
        {"created_at"=>"2013-11-19","position_f"=>45}
    ]
]

nwar = arr.flatten.group_by{|h| h['created_at']}.map do |k,v|
  {"created_at" => k,"position_f" => v.reduce(0){|sum,h| sum += h['position_f']}}
end

ap nwar,:index => false

输出

[
    {
        "created_at" => "2013-11-16",
        "position_f" => 3
    },
    {
        "created_at" => "2013-11-17",
        "position_f" => 6
    },
    {
        "created_at" => "2013-11-18",
        "position_f" => 11
    },
    {
        "created_at" => "2013-11-19",
        "position_f" => 46
    }
]

Answer 2

假设您想要对所有数组中的所有position_f值求和，我会这样做：

sum = 0
outer_array.each do |inner_array|
  inner_array.each do |inner_hash|
    sum += inner_hash['position_f']
  end
end

Answer 3

result = data.flatten.group_by{|x|x[:created_at]}.map do |x|  
  val = x.first
  sum = x.last.map{|x| x[:position_f]}.inject(:+)
  {created_at: val, position_f: sum} 
end

p result #=>
    [{:created_at=>"2013-11-16", :position_f=>0},
     {:created_at=>"2013-11-17", :position_f=>0},
     {:created_at=>"2013-11-18", :position_f=>11},
     {:created_at=>"2013-11-19", :position_f=>46}]

或者，如果您不关心最终演示文稿：

result = data.flatten.each_with_object(Hash.new(0)) do |item,hash|
  hash[item[:created_at]] += item[:position_f]
end

p result #=> {"2013-11-16"=>0, "2013-11-17"=>0, "2013-11-18"=>11, "2013-11-19"=>46}

Answer 4

我认为这就是你想要的：

data = [
  [
    {"created_at"=>"2013-11-16","position_f"=>0},
    {"created_at"=>"2013-11-17","position_f"=>0},
    {"created_at"=>"2013-11-18","position_f"=>0},
    {"created_at"=>"2013-11-19","position_f"=>1}
  ],
  [
    {"created_at"=>"2013-11-16","position_f"=>0},
    {"created_at"=>"2013-11-17","position_f"=>0},
    {"created_at"=>"2013-11-18","position_f"=>11},
    {"created_at"=>"2013-11-19","position_f"=>45}
  ]
]

@hash = Hash.new(0)

data.each do |arr_of_hashes|
  arr_of_hashes.each_with_object(@hash) {|arr_hash, master_hash| master_hash[arr_hash["created_at"]] += arr_hash["position_f"]}
end

@hash

这是IRB的一次运行：

1.9.3p448 :020 > data = [
1.9.3p448 :021 >       [
1.9.3p448 :022 >           {"created_at"=>"2013-11-16","position_f"=>0},
1.9.3p448 :023 >           {"created_at"=>"2013-11-17","position_f"=>0},
1.9.3p448 :024 >           {"created_at"=>"2013-11-18","position_f"=>0},
1.9.3p448 :025 >           {"created_at"=>"2013-11-19","position_f"=>1}
1.9.3p448 :026?>       ],
1.9.3p448 :027 >       [
1.9.3p448 :028 >           {"created_at"=>"2013-11-16","position_f"=>0},
1.9.3p448 :029 >           {"created_at"=>"2013-11-17","position_f"=>0},
1.9.3p448 :030 >           {"created_at"=>"2013-11-18","position_f"=>11},
1.9.3p448 :031 >           {"created_at"=>"2013-11-19","position_f"=>45}
1.9.3p448 :032?>       ]
1.9.3p448 :033?>   ]
 => [[{"created_at"=>"2013-11-16", "position_f"=>0}, {"created_at"=>"2013-11-17", "position_f"=>0}, {"created_at"=>"2013-11-18", "position_f"=>0}, {"created_at"=>"2013-11-19", "position_f"=>1}], [{"created_at"=>"2013-11-16", "position_f"=>0}, {"created_at"=>"2013-11-17", "position_f"=>0}, {"created_at"=>"2013-11-18", "position_f"=>11}, {"created_at"=>"2013-11-19", "position_f"=>45}]] 
1.9.3p448 :034 > 
1.9.3p448 :035 >   @hash = Hash.new(0)
 => {} 
1.9.3p448 :036 > 
1.9.3p448 :037 >   data.each do |arr_of_hashes|
1.9.3p448 :038 >       arr_of_hashes.each_with_object(@hash) {|arr_hash, master_hash| master_hash[arr_hash["created_at"]] += arr_hash["position_f"]}
1.9.3p448 :039?>   end
 => [[{"created_at"=>"2013-11-16", "position_f"=>0}, {"created_at"=>"2013-11-17", "position_f"=>0}, {"created_at"=>"2013-11-18", "position_f"=>0}, {"created_at"=>"2013-11-19", "position_f"=>1}], [{"created_at"=>"2013-11-16", "position_f"=>0}, {"created_at"=>"2013-11-17", "position_f"=>0}, {"created_at"=>"2013-11-18", "position_f"=>11}, {"created_at"=>"2013-11-19", "position_f"=>45}]] 
1.9.3p448 :040 > 
1.9.3p448 :041 >   @hash
 => {"2013-11-16"=>0, "2013-11-17"=>0, "2013-11-18"=>11, "2013-11-19"=>46}

然后，如果你想要总数，你可以这样做：

1.9.3p448 :081 > @hash.values.sum
 => 57

如何使用哈希对数组求和

4 个答案: