合併基於Ruby的

-4

我有一個數組，我想，以配合在多維數組的散列的其他陣列id鍵的值哈希鍵和值哈希的多維數組，合併基於Ruby的

input = [ 
    [ {"id"=>"1","name"=>"a"}, 
    {"id"=>"2","name"=>"b"}, 
    {"id"=>"3","name"=>"c"}, 
    {"id"=>"4","name"=>"d"}, 
    {"id"=>"5","name"=>"e"}, 
    {"id"=>"6","name"=>"f"} 
    ], 
    [ {"id"=>"3","hoby"=>"AA"}, 
    {"id"=>"3","hoby"=>"BB"}, 
    {"id"=>"1","hoby"=>"CC"}, 
    {"id"=>"1","hoby"=>"DD"}, 
    {"id"=>"4","hoby"=>"EE"} 
    ], 
    [ {"id"=>"1","language"=>"A"}, 
    {"id"=>"1","language"=>"B"}, 
    {"id"=>"2","language"=>"B"}, 
    {"id"=>"2","language"=>"C"}, 
    {"id"=>"6","language"=>"D"} 
    ] 
]

我需要一個像陣列輸出，

output = [ 
    {"id"=>"1","name"=>"a","id"=>"1","hoby"=>"CC","id"=>"1","language"=>"A","id"=>"1","language"=>"B"}, 
    {"id"=>"2","name"=>"b","id"=>"2","language"=>"B"}, 
    {"id"=>"3","name"=>"c","id"=>"3","hoby"=>"AA","id"=>"3","hoby"=>"BB"}, 
    {"id"=>"4","name"=>"d","id"=>"4","hoby"=>"EE"}, 
    {"id"=>"5","name"=>"e"}, 
    {"id"=>"6","name"=>"f","id"=>"6","language"=>"D"} 
]

我已經寫了代碼對於這一點，

len = input.length - 1 
output = [] 
input[0].each do |value,index| 
    for i in 1..len 
    input[i].each do |j| 
     if value["id"] == j["id"] 
     output << value.merge(j) 
     end 
    end 
    end 
end

但我我得到了錯誤的輸出數組。在多維數組中可能有任何數量的子數組。

感謝，

來源

2015-04-08 rick

用於'ID = 1'存在多個嗜好。哪個「愛好」會優先？ – shivam

您需要添加更多關於您的需求的信息：人們只是猜測你想要什麼，因爲你所要求的不是有效的數據結構。如果我是你，我會後退一步，描述你正試圖解決的問題，而不是說「如何將這些數據轉化爲這些數據？」。 –

也許這可以幫助你。

input = [ 
    [ 
    {"id"=>"1","name"=>"a"}, 
    {"id"=>"2","name"=>"b"}, 
    {"id"=>"3","name"=>"c"}, 
    {"id"=>"4","name"=>"d"}, 
    {"id"=>"5","name"=>"e"}, 
    {"id"=>"6","name"=>"f"} 
    ], 
    [ 
    {"id"=>"3","hoby"=>"AA"}, 
    {"id"=>"3","hoby"=>"BB"}, 
    {"id"=>"1","hoby"=>"CC"}, 
    {"id"=>"1","hoby"=>"DD"}, 
    {"id"=>"4","hoby"=>"EE"} 
    ], 
    [ 
    {"id"=>"1","language"=>"A"}, 
    {"id"=>"1","language"=>"B"}, 
    {"id"=>"2","language"=>"B"}, 
    {"id"=>"2","language"=>"C"}, 
    {"id"=>"6","language"=>"D"} 
    ] 
]

這樣就可以使您的「排序」結果。

output = {} 
input.flatten.each do |h| 
    output[h["id"]] = {} unless output[h["id"]] 
    output[h["id"]].merge!(h) 
end 

output.values 
# => [ 
# => {"id"=>"1", "name"=>"a", "hoby"=>"DD", "language"=>"B"}, 
# => {"id"=>"2", "name"=>"b", "language"=>"C"}, 
# => {"id"=>"3", "name"=>"c", "hoby"=>"BB"}, 
# => {"id"=>"4", "name"=>"d", "hoby"=>"EE"}, 
# => {"id"=>"5", "name"=>"e"}, 
# => {"id"=>"6", "name"=>"f", "language"=>"D"} 
# => ]

但更好的方法是在輸入使用散列。你可以像輸入密鑰一樣定義輸入，如「hash」和「id」，所以如果你生成了數據，你沒有問題來排序。

成才這樣

{ 
    "1" => {"name" => "a", "hoby" => "DD", "language" => "B"} 
}

來源

2015-04-08 07:58:10

首先 - 它是不可能有一個哈希兩個元素使用相同的密鑰。將值分配給某個鍵時，將使新值的下一個相同鍵的分配覆蓋上一個鍵。

讓我們考慮例如：

hash = {} 
hash["id"] = 1 
hash["id"] = 3 
hash["id"] = 5

你會期望hash["id"]輸出什麼？ 1,3,5或者也許[1, 3, 5]？它會輸出5，因爲這是唯一鍵的最後一個分配。

話雖如此，這是不可能出現多次存儲在您的哈希值，但你可以嘗試處理它的東西，如：

input.flatten 
    .group_by { |h| h["id"] } 
    .map do |k, a| 
     a.each_with_object({}) { |in_h, out_h| out_h.merge!(in_h) } 
    end

這將導致與像哈希：

[{"id"=>"1", "name"=>"a", "hoby"=>"DD", "language"=>"B"}, 
{"id"=>"2", "name"=>"b", "language"=>"C"}, 
{"id"=>"3", "name"=>"c", "hoby"=>"BB"}, 
{"id"=>"4", "name"=>"d", "hoby"=>"EE"}, 
{"id"=>"5", "name"=>"e"}, 
{"id"=>"6", "name"=>"f", "language"=>"D"}]

那麼，它不是你所期望的哈希，但至少它可能會讓你朝某個方向發展。

希望有幫助！

來源

2015-04-08 07:58:04

合併基於Ruby的

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