紅寶石 - 組通過重複多次哈希

Question

我有以下陣列

t = [ 
    {nil => 1, 10 => 2, 16 => 4, 5=> 10}, 
    {nil => 9, 5 => 2, 17 => 3, 10 => 2}, 
    {10 => 4, 5 => 9, 17 => 1} 
] 

我怎麼能得到這個作爲結果的關鍵？

{nil => [1,9,0],10 => [2,2,4], 16 => [4,0,0], 5 => [10,2,9], 17=>[0,3,1]} 

我已經看到了，我可以用這樣的

t.group_by{|h| h['key']} 

，但我不知道我是否可以把一個正則表達式的括號內

在此先感謝

哈維爾

編輯：

只是想組由陣列內的每個散列的每個鍵，如果該鍵不存在，則該值爲0的散列

Answer 1

不是我曾經寫過的最優雅的代碼嘗試，但它的工作，很容易理解：

def jqq(a) 
    keys = [] 
    result = {} 

    a.each do |h| 
    keys += h.keys 
    end 

    keys.uniq.each do |key| 
    result[key] = [] 
    a.each do |h| 
     h.default = 0 
     result[key] << h[key] 
    end 
    end 

    result 
end 

t = [ 
    {nil => 1, 10 => 2, 16 => 4, 5=> 10}, 
    {nil => 9, 5 => 2, 17 => 3, 10 => 2}, 
    {10 => 4, 5 => 9, 17 => 1} 
] 

puts jqq(t) 
# {nil=>[1, 9, 0], 10=>[2, 2, 4], 16=>[4, 0, 0], 5=>[10, 2, 9], 17=>[0, 3, 1]} 

Answer 2

我不認爲有任何的任何功能可 只是給了一個與哈希

def do_my_work(data) 
    hash = {} 
    #get all keys first 
    arr.map{|m| m.keys}.flatten.uniq.each {|a| hash[a]=[]} 
    # Now iterate and fill the values 
    arr.each do |elm| 
     hash.each do |k,v| 
     hash[k] << (elm[k].nil? ? 0 : elm[k]) 
     end 
    end 
    end 

    hash = do_my_work(t) 

    puts hash 
    # => {nil=>[1, 9, 0], 10=>[2, 2, 4], 16=>[4, 0, 0], 5=>[10, 2, 9], 17=>[0, 3, 1]} 

Answer 3

怎麼樣這一個模糊：

t = [ 
    {nil => 1, 10 => 2, 16 => 4, 5=> 10}, 
    {nil => 9, 5 => 2, 17 => 3, 10 => 2}, 
    {10 => 4, 5 => 9, 17 => 1} 
] 

# Create hash of possible keys 
keys = t.reduce({}) { |m, h| h.each_key { |k| m[k] = [] }; m } 

# Iterate through array, for each hash, for each key, append the 
# value if key is in hash or zero otherwise 
t.reduce(keys) { |m, h| m.each_key { |k| m[k] << (h[k] || 0) }; m } 

puts keys 
#=> {nil=>[1, 9, 0], 10=>[2, 2, 4], 16=>[4, 0, 0], 5=>[10, 2, 9], 17=>[0, 3, 1]}