如果密鑰相同，如何將哈希數組分成不同的數組？

Question

我有哈希以下形式的數組：

{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>5.0} 
{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.8499999999999996} 
{"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>0.0} 
{"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.4} 

基本上我想是數組包含的關鍵beneficiary_document相同的值陣列分開的，所以在這個例子中我會人希望兩個陣列，一個包含：

{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>5.0} 
{"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>0.0} 

和含

{"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.4} 
{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.8499999999999996} 

我怎樣授予這一個又一個？

非常感謝您的閱讀。

Answer 1

考慮：

tst=[ 
{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>5.0}, 
{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.84}, 
{"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>0.0}, 
{"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.4} 
] 

您可以使用.group_by獲得通過的關鍵要素的哈希值。在這種情況下，使用傳遞給塊的密鑰["beneficiary_document"]，您將通過該密鑰獲得數組散列值 - 在這種情況下爲兩個。

你可以這樣做：

tst.group_by { |h| h["beneficiary_document"] } 
# {"43991028"=>[{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>5.0}, {"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>0.0}], "71730550"=>[{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.84}, {"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.4}]} 

要看到它漂亮的印刷：

require "pp" 
PP.pp(tst.group_by {|h| h["beneficiary_document"] },$>,120) 
{"43991028"=> 
    [{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>5.0}, 
    {"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"43991028", "calification_by_qualifier"=>0.0}], 
"71730550"=> 
    [{"user_id"=>2, "user_name"=>"Pepo", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.84}, 
    {"user_id"=>3, "user_name"=>"Carlos", "beneficiary_document"=>"71730550", "calification_by_qualifier"=>3.4}]} 

您也可以實現與返回數組作爲default procedure哈希相同的結果，然後調用.map高於tst並通過該密鑰將散列推入陣列中：

h=Hash.new { |h,k| h[k]=[] } 
tst.map { |eh| h[eh["beneficiary_document"]].push(eh) } 

或者是合併成一個單一的語句：

tst.each_with_object(Hash.new { |h,k| h[k]=[] }) { |g,h| 
    h[g["beneficiary_document"]].push(g)} 

這三種方法創建相同的哈希值。第一個，.group_by，是最簡單的。

Answer 2

這裏有三種方法通過構造一個散列並提取值來獲得所需的結果。

arr = [{"id"=>2, "name"=>"Pepo", "doc"=>"43991028", "cal"=>5.0}, 
     {"id"=>2, "name"=>"Pepo", "doc"=>"71730550", "cal"=>3.8}, 
     {"id"=>3, "name"=>"Carlos", "doc"=>"43991028", "cal"=>0.0}, 
     {"id"=>3, "name"=>"Carlos", "doc"=>"71730550", "cal"=>3.4}] 

＃1

這使用的Hash::new，其包括當被執行h[k]時調用，對於沒有密鑰k散列h的塊的形式。

arr.each_with_object(Hash.new { |h,k| h[k]=[] }) { |g,h| h[g["doc"]] << g }. 
    values 
    #=> [[{"id"=>2, "name"=>"Pepo", "doc"=>"43991028", "cal"=>5.0}, 
    #  {"id"=>3, "name"=>"Carlos", "doc"=>"43991028", "cal"=>0.0}], 
    # [{"id"=>2, "name"=>"Pepo", "doc"=>"71730550", "cal"=>3.8}, 
    #  {"id"=>3, "name"=>"Carlos", "doc"=>"71730550", "cal"=>3.4}]] 

＃2

這相當於＃1。

arr.each_with_object({}) { |g,h| (h[g["doc"]] ||= []) << g }. 
    values 

＃3

這使用的Hash#update（又名merge!）的形式，其使用的塊（這裏{ |_,o,n| o+n }），以確定存在於被合併兩個散列密鑰的值。請參閱文檔瞭解確定值塊的三個變量的定義。

arr.each_with_object({}) { |g,h| h.update(g["doc"]=>[g]) { |_,o,n| o+n } }. 
    values