如果密钥相同，如何将哈希数组分成不同的数组？

Question

我有哈希以下形式的数组：

{"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} 

基本上我想是数组包含的关键beneficiary_document相同的值阵列分开的，所以在这个例子中我会人希望两个阵列，一个包含：

{"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} 

我怎样授予这一个又一个？

非常感谢您的阅读。

Answer 1

考虑：

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获得通过的关键要素的哈希值。在这种情况下，使用传递给块的密钥["beneficiary_document"]，您将通过该密钥获得数组散列值 - 在这种情况下为两个。

你可以这样做：

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}]} 

您也可以实现与返回数组作为default procedure哈希相同的结果，然后调用.map高于tst并通过该密钥将散列推入阵列中：

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

或者是合并成一个单一的语句：

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

这三种方法创建相同的哈希值。第一个，.group_by，是最简单的。

Answer 2

这里有三种方法通过构造一个散列并提取值来获得所需的结果。

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

这使用的Hash::new，其包括当被执行h[k]时调用，对于没有密钥k散列h的块的形式。

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

这相当于＃1。

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

＃3

这使用的Hash#update（又名merge!）的形式，其使用的块（这里{ |_,o,n| o+n }），以确定存在于被合并两个散列密钥的值。请参阅文档了解确定值块的三个变量的定义。

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