紅寶石：刪除特定鍵的所有實例從哈希

Question

的哈希我有一個像

 
h = {1 => {"inner" => 45}, 2 => {"inner" => 46}, "inner" => 47} 

哈希如何刪除每對包含密鑰「內部」？
你可以看到一些「內部」對直接出現在h而其他h

注意成對出現的，我只想要刪除的「內部」對，所以如果我打電話給我的質量上刪除方法上述散，我應該得到

h = {1 => {}, 2 => {}}

由於這些對不具備一鍵== 「內部」

Answer 1

def f x 
    x.inject({}) do |m, (k, v)| 
    v = f v if v.is_a? Hash # note, arbitrarily recursive 
    m[k] = v unless k == 'inner' 
    m 
    end 
end 

p f h 

更新：略有改善...

def f x 
    x.is_a?(Hash) ? x.inject({}) do |m, (k, v)| 
    m[k] = f v unless k == 'inner' 
    m 
    end : x 
end 

Answer 2

真的，這是什麼拒絕！是：

def f! x 
    x.reject!{|k,v| 'inner' == k} if x.is_a? Hash 
    x.each{|k,v| f! x[k]} 
end 

Answer 3

這裏是我想出了：

class Hash 
    def deep_reject_key!(key) 
    keys.each {|k| delete(k) if k == key || self[k] == self[key] } 

    values.each {|v| v.deep_reject_key!(key) if v.is_a? Hash } 
    self 
    end 
end 

工程的哈希或HashWithIndifferentAccess

> x = {'1' => 'cat', '2' => { '1' => 'dog', '2' => 'elephant' }} 
=> {"1"=>"cat", "2"=>{"1"=>"dog", "2"=>"elephant"}} 

> y = x.with_indifferent_access 
=> {"1"=>"cat", "2"=>{"1"=>"dog", "2"=>"elephant"}} 

> x.deep_reject_key!(:"1") 
=> {"1"=>"cat", "2"=>{"1"=>"dog", "2"=>"elephant"}} 

> x.deep_reject_key!("1") 
=> {"2"=>{"2"=>"elephant"}} 

> y.deep_reject_key!(:"1") 
=> {"2"=>{"2"=>"elephant"}} 

Answer 4

類似的答案，但它是一個白名單類型的方法。對於紅寶石1.9+

# recursive remove keys 
def deep_simplify_record(hsh, keep) 
    hsh.keep_if do |h, v| 
    if v.is_a?(Hash) 
     deep_simplify_record(v, keep) 
    else 
     keep.include?(h) 
    end 
    end 
end 

hash = {:a => 1, :b => 2, :c => {:a => 1, :b => 2, :c => {:a => 1, :b => 2, :c => 4}} } 
deep_simplify_record(hash, [:b, :c]) 
# => {:b=>2, :c=>{:b=>2, :c=>{:b=>2, :c=>4}}} 

此外，這裏還有一些我喜歡用於散列的其他方法。 https://gist.github.com/earlonrails/2048705

Answer 5

def except_nested(x,key) 
    case x 
    when Hash then x = x.inject({}) {|m, (k, v)| m[k] = except_nested(v,key) unless k == key ; m } 
    when Array then x.map! {|e| except_nested(e,key)} 
    end 
    x 
end