Question

如果我有像这样的红宝石哈希，

first = {：a =＆gt; 1，：b =＆gt; 2，：c =＆gt; 3，：d =＆gt; 4，：e =＆gt; 5}

如何通过单行脚本

实现这一目标

second = {：a =＆gt; 1，：c =＆gt; 3，：e =＆gt; 5}

非常感谢你。

Answer 1

你想要的是一个漂亮的代码行

second = first.slice(:a, :c, :e) # => {:a=>1, :c=>3, :e=>5}

编辑：之前的回答是使用Rails。这是一个使用Ruby的解决方案

second = first.delete_if {|k,v| ![:a, :c, :e].include?(k) } # => {:a=>1, :c=>3, :e=>5}

Answer 2

first.keep_if {|键| [：一，：C，：d] .INCLUDE（键）}

Answer 3

尝试delete_if或keep_if，它们都是Ruby核心的一部分。它们都在当前哈希上操作。 slice也已经是核心Ruby的一部分。

first =  {:a=>1,:b=>2,:c=>3,:d=>4,:e=>5}
first_clone = first.clone
p first.keep_if { |key| [:a, :c, :e].include?(key) } # =>  {:a=>1,:c=>3,:e=>5}
p first_clone.delete_if { |key, value| [:b, :d, :f].include?(key) } # =>  {:a=>1,:c=>3,:e=>5}

文档：

Answer 4

方法1

first = {:a=>1,:b=>2,:c=>3,:d=>4,:e=>5}
first.delete(:b)
first.delete(:d)
second = first

方法2

first = {:a=>1,:b=>2,:c=>3,:d=>4,:e=>5}
second = first.delete_if {|key, value| key == :b || key == :d }

Answer 5

假设您不知道要保留/删除的密钥......您可以这样做：

first = {:a=>1,:b=>2,:c=>3,:d=>4,:e=>5}
iterator = 0
second = {}
first.each_pair do |key, value|
  second[key] = value if iterator % 2 == 0
  iterator += 1
end

second # is now {:a=>1,:c=>3,:e=>5}

Ruby如何获得部分哈希

5 个答案: