使用克隆时如何消除 ruby​​ 中的重复？

Question

此方法只是更改找到的任何player_id 的id。 Reek 讨厌它，但我找不到一种方法以有意义的方式重构它。

(1..9).each { |n|
    n = n.to_s
    self.player_ids[n] = self.site_id.clone << "_" << self.player_ids[n].clone if self.player_ids[n]        
}

我是否必须忍受这种重复，因为克隆功能不允许我：

player_id = self.player_ids[n]
player_id = self.site_id.clone << "_" << player_id.clone if player_id

示例输入：

{:player_ids => {"2" => "player_name1", "6" => "player_name4", "9" => "player_name9"}

输出：

{:player_ids => {"2" => "PRE_player_name1", "6" => "PRE_player_name4", "9" => "PRE_player_name9"}

Answer 1

根据您的示例输入和输出，这里有两种可能的技术：

site_id = 'PRE'
prefix  = "#{site_id}_"
h = {:player_ids => {"2" => "player_name1", "6" => "player_name4", "9" => "player_name9"}}

# If mutating the original hash is not OK
h2 = h.dup
h2[:player_ids] = Hash[ h[:player_ids].map{ |s,n| [s, n.sub(/^/,prefix)] } ]
p h, h2
#=> {:player_ids=>{"2"=>"player_name1", "6"=>"player_name4", "9"=>"player_name9"}}
#=> {:player_ids=>{"2"=>"PRE_player_name1", "6"=>"PRE_player_name4", "9"=>"PRE_player_name9"}}

# If mutating the original hash is OK
h[:player_ids].each{ |id_string,name| name.sub! /^/, prefix }
p h
#=> {:player_ids=>{"2"=>"PRE_player_name1", "6"=>"PRE_player_name4", "9"=>"PRE_player_name9"}}

如果这不是您想要的，请编辑您问题的示例输入/输出并发表评论以澄清您的需求。

Answer 2

仅当您在字符串上使用变异操作 (<<) 时才需要 #clone。使用字符串插值可以防止不必要的克隆，也使您的代码更加惯用。

player_ids[n] &&= "#{site_id.clone}_#{player_ids[n]}"

作为奖励，我删除了对 self 的不必要引用，并将 if 替换为 and-equals。