如果hash ['a']不存在，如何分配hash ['a'] ['b'] ='c'？ [英] How to assign hash['a']['b']= 'c' if hash['a'] doesn't exist?

问题描述

有没有比

简单的方法

 如果hash.key？（'a'）
 hash [' a'] ['b'] ='c'
 else 
 hash ['a'] = {} $ b $ hash ['a'] ['b'] ='c'
 end 
  

解决方案

最简单的方法是用块参数构建你的Hash ：

  hash = Hash.new {| h，k | h [k] = {}} 
 hash ['a'] ['b'] = 1 
 hash ['a'] ['c'] = 1 
 hash ['b '] ['c'] = 1 
 puts hash.inspect 
＃{a=> {b=> 1，c=> 1}，b => {c=> 1}}
  

$ c> new 创建一个新的空Hash作为默认值。您不希望这样：

  hash = Hash.new（{}）
  

，因为它将为所有默认条目使用完全相同哈希。

$ b $另外，正如Phrogz所指出的那样，你可以使用 default_proc ：

  hash = Hash.new {| h，k | h [k] = Hash.new（& h.default_proc）} 
  

UPDATE ：我想我应该对 Hash.new（{}）进行澄清。当你这样说：

  h = Hash.new（{}）
  

这就好像这样说：

  h = Hash.new 
 h.default = {} 
  

然后，当您访问 h 将 h [：k] [：m] = y 赋值，它的行为就像您这样做：

  if（h.has_key？（：k））
h [：k] [：m] = y 
 else 
 h.default [：m] = y 
 end 
  

然后，如果你 h [：k2] [：n] = z ，你最终将分配 h.default [：n] = ž。请注意， h 仍然表示 h.has_key？（：k）为false。

然而，当你这样说的时候：

  h = Hash.new（0）
  

一切都会好起来的，因为你永远不会修改 h [k] 在这里，你只能从 h （它将使用默认值，如果需要的话）读取一个值，或者给赋值一个新值。 h 。

Is there any way simpler than

if hash.key?('a')
  hash['a']['b'] = 'c' 
else  
  hash['a'] = {}
  hash['a']['b'] = 'c' 
end

解决方案

The easiest way is to construct your Hash with a block argument:

hash = Hash.new { |h, k| h[k] = { } }
hash['a']['b'] = 1
hash['a']['c'] = 1
hash['b']['c'] = 1
puts hash.inspect
# "{"a"=>{"b"=>1, "c"=>1}, "b"=>{"c"=>1}}"

This form for new creates a new empty Hash as the default value. You don't want this:

hash = Hash.new({ })

as that will use the exact same hash for all default entries.

Also, as Phrogz notes, you can make the auto-vivified hashes auto-vivify using default_proc:

hash = Hash.new { |h, k| h[k] = Hash.new(&h.default_proc) }

UPDATE: I think I should clarify my warning against Hash.new({ }). When you say this:

h = Hash.new({ })

That's pretty much like saying this:

h = Hash.new
h.default = { }

And then, when you access h to assign something as h[:k][:m] = y, it behaves as though you did this:

if(h.has_key?(:k))
    h[:k][:m] = y
else
    h.default[:m] = y
end

And then, if you h[:k2][:n] = z, you'll end up assigning h.default[:n] = z. Note that h still says that h.has_key?(:k) is false.

However, when you say this:

h = Hash.new(0)

Everything will work out okay because you will never modified h[k] in place here, you'll only read a value from h (which will use the default if necessary) or assign a new value to h.

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