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

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

推荐答案

最简单的方法是 使用块参数构建您的哈希:

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}}"

new 的这个表单创建一个新的空哈希作为默认值.你不想要这个:

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

hash = Hash.new({ })

因为这将对所有默认条目使用完全相同的哈希.

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

此外，正如 Phrogz 所指出的，您可以使用 default_proc:

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:我想我应该澄清我对 Hash.new({ }) 的警告.当你这样说时:

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

h = Hash.new({ })

这很像这样说:

h = Hash.new
h.default = { }

然后，当您访问 h 以将某些内容分配为 h[:k][:m] = y 时，它的行为就好像您这样做了:

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

然后，如果你h[:k2][:n] = z，你最终会分配h.default[:n] = z.请注意，h 仍然表示 h.has_key?(:k) 是假的.

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.

但是，当你这样说时:

h = Hash.new(0)

一切都会好起来的，因为你永远不会在这里修改 h[k]，你只会从 h 读取一个值(它将使用默认的如有必要)或为 h 分配一个新值.

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.

这篇关于如果 hash['a'] 不存在，如何分配 hash['a']['b']='c'?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！