How to handle combination []+= for auto-vivifying hash in Ruby?

二次信任 提交于 2019-12-19 11:01:23

问题


In order to implement auto-vivification of Ruby hash, one can employ the following class

class AutoHash < Hash
  def initialize(*args)
    super()
    @update, @update_index = args[0][:update], args[0][:update_key] unless 
args.empty?
  end

  def [](k)
    if self.has_key?k
      super(k)
    else
      AutoHash.new(:update => self, :update_key => k)
    end
  end

  def []=(k, v)
    @update[@update_index] = self if @update and @update_index
    super
  end

  def few(n=0)
    Array.new(n) { AutoHash.new }
  end
end

This class allows to do the following things

a = AutoHash.new
a[:a][:b] = 1
p a[:c] # => {}             # key :c has not been created
p a     # => {:a=>{:b=>1}}  # note, that it does not have key :c

a,b,c = AutoHash.new.few 3
b[:d] = 1
p [a,b,c] # => [{}, {:d=>1}, {}]  # hashes are independent

There is a bit more advanced definition of this class proposed by Joshua, which is a bit hard for me to understand.

Problem

There is one situation, where I think the new class can be improved. The following code fails with the error message NoMethodError: undefined method '+' for {}:AutoHash

a = AutoHash.new
5.times { a[:sum] += 10 }

What would you do to handle it? Can one define []+= operator?


Related questions

  1. Is auto-initialization of multi-dimensional hash array possible in Ruby, as it is in PHP?
  2. Multiple initialization of auto-vivifying hashes using a new operator in Ruby ruby hash initialization r
  3. still open: How to create an operator for deep copy/cloning of objects in Ruby?

回答1:


There is no way to define a []+= method in ruby. What happens when you type

x[y] += z

is

x[y] = x[y] + z

so both the [] and []= methods are called on x (and + is called on x[y], which in this case is an AutoHash). I think that the best way to handle this problem would be to define a + method on AutoHash, which will just return it's argument. This will make AutoHash.new[:x] += y work for just about any type of y, because the "empty" version of y.class ('' for strings, 0 for numbers, ...) plus y will almost always equal y.

class AutoHash
  def +(x); x; end
end

Adding that method will make both of these work:

# Numbers:
a = AutoHash.new
5.times { a[:sum] += 10 }
a[:sum] #=> 50

# Strings:
a = AutoHash.new
5.times { a[:sum] += 'a string ' }
a[:sum] #=> "a string a string a string a string a string "

And by the way, here is a cleaner version of your code:

class AutoHash < Hash
  def initialize(args={})
    super
    @update, @update_index = args[:update], args[:update_key]
  end

  def [](k)
    if has_key? k
      super(k)
    else
      AutoHash.new :update => self, :update_key => k
    end
  end

  def []=(k, v)
    @update[@update_index] = self if @update and @update_index
    super
  end

  def +(x); x; end

  def self.few(n)
    Array.new(n) { AutoHash.new }
  end
end

:)




回答2:


What I think you want is this:

hash = Hash.new { |h, k| h[k] = 0 }

hash['foo'] += 3 # => 3

That will return 3, then 6, etc. without an error, because the the new value is default assigned 0.




回答3:


require 'xkeys' # on rubygems.org

a = {}.extend XKeys::Hash
a[:a, :b] = 1
p a[:c] # => nil (key :c has not been created)
p a # => { :a => { :b => 1 } }

a.clear
5.times { a[:sum, :else => 0] += 10 }
p a # => { :sum => 50 }


来源:https://stackoverflow.com/questions/3172342/how-to-handle-combination-for-auto-vivifying-hash-in-ruby

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!