在散列数组中查找值

Question

taglist = [{:name=>"Daniel_Xu_Forever", :tag=>["helo", "world"]},   
{:name=>"kcuf", :tag=>["hhe"]},  
{:name=>"fine", :tag=>[]},   
{:name=>"how hare you", :tag=>[]},  
{:name=>"heki", :tag=>["1", "2", "3"]}, 
{:name=>"railsgirls", :tag=>[]},  
{:name=>"_byoy", :tag=>[]},   
{:name=>"ajha", :tag=>[]},  
{:name=>"nimei", :tag=>[]}]

How to get specified name's tag from taglist

For example , I want to extract user "fine" 's tag?

Could this be achieved without do iterator?

Answer 1

This will return the contents of the :tag key for any users name which == 'fine'

taglist.select { |x| x[:name] == 'fine' }.map { |u| u[:tag] }

First you select out only the users you are interested with .select .

And then use .map to collect an array of only what you want.

In this case the end result will be: []

Answer 2

Is do really an iterator?

taglist.find{|tl| tl[:name] == 'fine'}[:tag]

Just to be silly how about:

eval taglist.to_s[/:name=>"fine", :tag=>(.*?)}/, 1]
#=> []

Answer 3

No, it cannot be done without a loop.

And even if you find a solution where your code avoids a loop, for sure the library function that you're calling will include a loop. Finding an element in an array requires a loop. Period.

For example, take this (contrived) example

  pattern = "fine"
  def pattern.===(h); self == h[:name]; end
  taglist.grep(pattern)

which does not seem to use a loop, but calls grep which is implemented using a loop.

Or another, equally contrived, example

  class Hash; def method_missing(sym); self[sym]; end; end
  taglist.group_by(&:name)["fine"]

which again does seem to call group_by without a loop, but actually it does.

So the answer is, no.

Answer 4

So my first answer missed the no do rule. Here is an answer that doesn't use a do block.

i=0
begin
  if taglist[i][:name] == 'fine'
    tag = taglist[i][:tag]
    break
  end
  i+=1
end while i < taglist.length - 0

Technically I think this is still using a block. But probably satisfies the restriction.