how can i delete a hash based on multiple values

Question

I have the following list:

hash_list = 
{ "a"=>{"unit_id"=>"43", "dep_id"=>"153","_destroy"=>"false"},
  "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
  "c"=>{"unit_id"=>"43", "dep_id"=>"154", "_destroy"=>"false"},
  "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"}
}

I need the result as:

{
 "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
 "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
 "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"}
}

Basically , the unit_id should not repeat. But, all _destroy=="1", entries can appear in the list.

Please help.

Based on what multiple values?? What is the logic behind removing "c" from the result hash? — Surya
– Surya, Commented Jun 8, 2015 at 9:15

Gagan Gami · Accepted Answer · 2015-06-08 09:37:25Z

5

Try this:

> hash_list.to_a.uniq{|_,v|v["unit_id"] unless v["_destroy"] == "1"}.to_h

#=> {
      "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
      "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
      "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"}
    }

This will check for unit_id as uniq and also let appear "_destory" == "1" entries as you have mention.

edited Jun 8, 2015 at 9:37

answered Jun 8, 2015 at 9:28

Gagan Gami

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6 Comments

Wayne Conrad Over a year ago

There's a bug here: Only one element having _destroy == 1 will be retained. The others will be deleted.

Gagan Gami Over a year ago

@WayneConrad : let me check

Cary Swoveland Over a year ago

@WayneConrad is well-known as a spoil-sport. He's saying that v["unit_id"] unless v["_destroy"] == "1" #=> nil for all values h for which h["_destroy"] #=> "1", so uniq will keep only the first of those. Regretfully, I don't see a fix.

Cary Swoveland Over a year ago

@WayneConrad, anyone who spotted the problem would have been the spoil-sport, considering that Gagan had what seemed like an admirable answer. It was a good catch. I was only making jest. :-)

Wayne Conrad Over a year ago

@CarySwoveland Oh, humor! Earth creatures always get me with humor. :)

|

Wayne Conrad · Accepted Answer · 2015-06-08 19:13:36Z

1

This code:

keepers = hash_list.select { |_k, v| v["_destroy"] == "1" }.to_h
new_hash_list = hash_list.to_a.uniq { |_k, v| v["unit_id"] }.to_h
new_hash_list.merge!(keepers)

When run against this data:

hash_list = { 
  "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
  "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
  "c"=>{"unit_id"=>"43", "dep_id"=>"154", "_destroy"=>"false"},
  "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"},
  "e"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"1"},
}

produces this result:

{
  "a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
  "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"},
  "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"},
  "e"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"1"},
}

edited Jun 8, 2015 at 19:13

answered Jun 8, 2015 at 12:24

Wayne Conrad

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2 Comments

Cary Swoveland Over a year ago

I almost didn't upvote because of that misaligned "_destroy"=>"false". The outer braces are also unconventional. In future, please show more respect for readers' sensibilities.

Cary Swoveland Over a year ago

Alternatively,

keepers, no_dups = new_hash_list.partition { |_k, v| v["_destroy"] == "1" }; no_dups.uniq! { |_k,h| h["unit_id"] }; keepers.concat(no_dups).to_h

. The last line could instead be keepers.to_h.update(no_dups.to_h).

Cary Swoveland · Accepted Answer · 2015-06-09 14:34:44Z

0

Another way:

new_hash_list =
  { "a"=>{ "unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false" },
    "b"=>{ "unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"     },
    "c"=>{ "unit_id"=>"43", "dep_id"=>"154", "_destroy"=>"false" },
    "d"=>{ "unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false" },
    "e"=>{ "unit_id"=>"43", "dep_id"=>"cat", "_destroy"=>"1"     }}

require 'set'

s = Set.new
new_hash_list.each_with_object({}) do |(k,v),h|
  (h[k] = v) if v["_destroy"] == "1" || s.add?(v["unit_id"])
end
  #=> {"a"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"false"},
  #    "b"=>{"unit_id"=>"43", "dep_id"=>"153", "_destroy"=>"1"    },
  #    "d"=>{"unit_id"=>"42", "dep_id"=>"154", "_destroy"=>"false"},
  #    "e"=>{"unit_id"=>"43", "dep_id"=>"cat", "_destroy"=>"1"    }}

edited Jun 9, 2015 at 14:34

answered Jun 8, 2015 at 17:24

Cary Swoveland

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