How to Parse and find JSON element in Ruby?

Question

I'm trying to return the lowest result number that it's name is my.name:

{
    "total_results"=>"73", 
    "results"=>
        {
        "28"=> {"description"=>"description_here", "name"=> "my.name"},
        "25"=> {"description"=>"description_here", "name"=> "other1.name"},
        "24"=> {"description"=>"description_here", "name"=> "my.name"},
        "21"=> {"description"=>"description_here", "name"=> "other2.name"}
        }
}

So in the example above it will return 24. I'm fairly new to Ruby so I'm struggling to find a good way to do this.

Van Huy · Accepted Answer · 2016-02-04 05:57:06Z

2

Your example is a hash in Ruby

h = {
"total_results"=>"73", 
"results"=>
    {
    "28"=> {"description"=>"description_here", "name"=> "my.name"},
    "25"=> {"description"=>"description_here", "name"=> "other1.name"},
    "24"=> {"description"=>"description_here", "name"=> "my.name"},
    "21"=> {"description"=>"description_here", "name"=> "other2.name"}
    }
}

You can use select to get all matching values, then get min key of them like

h["results"].select{|k,v| v["name"] == "my.name"}.keys.map(&:to_i).min
# => 24

edited Feb 4, 2016 at 5:57

answered Feb 4, 2016 at 5:08

Van Huy

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3 Comments

Todd A. Jacobs Over a year ago

This is a good answer. However, you can shorten your method chain with min_by(&:to_i) insteap of #map. It will be faster, too.

Van Huy Over a year ago

min_by(&:to_i) returns a string ("24"), that is not what I expected

Todd A. Jacobs Over a year ago

That's because it converts it for the comparison, but doesn't change the original format. This is usually the Right Thing™ to do, especially when dealing with JSON strings. I'm not trying to bikeshed; I'm just pointing out that you aren't really returning the matching key, you're returning an integer representation of it. I try to avoid unintentional coercions in results, as that might matter to future visitors with similar issues. I'll just leave it at that, since your answer clearly addressed the OPs specific need. Well done!

philip yoo · Accepted Answer · 2016-02-04 04:58:46Z

0

This is a solution. Definitely not the best, but I can at least show you how you can traverse through that hash..

foo = {
"total_results"=>"73", 
"results"=>
    {
    "28"=> {"description"=>"description_here", "name"=> "my.name"},
    "25"=> {"description"=>"description_here", "name"=> "other1.name"},
    "24"=> {"description"=>"description_here", "name"=> "my.name"},
    "21"=> {"description"=>"description_here", "name"=> "other2.name"}
    }
}

bar = nil

foo['results'].each do |k, v|
  if v['name'] == "my.name"
    if bar.nil? || k < bar
      bar = k
    end
  end
end


puts bar  #=> "24"

answered Feb 4, 2016 at 4:58

philip yoo

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Comments

Abhi · Accepted Answer · 2016-02-04 04:58:54Z

0

Try this:

h = {
    "total_results"=>"73", 
    "results"=>
        {
        "28"=> {"description"=>"description_here", "name"=> "my.name"},
        "25"=> {"description"=>"description_here", "name"=> "other1.name"},
        "24"=> {"description"=>"description_here", "name"=> "my.name"},
        "21"=> {"description"=>"description_here", "name"=> "other2.name"}
        }
}

To get the minimum key:

min_key = h["results"].keys.min

To get the min_key value

h["results"][min_key]

answered Feb 4, 2016 at 4:58

Abhi

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Comments

Nirajan Pokharel · Accepted Answer · 2016-02-04 05:45:31Z

0

h = {
    "total_results"=>"73", 
    "results"=>
        {
        "28"=> {"description"=>"description_here", "name"=> "my.name"},
        "25"=> {"description"=>"description_here", "name"=> "other1.name"},
        "24"=> {"description"=>"description_here", "name"=> "my.name"},
        "21"=> {"description"=>"description_here", "name"=> "other2.name"}
        }
} 

h["results"].select{|key,hash| [key, hash] if hash["name"]=="my.name"}.keys.map(&:to_i).min

=> 24

edited Feb 4, 2016 at 5:45

answered Feb 4, 2016 at 5:20

Nirajan Pokharel

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4 Comments

Gourav Over a year ago

Your answer is correct but a small change is required, .keys will return array of strings like ["28","24"] and you cannot apply min on string it doesn't give minimum values, If the array consisted of ["28","201","24"].min will return "201". so h["results"].select{|key,hash| [key, hash] if hash["name"]=="my.name"}.keys.map(&:to_i).min

Todd A. Jacobs Over a year ago

@GouravNaik You are wrong, at least on Ruby 2.3.0. ["28", "24"].min #=> "24" works just fine.

Nirajan Pokharel Over a year ago

How about ["28", "201", "24"].min ? @CodeGnome

Todd A. Jacobs Over a year ago

@NirajanPokharel Fair enough. If you're going to consider data sets other than what the OP provided, then you're going to have to coerce to integers for the comparison. But you can still make it shorter and more expressive with min_by(&:to_i) rather than mapping and then calling min on the interim array.

Todd A. Jacobs · Accepted Answer · 2016-02-04 06:14:38Z

0

A One-Line Solution

Given your posted hash stored in h, you can do the following:

h['results'].select { |k,v| v['name'] == 'my.name' }.keys.min_by(&:to_i)
#=> "24"

Explanation

While this one-liner trades compactness for clarity, it's doing the following:

Sifting results with Enumerable#select for values with a nested hash key of 'name'.
Returning an interim hash containing hashes where the value assigned to the 'name' key contains your desired string value. For example:
```
{"28"=>{"description"=>"description_here", "name"=>"my.name"},
 "24"=>{"description"=>"description_here", "name"=>"my.name"}}
```
Using Enumerable#min_by to coerce the keys to integers for the comparison, then find the smallest value and return it in the original format (e.g. as a string).

It only looks like magic. It's just Ruby. :)

edited Feb 4, 2016 at 6:14

answered Feb 4, 2016 at 5:16

Todd A. Jacobs

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2 Comments

Tom Over a year ago

syntax error, unexpected '.' .min&.shift ... ^

Todd A. Jacobs Over a year ago

@Tom I refactored my answer as there was a separate edge case that I hadn't considered, so I've removed the safe navigation operator as the remaining chains don't require them. So, while the original code did work if you had the necessary Ruby version (e.g. MRI Ruby >= 2.3.0), it no longer matters. Happy coding!

Hieu Pham · Accepted Answer · 2016-02-04 09:07:18Z

0

Find min_key just by a single line of code:

min_key = your_hash['results'].sort.to_h.invert['my.name']

answered Feb 4, 2016 at 9:07

Hieu Pham

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