How can I filter an array of hashes to get only th

Question

I'm trying get a subset of keys for each hash in an array.

The hashes are actually much larger, but I figured this is easier to understand:

[
  {
    id:2,
    start: "3:30",
    break: 30,
    num_attendees: 14
  },
  {
    id: 3,
    start: "3: 40",
    break: 40,
    num_attendees: 4
  },
  {
    id: 4,
    start: "4: 40",
    break: 10,
    num_attendees: 40
  }
]

I want to get only the id and start values.

I've tried:

return_keys = ['id','start']
return_array = events.select{|key,val|  key.to_s.in? return_keys}

but this returns an empty array.

Answer 1

If you happen to be using Rails (or don't mind pulling in all or part of ActiveSupport) then you could use Hash#slice:

return_array = events.map { |h| h.slice(:id, :start) }

Hash#slice does some extra work under the covers but it is probably fast enough that you won't notice it for small hashes and the clarity is quite nice.

Answer 2

Considering that efficiency appears to be a concern, I would suggest the following.

Code

require 'set'

def keep_keys(arr, keeper_keys)
  keepers = keeper_keys.to_set
  arr.map { |h| h.select { |k,_| keepers.include?(k) } }
end

This uses Hash#select, which, unlike Enumerable#select, returns a hash. I've converted keeper_keys to a set for fast lookups.

Examples

arr = [{ id:2, start: "3:30", break: 30 },
       { id: 3, break: 40, num_attendees: 4 },
       { break: 10, num_attendees: 40 }]

keep_keys arr, [:id, :start]
  #=> [{:id=>2, :start=>"3:30"}, {:id=>3}, {}] 
keep_keys arr, [:start, :break]
  #=> [{:start=>"3:30", :break=>30}, {:break=>40}, {:break=>10}] 
keep_keys arr, [:id, :start, :cat]
  #=> [{:id=>2, :start=>"3:30"}, {:id=>3}, {}] 
keep_keys arr, [:start]
  #=> [{:start=>"3:30"}, {}, {}] 
keep_keys arr, [:cat, :dog]

Answer 3

This should do what you want:

events.map do |hash|
  hash.select do |key, value|
    [:id, :start].include? key
  end
end

Potentially faster (but somewhat less pretty) solution:

events.map do |hash|
  { id: hash[:id], start: hash[:start] }
end

If you need return_keys to be dynamic:

return_keys = [:id, :start]
events.map do |hash|
  {}.tap do |new_hash|
    return_keys.each do |key|
      new_hash[key] = hash[key]
    end
  end
end

Note that, in your code, select picks out elements from the array, since that's what you called it on, but doesn't change the hashes contained within the array.

If you're concerned about performance, I've benchmarked all of the solutions listed here (code):

                user     system      total        real
amarshall 1  0.140000   0.000000   0.140000 (  0.140316)
amarshall 2  0.060000   0.000000   0.060000 (  0.066409)
amarshall 3  0.100000   0.000000   0.100000 (  0.101469)
tadman 1     0.140000   0.010000   0.150000 (  0.145489)
tadman 2     0.110000   0.000000   0.110000 (  0.111838)
mu           0.130000   0.000000   0.130000 (  0.128688)

Answer 4

A better solution is to use a hash as your index instead of doing a linear array lookup for each key:

events = [{id:2, start:"3:30",break:30,num_attendees:14},{id:3, start:"3:40",break:40,num_attendees:4},{id:4, start:"4:40",break:10,num_attendees:40}]

return_keys = [ :id, :start ]

# Compute a quick hash to extract the right values: { key => true }
key_index = Hash[return_keys.collect { |key| [ key, true ] }]

return_array = events.collect do |event|
  event.select do |key, value|
    key_index[key]
  end
end

# => [{:id=>2, :start=>"3:30"}, {:id=>3, :start=>"3:40"}, {:id=>4, :start=>"4:40"}]

I've adjusted this to use symbols as the key names to match your definition of events.

This can be further improved by using the return_keys as a direct driver:

events = [{id:2, start:"3:30",break:30,num_attendees:14},{id:3, start:"3:40",break:40,num_attendees:4},{id:4, start:"4:40",break:10,num_attendees:40}]

return_keys = [ :id, :start ]

return_array = events.collect do |event|
  Hash[
    return_keys.collect do |key|
      [ key, event[key] ]
    end
  ]
end

The result is the same. If the subset you're extracting tends to be much smaller than the original, this might be the best approach.