How can I improve this Array.select in Ruby?

Question

I just wrote this... horrible line of Ruby code, and I'm wondering if there isn't a better way of doing it.

I have an array of hashes, of which I want to fetch one. The "conditions" of what I want to select is in two separate arrays--one with keys and one with values. If a hash in the array has the correct key == value for every pair in keys and values, I want to return that hash.

Any pointers to make the code more readable?

arr = [
  {:foo => 'foo', :bar => 'bar', :baz => 'baz'},
  {:foo => 'bar', :bar => 'bar', :baz => 'bar'},
  {:foo => 'foo', :bar => 'foo', :baz => 'foo'},
]

keys = [:foo, :bar]
values  = ['foo', 'bar']

arr.select{|c| keys.map{|k| i = keys.index(k); c[keys[i]] == values[i]}.delete(false) != false}.first
# => {:foo => 'foo', :bar => 'bar', :baz => 'baz'}

I almost never use an array if I'm going to be looking for elements in the collection. I think arrays are great for queues and stacks but I'll use a Hash if I have to find something, or a Hash or a Set if I only want unique values. — the Tin Man
– the Tin Man, Commented Nov 24, 2010 at 2:54
@Greg, if you never had a deal with smth, it doesn't mean, that it have no sense. — Nakilon
– Nakilon, Commented Nov 24, 2010 at 3:03
Does your last line have to be such a long line? This isn't perl! — Andrew Grimm
– Andrew Grimm, Commented Nov 24, 2010 at 23:15

Paige Ruten · Accepted Answer · 2010-11-24 02:51:10Z

2

Do you have to specify what you're looking for as an array of keys and an array of values? If you do, then convert them to a Hash like this:

hsh = Hash[*keys.zip(values).flatten]  #=> {:foo=>"foo", :bar=>"bar"}

And then select like this:

arr.select { |c| c == c.merge(hsh) }   #=> [{:foo=>"foo", :bar=>"bar", :baz=>"baz"}]

If you can specify what you're looking for as a Hash in the first place, you don't need the first line.

answered Nov 24, 2010 at 2:51

Paige Ruten

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4 Comments

vonconrad Over a year ago

Nice! I really like this solution--a lot more elegant than what I've currently got.

Nakilon Over a year ago

Convert them to hash to make solution 2 times shorter and 200 times slower

vonconrad Over a year ago

You're right--the Hash method is slower, but 200x is a bit of an overexaggeration. My benchmarks show about 3x, regardless of size.

Wayne Conrad Over a year ago

@Nakilon, The speed of the program only matters when the program is not fast enough. Until then, it is the speed of the human that matters.

Nakilon · Accepted Answer · 2010-11-24 02:58:25Z

2

arr = [
  {foo:'foo', bar:'bar', baz:'baz'},
  {foo:'bar', bar:'bar', baz:'bar'},
  {foo:'foo', bar:'foo', baz:'foo'},
]

keys = [:foo, :bar]
values = ['foo', 'bar']

p arr.find { |c|
    keys.zip(values).all? { |k,v|
        c[k] == v
    }
}

You can use {foo:'bar'} syntax to declare hashes if you have a symbol as key.
Use Enumerable#find to find first occurence.

answered Nov 24, 2010 at 2:58

Nakilon

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