Ruby hash: return the first key value under which is not nil

Question

Say I have a hash

hash = {a:1, b:false, c:nil}

& a series of keys somewhere: [:c, :b, :a]. Is there a Ruby idiom for returning such a key value under which != nil?

The obv

[:c, :b, :a].select {|key| hash[key] != nil}.first     # returns :b

seems too long.

Answer 1

For that I think Enumerable#find might work:

find(ifnone = nil) { |obj| block } → obj or nil
find(ifnone = nil) → an_enumerator

Passes each entry in enum to block. Returns the first for which block is not false. If no object matches, calls ifnone and returns its result when it is specified, or returns nil otherwise.

If no block is given, an enumerator is returned instead.

In your case it'd return the first for which block is not nil:

p %i[c b a].find { |key| !{ a: 1, b: nil, c: nil }[key].nil? } # :a
p %i[c b a].find { |key| !{ a: 1, b: 1, c: nil }[key].nil? }   # :b

Answer 2

If you want to filter elements with falsy values, you can use the following expressions.

keys = [:d, :c, :b, :a]
hash = { a: 1, b: nil, c: nil, d: 2 }
keys.select(&hash)
#  => [:d, :a]

If you want to filter elements with exactly nil as a value, it is not correct, as Mr. Ilya wrote.