find first element in array for which block returns true and return the block's return value

Question

I need to iterate over an array and apply a supplied block to each element, and return the first true value returned by the block, which implies that I need to stop immediately as soon as I get a true value.

below is my code. I am a ruby newbie, and I am not sure if this code is reinventing the wheel. Maybe there is a library method or methods that can do that already? or may be this code can be simplified?

RS = {
  :x => %w(\d+ a\d+ bb\d+ ccc\d+).map{|x| /^#{x}$/},
  :y => %w(\w+ 1\w+ 22\w+ 333\w+).map{|x| /^#{x}$/}
}.freeze

def find s, t
  r = RS[s]
  if r
    r.each do |p|
      m = p.match t
      return m if m
    end
    nil
  end
end

p find :x, 'bb12345'

You're looking for something like Enumerable#find that returns what the block returned rather than which element caused the block to return the truthy value? — mu is too short
– mu is too short, Commented Mar 14, 2013 at 21:24
I think what you're doing is about all there is. You could pretty it up a bit and monkey patch it into Enumerable but that's about it. — mu is too short
– mu is too short, Commented Mar 14, 2013 at 23:20
possible duplicate of Functionally find mapping of first value that passes a test — Dave Schweisguth
– Dave Schweisguth, Commented Dec 4, 2013 at 0:37

Alex.Bullard · Accepted Answer · 2013-03-15 00:12:08Z

2

If you want the result of the block you could do it this way. This will iterate over the whole array, but wont evaluate any matches after the first one.

def find(s,t)
  RS[s].inject(nil) {|m, p| m || p.match(t)}
end

You can break out early doing something like this

RS[s].inject(nil) {|m, p| (m && (break m)) || p.match(t)}

edited Mar 15, 2013 at 0:12

answered Mar 14, 2013 at 20:48

Alex.Bullard

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1 Comment

akonsu Over a year ago

thanks. I need to return true in this case because this is the value that the block returns.

ribamar · Accepted Answer · 2017-10-17 12:07:42Z

2

This is duplicated with: Ruby - Array.find, but return the value the block

You want a lazy map:

[nil, 1, 2, 3].lazy.map{|i| i && i.to_s}.find{|i| i}    
# => "1"

answered Oct 17, 2017 at 12:07

ribamar

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Comments

peter · Accepted Answer · 2014-03-23 16:49:22Z

1

Hopefully still actual: here a solution using detect, i made it possible to verbose the output so you can see which expressions are evaluated before returning a hit.

def find_match symbol, string , verbose = false, match = nil
  if verbose
    RS.detect{|x,v|x==symbol;v.detect{|re|puts re;match=string.match(/#{re}/)}}
  else
    RS.detect{|x,v|x==symbol;v.detect{|re|match=string.match(/#{re}/)}}
  end
  match
end

p find_match :x, 'bb12345'
p find_match :x, 'ee12345' , true #verbose output
p find_match :x, '12345'
p find_match :y, '22abcd'

#<MatchData "bb12345">
(?-mix:^\d+$)
(?-mix:^a\d+$)
(?-mix:^bb\d+$)
(?-mix:^ccc\d+$)
(?-mix:^\w+$)
#<MatchData "ee12345">
#<MatchData "12345">
#<MatchData "22abcd">

answered Mar 23, 2014 at 16:49

peter

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Comments

haejeong87 · Accepted Answer · 2013-03-15 05:58:36Z

0

If your regex patterns are simple, you can just apply the regex again at the end, maybe.

Something like:

def find(s,t)
  r = RS[s] and r.find{|p| p.match(t)}.try(:match, t)
end

Although it makes one redundant call to match, it is easier to understand.

First, find the pattern you want, then use that pattern.

answered Mar 15, 2013 at 5:58

haejeong87

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