Ruby: next / previous values in array, loop array, array position

Question

Let's say I have an array of random numbers in no particular order. Let's say these are the ID #'s for people who ran in a marathon, and they're added to the array in the order they finish, something like:

race1 = [8, 102, 67, 58, 91, 16, 27]
race2 = [51, 31, 7, 15, 99, 58, 22]

This is a simplified and somewhat contrived example, but I think it conveys the basic idea.

Now a couple questions:

First, how could I get the IDs that are before and after a particular entry? Let's say I am looking at runner 58 and I want to know who finished before and after him.

race1, runner58: previous finisher=67, next finisher=91
race2, runner58: previous finisher=99, next finisher=22

Second, if I'm looking at the person who finished first or last, how can I have the "next" or "previous" loop back around the array?

race1, runner8: previous finisher=27, next finisher=102
race2, runner22: previous finisher=58, next finisher=51

Lastly, I'd like to show what position each runner finished in. Given just the array and a value in it, how can I find out what 'ordinal' position it is in the array? Ie:

race1: runner8=1st, runner102=2nd, runner67=3rd ... runner27=last

Thanks very much!

Jonas Elfström · Accepted Answer · 2011-03-28 20:51:22Z

7

First & Second:

index = race1.find_index(58)
if !index.nil?
  puts "#{race1[index-1], #{race1[index+1] || race1[0]}"
end

Lastly:

gem install linguistics

then

require 'linguistics'
Linguistics::use( :en )
race1.each_with_index {|runner, i| puts "runner#{runner}=#{(i+1).en.ordinal}"}

edited Mar 28, 2011 at 20:51

answered Mar 28, 2011 at 20:06

Jonas Elfström

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2 Comments

Andy Over a year ago

What condition is the nil guard in race1[index-1] || race1[-1] supposed to catch?

Andrew Over a year ago

Nice tip on the linguistics gem. Thanks!

maerics · Accepted Answer · 2011-03-29 04:39:15Z

5

For items #1 and #2 I would create a method which returns the previous and next elements by id, wrapping automatically for the first and last, e.g.:

class Array
  def prev_next(id)
    idx = self.index(id)
    raise Error.new("no racer with id #{id}") unless idx
    [self[idx-1], self[(idx+1)%self.size]]
  end
end
race1.prev_next(58) # => [67, 91]
race1.prev_next(8) # => [27, 102]

Note that the only possible negative index, -1, actually rolls to the end thanks to Ruby's array.slice and by using the modulus of the array size we can adjust for wrapping over the end. For the third item, ordinalization can be done like so:

class Integer
  def ordinalize
    s = self.to_s
    case s
      when /1[123]$/ then s + 'th'
      when /1$/ then s + 'st'
      when /2$/ then s + 'nd'
      when /3$/ then s + 'rd'
      else s + 'th'
    end
  end
end
race1.each_with_index {|x,i| puts "runner#{x}=#{(i+1).ordinalize}"}
# runner8=1st
# runner102=2nd
# runner67=3rd
# ...

edited Mar 29, 2011 at 4:39

answered Mar 28, 2011 at 20:09

maerics

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3 Comments

Jonas Elfström Over a year ago

Why a regular expression when you are dealing with a fixnum?!

maerics Over a year ago

@Jonas Elfström: but for integer 21 we want "21st" not "21th". I could do some integer arithmetic to accomplish the same thing but the solution is more clear to me when considering strings.

Jonas Elfström Over a year ago

Ah, I was completely off track.

Andy · Accepted Answer · 2011-03-28 20:01:27Z

2

The find_index method will probably your best bet here.

runner_index = race.find_index finisher_number #find the index of the racer
previous_runner = race[runner_index - 1] #a negative index will wrap around
next_runner = race[runner_index + 1] || race[0] #use a null guard to wrap to the front

answered Mar 28, 2011 at 20:01

Andy

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1 Comment

fl00r Over a year ago

it is synonym of index method

tranquiliste · Accepted Answer · 2011-03-28 19:59:22Z

0

Hello I don't know exactly what you want to do but if you are using ruby 1.9 you should look at the rotate method and if you are using ruby 1.8.7, you should try to build something around the index method like

race1[race1.index(58) +1] will give 98 (the next one)
race1[race1.index(58) -1] will give 67 (the previous one)

You need to tweak a little bit to simulate the rotate (by testing the value of the returned indice compared to the size of the array or to 0)

answered Mar 28, 2011 at 19:59

tranquiliste

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Comments

jjnevis · Accepted Answer · 2015-03-26 09:33:22Z

0

First & Second:

race1.rotate(race1.index(58) + 1).first # next
race1.rotate(race1.index(58) - 1).first # previous

Credit: saw this on this post: http://jamonholmgren.com/rubymotion-react-pattern/

answered Mar 26, 2015 at 9:33

jjnevis

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