Ruby, convert array key to value

Question

I have a simple array like this:

stops = ["red_stop", "green_stop", "blue_stop", "yellow_stop", "purple_stop"]

I want to determine the distance (number of stops) between two 'stops'. I understand:

stops.index("purple_stop")

will return 4, but I don't understand how to write a Ruby method that will take any two stops as arguments and calculate the number of stops between them (e.g. the "red_stop" is 3 stops to the "yellow_stop"

You mean how to calculate stops.index("purple_stop") - stops.index("red_stop") ? — Yu Hao
– Yu Hao, Commented Sep 25, 2014 at 11:07
Yes, a = stop.index("purple_stop) b = stop.index("red_stop") puts a - b (but I want to write that as a Ruby method — mdror
– mdror, Commented Sep 25, 2014 at 11:10

BroiSatse · Accepted Answer · 2014-09-25 11:08:24Z

2

It would be simply:

class Array
  def dist(a,b)
    (index(b) - index(a)).abs
  end
end

stops = ["red_stop", "green_stop", "blue_stop", "yellow_stop", "purple_stop"]
stops.dist('red_stop', 'blue_stop') #=> 2

Note however it will not work if your array has duplicates.

answered Sep 25, 2014 at 11:08

BroiSatse

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2 Comments

Travis Hohl Over a year ago

If anyone else is wondering what the difference is between Array#index and Array#find_index, I can't tell that there is one. They appear to be identical. ruby-doc.org/core-2.1.3/Array.html#method-i-index

Stefan Over a year ago

@thohl they are: Array.instance_method(:index) == Array.instance_method(:find_index) #=> true

aelor · Accepted Answer · 2014-09-25 11:08:04Z

1

use this:

stops = ["red_stop", "green_stop", "blue_stop", "yellow_stop", "purple_stop"]

def getdiff(source, destination, stops)
 (stops.index(destination) - stops.index(source)).abs
end

getdiff("red_stop", "yellow_stop", stops)

answered Sep 25, 2014 at 11:08

aelor

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2 Comments

Jesper Over a year ago

Perhaps because before you edited your answer, your method took too few arguments.

aelor Over a year ago

that was a mistake while I typed, I hve edited it now. Hopefully he removes his downvote

Cary Swoveland · Accepted Answer · 2014-09-25 17:41:43Z

If you need all combinations:

stops = ["red_stop", "green_stop", "blue_stop", "yellow_stop", "purple_stop"]

stops.map.with_index { |s,i| [s,i] }
         .combination(2).each_with_object({}) { |((s1,d1),(s2,d2)),h|
           h[[s2,s1]] = h[[s1,s2]] = (d1-d2).abs }
  #=> {["red_stop",    "green_stop"] =>1, ["green_stop",  "red_stop"]   =>1,
  #    ["red_stop",    "blue_stop"]  =>2, ["blue_stop",   "red_stop"]   =>2,
  #    ["red_stop",    "yellow_stop"]=>3, ["yellow_stop", "red_stop"]   =>3,
  #    ["red_stop",    "purple_stop"]=>4, ["purple_stop", "red_stop"]   =>4,
  #    ["green_stop",  "blue_stop"]  =>1, ["blue_stop",   "green_stop"] =>1, 
  #    ["green_stop",  "yellow_stop"]=>2, ["yellow_stop", "green_stop"] =>2,
  #    ["green_stop",  "purple_stop"]=>3, ["purple_stop", "green_stop"] =>3, 
  #    ["blue_stop",   "yellow_stop"]=>1, ["yellow_stop", "blue_stop"]  =>1,
  #    ["blue_stop",   "purple_stop"]=>2, ["purple_stop", "blue_stop"]  =>2,
  #    ["yellow_stop", "purple_stop"]=>1, ["purple_stop", "yellow_stop"]=>1}

The steps are as follows:

a = stops.map.with_index { |s,i| [s,i] }
  #=> [["red_stop",    0], ["green_stop",  1], ["blue_stop", 2],
  #    ["yellow_stop", 3], ["purple_stop", 4]]

b = a.combination(2)
  #=> #<Enumerator: [["red_stop", 0], ["green_stop", 1], ["blue_stop", 2],
  #                  ["yellow_stop", 3], ["purple_stop", 4]]:combination(2)>

We can convert the enumerator b to an array to see its elements:

b.to_a
  #=> [[["red_stop",    0], ["green_stop",  1]],
  #    [["red_stop",    0], ["blue_stop",   2]],
  #    [["red_stop",    0], ["yellow_stop", 3]],
  #    [["red_stop",    0], ["purple_stop", 4]],
  #    [["green_stop",  1], ["blue_stop",   2]],
  #    [["green_stop",  1], ["yellow_stop", 3]],
  #    [["green_stop",  1], ["purple_stop", 4]],
  #    [["blue_stop",   2], ["yellow_stop", 3]],
  #    [["blue_stop",   2], ["purple_stop", 4]],
  #    [["yellow_stop", 3], ["purple_stop", 4]]]

Create a hash with the distances for each pair (not the decomposition of the block variables):

b.each_with_object({}) { |((s1,d1),(s2,d2)),h|
  h[[s2,s1]] = h[[s1,s2]] = (d1-d2).abs }
  #=> {["red_stop", "green_stop"]=>1, ["green_stop", "red_stop"]=>1,
  #    ...
  #    ["yellow_stop", "purple_stop"]=>1, ["purple_stop", "yellow_stop"]=>1}

Darren Stone · Accepted Answer · 2014-09-25 23:02:24Z

In your example of stops, if your entries are considered to be bidirectional then the assumption is stops.dist(a,b) == stops.dist(b,a). If so, then you're getting good answers already, such as @BroiSatse above.

If your stops array represents nodes or states that transition directionally (e.g. in a loop or digraph) then you may need to have a dist method to be respect that. In other words, your array is implied to wrap around.

class Array
  def dist(a,b)
    (index(b)-index(a)) % self.size
  end
end

stops = ["red_stop", "green_stop", "blue_stop", "yellow_stop", "purple_stop"]
stops.dist('red_stop', 'blue_stop')  #=> 2 (as expected)
stops.dist('blue_stop', 'red_stop')  #=> 3 (not 2, since it's directional)

An example of traffic light states (where transition is directional):

lights = ['green', 'yellow', 'red']
lights.dist('green', 'red')      #=> 2
lights.dist('red', 'green')      #=> 1 (green light follows a red light)

This depends on the domain you're trying to encode. If direction is unimportant, then please ignore this answer! ;-) Good luck.

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