5
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Given the task of having to return an array that alternately takes elements from two arrays, what would be most Ruby way of completing the task? As of now, I have this code, but it doesn't feel very elegant.

def alternateTake (a, b)
  raise ArgumentError, "Unequal length" if a.length != b.length
  build = []
  (0..a.length-1).each do |x|
    build << (x.even? ? a[x] : b[x])
  end
  build
end

Example input & output:

method([1, 4, 3, 7], [1, 4, 2, 33]) = [1, 4, 3, 33]
method([98, 12, 41], [35, 22, 14]) = [98, 22, 41]
method([12, 33], [66, 45, 3]) = ERROR
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1 Answer 1

5
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This one-liner should be what you're looking for:

def alternate_take(xs, ys) 
  raise ArgumentError, "Unequal length" if xs.length != ys.length
  xs.zip(ys).map.with_index{|(x,y),i| i.even? ? x : y} 
end

Follow Up

Tokland's answer to OP's followup about a solution without indexes is really clever. Here's another variation, which I don't like as much as the index version, but it answers the question:

def alternate_take(xs, ys) 
  raise ArgumentError, "Unequal length" if xs.length != ys.length
  xs.zip(ys).reduce([]){|m,(x,y)| m + (m.size.even? ? [x] : [y])} 
end
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7
  • \$\begingroup\$ Looks good. Some details: maybe use unless instead of if, give plural names to a and b (maybe: xs, ys) and unpack the pair |(x, y), idx| ... \$\endgroup\$
    – tokland
    May 16, 2016 at 22:55
  • \$\begingroup\$ Is there any way to use iterators or something like that and avoid using indices? \$\endgroup\$
    – Michael
    May 16, 2016 at 22:57
  • \$\begingroup\$ @tokland, Both good suggestions, edited. \$\endgroup\$
    – Jonah
    May 16, 2016 at 23:01
  • 2
    \$\begingroup\$ @Michael. A lot of ways, for example: xs.zip(ys, [true, false].cycle).map { |x, y, take_x| take_x ? x : y }. Not that there is anything wrong with using with_index. \$\endgroup\$
    – tokland
    May 16, 2016 at 23:01
  • \$\begingroup\$ Jonah, the second version is so horrible.. I don't think it's worth it. Children watching! :) There is nothing wrong with using with_index, it's not the same than accessing the array by indexes (which is indeed ugly). \$\endgroup\$
    – tokland
    May 17, 2016 at 7:23

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