Check if Ruby Array starts with another Array

Question

I am trying to check if an array starts with another array.

I managed to write some code that suits my use case:

def self.array_start_with?(ary, beginning)
  return true if beginning.empty?

  ary[0..(beginning.size - 1)] == beginning
end

but it's not very beautiful code, and I wonder if there is a more idiomatic way. The first code line within the method body is necessary when the beginning is empty but ary isn't. Otherwise the beginning will be compared to ary as a whole (ary[0..-1]). I could write:

ary[0..[beginning.size - 1, 0].max]

but that's uglier.

The second line of code doesn't honor the order of elements as [1, 2] == [2, 1] is true in Ruby. In my specific implementation, this isn't an issue as there can only be one order of the elements in the input but I'm still curious whether a more general solution exists.

Edit: Brain fart on [1, 2] == [2, 1] being true. Sorry.

What version of ruby are you using? [1, 2] == [2, 1] should be false. — Jacob Vanus
– Jacob Vanus, Commented Jan 4, 2018 at 17:44
Your code works fine. You can also say ary.first(beginning.length) == beginning, reads a little clearer. Or you could say arr[0...beginning.length] == beginning (triple periods means "up to but not including") — max pleaner
– max pleaner, Commented Jan 4, 2018 at 17:45
Jacob: You are absolutely right. I must have had a complete brain fart. — Eneroth3
– Eneroth3, Commented Jan 4, 2018 at 19:45

Mark Thomas · Accepted Answer · 2018-01-04 17:59:15Z

3

Here's one way to do it:

ary.take(beginning.size) == beginning

One advantage is that this works for empty arrays, so you don't need the extra check.

edited Jan 4, 2018 at 17:59

answered Jan 4, 2018 at 17:44

Mark Thomas

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Nu-hin · Accepted Answer · 2018-12-18 15:35:50Z

I needed to perform this calculation very fast, so I created a benchmark with 7 methods as candidates:

require 'benchmark'

def candidate1(a, b)
  return false if b.size > a.size

  b.each_with_index do |el, i|
    return false if a[i] != el
  end

  true
end

def candidate2(a, b)
  return false if b.size > a.size

  i = 0
  b.each do |el|
    return false if a[i] != el
    i += 1
  end

  true
end

def candidate3(a, b)
  return false if b.size > a.size

  0.upto(b.size) do |i|
    return false if a[i] != b[i]
  end

  true
end

def candidate4(a, b)
  0.upto(b.size) do |i|
    return false if a[i] != b[i]
  end

  true
end

def candidate5(a, b)
  i = 0
  b.each do |el|
    return false if a[i] != el
    i += 1
  end

  true
end

def candidate6(a, b)
  return false if b.size > a.size

  i = -1
  b.each do |el|
    return false if a[i+=1] != el
  end

  true
end

def candidate7(a, b)
  a.take(b.size) == b
end

TEST_CASES = [
  [[3,3,3,3], [3,3,3,3]],
  [[3,3], [3,3,3,3]],
  [[3,3,3,3], [3,3]],
  [[1,2,3,4], [5,6,7,8]],
  [[1,2,3,4], [5,6,7]],
  [[], []],
  [[1,2,3,4],[1,2,3]],
  [[1,2,3,4,5,6,7,8,9,10,11,12,13],[1,2,3]],
  [[1,2,3],[1,2,3,4,5,6,7,8,9,10,11,12,13]]
]

N = 1_000_000

Benchmark.bm do |x|
  puts "Running all testcases #{N} times on each candidate"

  i = 1.upto(7) do |i|
    x.report do
      method_name = "candidate#{i}"
      puts method_name

      N.times do
        TEST_CASES.each_with_index do |c, j|
          send(method_name, c[0], c[1])
        end
      end
    end
  end
end

Here's what I got:

candidate1
 4.713704   0.010974   4.724678 (  4.742682)
candidate2
 3.867821   0.009236   3.877057 (  3.888543)
candidate3
 5.521557   0.014552   5.536109 (  5.577994)
candidate4
 6.403158   0.010979   6.414137 (  6.428128)
candidate5
 4.730323   0.010388   4.740711 (  4.753335)
candidate6
 3.891551   0.021362   3.912913 (  3.964875)
candidate7
 3.942376   0.008003   3.950379 (  3.964634)

To summarise: candidates 2 and 6 were always outperforming others, sometimes one was winning, sometimes another:

def candidate2(a, b)
  return false if b.size > a.size

  i = 0
  b.each do |el|
    return false if a[i] != el
    i += 1
  end

  true
end

def candidate6(a, b)
  return false if b.size > a.size

  i = -1
  b.each do |el|
    return false if a[i+=1] != el
  end

  true
end

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