CODEWITHSUNDEEP
ruby-on-rails-3
Coderama
Coderama

Reputation: 11362

Determining the current element is the last element of a collection?

Is there a better way to find if the current element of a collection is the last element?

@oranges = Orange.all

@oranges.each_with_index do |o, current_index|
  puts @oranges.size == (current_index + 1) ? "Last element!" : "Get next element"
end

OR

@oranges.each do |o|
  puts @oranges.last == o ? "Last element!" : "Get next element"
end

Upvotes: 1

Views: 1601

Answers (2)

Mike Lewis
Mike Lewis

Reputation: 64177

Either looks fine, however if you were seriously considered about the perforamnce, I ran some benchmarks:

require 'benchmark'
a = (1..1000).to_a

def first(a)
  a.each_with_index do |o,i|
    a.size == (i + 1)
  end
end

def first_cached(a)
  a_size = a.size
  a.each_with_index do |o,i|
    a_size == (i + 1)
  end
end

def second(a)
  a.each do |e|
    a.last == e
  end
end

def second_cached(a)
  a_last = a.last
  a.each do |e|
    a_last == e
  end
end

Benchmark.bm(7) do |x|
x.report("first") {10000.times {first(a)}}
x.report("first_cached") {10000.times{first_cached(a)}}
x.report("second") {10000.times{second(a)}}
x.report("second_cached") {10000.times{second_cached(a)}}
end

which returned:

             user     system      total        real
first    2.020000   0.010000   2.030000 (  2.024102)
first_cached  1.930000   0.000000   1.930000 (  1.947230)
second   1.920000   0.010000   1.930000 (  1.922338)
second_cached  1.350000   0.000000   1.350000 (  1.352786)

So the second version, with a cached size yielded better results... however if these micro performances don't matter, it shouldn't be a problem.

Upvotes: 3

Behrang Saeedzadeh
Behrang Saeedzadeh

Reputation: 47971

Does the first solution even work? o is an orange and it doesn't have the size method.

You can instead do :

@oranges.each_with_index do |o, current_index|
  puts current_index == @oranges.size - 1 ? "Last element!" : "Get next element"
end

Apart from that both of the approaches are fine.

Upvotes: 1

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