CODEWITHSUNDEEP
rubyarraysenumerationidioms
Matt Brictson
Matt Brictson

Reputation: 11082

Iterating over each element of an array, except the first one

What is the idiomatic Ruby way to write this code?

Given an array, I would like to iterate through each element of that array, but skip the first one. I want to do this without allocating a new array.

Here are two ways I've come up with, but neither feels particularly elegant.

This works but seems way too verbose:

arr.each_with_index do |elem, i|
  next if i.zero? # skip the first
  ...
end

This works but allocates a new array:

arr[1..-1].each { ... }

Edit/clarification: I'd like to avoid allocating a second array. Originally I said I wanted to avoid "copying" the array, which was confusing.

Upvotes: 7

Views: 4160

Answers (3)

PinnyM
PinnyM

Reputation: 35533

Using the internal enumerator is certainly more intuitive, and you can do this fairly elegantly like so:

class Array
  def each_after(n)
    each_with_index do |elem, i|
      yield elem if i >= n
    end
  end
end

And now:

arr.each_after(1) do |elem|
  ...
end

Upvotes: 6

Matt Brictson
Matt Brictson

Reputation: 11082

OK, maybe this is bad form to answer my own question. But I've been racking my brain on this and poring over the Enumerable docs, and I think I've found a good solution:

arr.lazy.drop(1).each { ... }

Here's proof that it works :-)

>> [1,2,3].lazy.drop(1).each { |e| puts e }
2
3

Concise: yes. Idiomatic Ruby… maybe? What do you think?

Upvotes: 2

7stud
7stud

Reputation: 48599

I want to do this without creating a copy of the array.

1) Internal iterator:

arr = [1, 2, 3]
start_index = 1

(start_index...arr.size).each do |i|
  puts arr[i]
end

--output:--
2
3

2) External iterator:

arr = [1, 2, 3]
e = arr.each
e.next

loop do
  puts e.next
end

--output:--
2
3

Upvotes: 1

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