How to modify an iterator while iterating over an array

Question

I want to skip a loop x times according to a condition that is determined at runtime. How can I do this?

for i in (0..5)
  if i==0
    3.times {next} # i=i+3 also doesnt work
  end
  puts i
end

Expect to output

3
4
5

EDIT: To clarify, the question is both the condition (ie i==0) and skipping x times iteration are determined dynamically at runtime, more convoluted example:

condition = Array.new(rand(1..100)).map{|el| rand(1..10000)} #edge cases will bug out
condition.uniq!

for i in (0..10000)
  if condition.include? i
    rand(1..10).times {next} # will not work
  end
  puts i
end

Not sure what is the exact input and output of your function. In your example, why not just use (3..5).each {|i| puts i} — Lei Chen
– Lei Chen, Commented Dec 20, 2015 at 3:43
@LeiChen, that's clearly how it should be done, so why not post an answer? Maybe def skip_first(range, nbr_to_skip). — Cary Swoveland
– Cary Swoveland, Commented Dec 20, 2015 at 4:08
For your "convoluted" example, I would prefer to compute an array keepers in advance, so you could just write for i in keepers..... That might be written condition = Array.new(rand(1..4)).flat_map do |el| first = rand(1..20); [*first..first+rand(0..4)]; end.uniq # => [7, 8, 9, 18, 6]; keepers = [*1..20]-condition #=> [1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20]. I realize there will be other situations where you will need next if .... within the block. — Cary Swoveland
– Cary Swoveland, Commented Dec 20, 2015 at 6:57

Jerrÿ Chang · Accepted Answer · 2015-12-20 05:29:56Z

4

simple method to skip by a defined multiple.

array_list = (0..5).to_a
# Use a separate enum object to hold index position
enum = array_list.each 
multiple = 3

array_list.each do |value|
  if value.zero?     
    multiple.times { enum.next }
  end   
  begin puts enum.next rescue StopIteration end
end

answered Dec 20, 2015 at 5:29

Jerrÿ Chang

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1 Comment

Cary Swoveland Over a year ago

You can write that more succinctly: multiple = 3; enum = (0..5).each; loop do; multiple.times { enum.next } if enum.next.zero?; puts 'hi'; end. "hi" is printed thrice. This works because Enumerator#next raises a StopIteration exception when an attempt is made to go beyond the end of the enumeration, and Kernel#loop handles the exception by breaking out of the loop. You could instead write enum = (0..5).to_enum, which is arguably more descriptive.

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