So I'm iterating over a range like so:
(1..100).each do |n| # n = 1 # n = 2 # n = 3 # n = 4 # n = 5 end But what I'd like to do is iterate by 10's.
So in stead of increasing n by 1, the next n would actually be 10, then 20, 30, etc etc.
4 Answers
Basically you use the step() method. For example:
(10..100).step(10) do |n| # n = 10 # n = 20 # n = 30 # ... end 1You can use Numeric#step.
0.step(30,5) do |num| puts "number is #{num}" end # >> number is 0 # >> number is 5 # >> number is 10 # >> number is 15 # >> number is 20 # >> number is 25 # >> number is 30 Here's another, perhaps more familiar-looking way to do it:
for i in (0..10).step(2) do puts i end 3rng.step(n=1) {| obj | block } => rng Iterates over rng, passing each nth element to the block. If the range contains numbers or strings, natural ordering is used. Otherwise step invokes succ to iterate through range elements. The following code uses class Xs, which is defined in the class-level documentation.
range = Xs.new(1)..Xs.new(10) range.step(2) {|x| puts x} range.step(3) {|x| puts x} produces:
1 x 3 xxx 5 xxxxx 7 xxxxxxx 9 xxxxxxxxx 1 x 4 xxxx 7 xxxxxxx 10 xxxxxxxxxx ......