like this
range = (0..10)
how can I get number like this:
0 5 10
plus five every time but less than 10
if range = (0..20) then i should get this:
0 5 10 15 20
A:
The step method described in http://ruby-doc.org/core/classes/Range.html should do the job but seriously harms may harm the readability.
Just consider:
(0..20).step(5){|n| print ' first ', n }.each{|n| print ' second ',n }
You may think that step(5) kind of produces a new Range, like why_'s question initially intended. But the each is called on the (0..20) and has to be replaced by another step(5) if you want to "reuse" the 0-5-10-15-20 range.
Maybe you will be fine with something like (0..3).map{|i| i*5}?
But "persisting" the step method's results with .to_a should also work fine.
efi
2010-04-06 09:08:36
“seriously harms the readability” – *what*?! Why? How?
Konrad Rudolph
2010-04-06 09:11:15
I'd argue that step is more readable than a map of that nature, at least in this case.
Amber
2010-04-06 09:11:18
`(0..20).step(5).each{...}.each{...}` will do what you want it to do.
Marc-André Lafortune
2010-04-06 15:29:02
@Marc-Andre Lafortune: `(0..20).step(5).each{print '.'}.each{print '!'}`will print`.....!!!!!!!!!!!!!!!!!!!!!`and that's hard to figure out at first glance - at least for me.
efi
2010-04-06 20:00:41
@efi: no, it will print .....!!!!! Did you actually try it? The reason is that `step(5)` (with no block) returns an Enumerator that will yield 5 times. This Enumerator is returned by both `each`. What's confusing you is that `step(5){a block}` returns the original enumerable, so further `each` will operate on the full range.
Marc-André Lafortune
2010-04-06 21:16:45
@Marc-Andre Lafortune: I actually tried `(0..20).step(5).each{print '.'}.each{print '!'}` in Ruby 1.8.7 and JRuby 1.4.0. In both cases I got 5 times '.' but 20 times '!'. I tried it in IRB (and JIRB) as well as via the normal ruby interpreter. Under Linux and Windows. Strange considering your explanation...
efi
2010-04-07 11:02:00
@efi: I must have been day-dreaming. I'm very surprised. The fact that enumerator.each{} returns the base object is unintuitive for me. It's not even spec'ed in rubyspec. I'll check to see if it's only me!
Marc-André Lafortune
2010-04-07 16:33:09
+6
A:
Try using .step() to go through at a given step.
(0..20).step(5) do |n|
print n,' '
end
gives...
0 5 10 15 20
As mentioned by dominikh, you can add .to_a on the end to get a storable form of the list of numbers: (0..20).step(5).to_a
Amber
2010-04-06 09:10:29
and if you want to store the numbers you can use (0..10).step(5).to_a -- this requires ruby >= 1.8.7 though. An alternative is 0.step(10, 5).to_a, which also required ruby >= 1.8.7
dominikh
2010-04-06 09:12:07
+2
A:
Like Dav said, but add to_a:
(0..20).step(5).to_a # [0, 5, 10, 15, 20]
name
2010-04-06 09:12:58
You generally don’t need to add `to_a` unless you really *do* need an array.
Konrad Rudolph
2010-04-06 09:16:19