Trying to do something like:
def action(number)
nums = [1...number]
code_here...
end
Do I need to write up a "count" snippet for this to work? Or is something similar to what I'm trying to do possible?
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"that counts to a given value" sounds more like a loop or maybe a method that yields those values.Stefan– Stefan2021-01-26 19:06:26 +00:00Commented Jan 26, 2021 at 19:06
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@Stefan It does, and I'd be able to do it in a loop, but I'm just trying to declare a variable whose value is the number and all the numbers before it starting at 1. Felt declaring it as an array off rip, as opposed to looping for those values and then pushing them into an array, would result in a faster log time. It's for leetcode and I like seeing "100%" on speed, though sometimes that means coming up with "hacky" solutions.CoderDude– CoderDude2021-01-27 20:17:25 +00:00Commented Jan 27, 2021 at 20:17
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2 Answers
There's another option to use enumerators:
1.upto(10).to_a #=> [1,2,3,4,5,6,7,8,9,10]
There are quite a few useful ones. For example, if you only need every other number up to certain number you can use:
0.step(100, 2).to_a #=> [0,2,4,6,8,..., 98,100]
Another option would be to use Array constructor:
Array.new(10) { |i| i + 1 } #=> [1,2,3,4,5,6,7,8,9,10]
which is a bit more verbose, but gives you much more freedom:
Array.new(10) { |i| 2 ** i } #=> [1,2,4,8,16,32,64,256,512]
4 Comments
Cary Swoveland
Hey, bro, long time no see. Still banging away on that piano, eh? How about something to represent the weird category, like
(1...).first(10)?
engineersmnky
@CarySwoveland weird it is
Enumerator.produce(1,&:succ) (less than 2.7 Enumerator.new {|y| i = 0; loop {y << i+=1}})Cary Swoveland
@engineersmnky, maybe add
.first(10).
Stefan
1.step.take(10) is another optionThere are two solutions for this as following:
If you want to add number in the array than:
nums = [*1..number]
# or
nums = (1..number).to_a
Ignoring last number than:
nums = [*1...number]
# or
nums = (1...number).to_a
2 Comments
CoderDude
It actually kinda hurts to be this close yet completely miss the target.
engineersmnky
Well maybe a few more than 2 options
1.upto(number) or Array.new(number) {|i| i + 1} or to start at 0 number.times or Array.new(number).each_index etc.