Get rand number from range

Question

I want to take a sample number from this defined range of numbers.

def self.ipg_amount_range
  (1..1000000000000000).to_a.sample
end

But when I load the code it takes a lot of time to load the code. Is there some way to speed up this code execution?

not a ruby guy but isn't (1..1000000000000000) an array? if so what do you expect? — nafas, Dec 01 '17 at 09:35
@nafas: no, it's a range. But adding `.to_a` materializes it into an array. — Sergio Tulentsev, Dec 01 '17 at 09:43
One may prefer to write `1_000_000_000_000_000` for readability or for brevity either `1e15` or `10**15`. — Sagar Pandya, Dec 01 '17 at 16:47

Yu Hao · Accepted Answer · 2017-12-01T09:40:54.037

5

The to_a method takes a lot of time to generate the array, which you don't need.

Just use:

rand(1..1000000000000000)

edited Dec 01 '17 at 09:40

answered Dec 01 '17 at 09:35

Yu Hao

2

Or even just `rand(1000000000000000)`. – Tom Lord Dec 01 '17 at 09:36
@TomLord or `(1000000000000000 * rand).round` :) – Aleksei Matiushkin Dec 01 '17 at 09:37
3

@mudasobwa: I don't see why would anyone ever want to use this form, when we have the former. ¯\\_(ツ)_/¯. In, say, javascript you don't have a choice, but here we do :) – Sergio Tulentsev Dec 01 '17 at 09:45
3

@TomLord Technically it should be `rand(1000000000000000) + 1` – Yu Hao Dec 01 '17 at 09:45

1 Answers1