2
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I have an array of 1000 numbers, randomly generated. 10 of those numbers must be between 0-9 (including 0 and 9), and 10 of those numbers must be between 991 and 1000 (including 991 and 1000). This is what I came up with:

arr = []
980.times do
  arr << (11..989).to_a.sample
end

arr2 = []
10.times do 
  arr2 << rand(10)
end

arr3 = []
10.times do
  arr3 << (990..1000).to_a.sample
end

arr4 = []
arr4 = arr + arr2 + arr3

arr4.shuffle

Is there a more elegant way to do this in ruby?

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  • \$\begingroup\$ Is it only 10 for ranges at the end or "at least 10"? \$\endgroup\$ – Fred the Magic Wonder Dog Dec 14 '13 at 17:45
5
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Some notes:

  • Try to learn some functional programming, you shouldn't write Ruby as it were a low-level language (like C). Favor expressions over statements. My 2-cents on the matter, I hope it helps: http://www.slideshare.net/tokland/functional-programming-with-ruby-9975242

  • Let me show a simple example of using FP. The first 4 lines of your code could be written like this: arr = 980.times.map { (11..989).to_a.sample }

  • arr4 = []: Why? the next line assigns it to another expression.

  • range.to_a.sample. That's very inefficient, better rand(range).

I'd write:

ranges = [0..9, 11..989, 990..1000]
output = ranges.flat_map do |range|
  range.map { rand(range) }
end.shuffle
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  • \$\begingroup\$ I can't improve on your "I'd write:", but here's a variant: ranges.reduce([]) {|arr, range| arr + Array.new(range.size) {rand(range)}}.shuffle \$\endgroup\$ – Cary Swoveland Jan 8 '14 at 7:54
  • \$\begingroup\$ @CarySwoveland: Not that efficiency is important here (those are small arrays), but note that reduce with + leads to O(n^2) code. \$\endgroup\$ – tokland Jan 8 '14 at 9:38
1
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Be careful at the boundaries — you didn't implement what you wanted. The output will never contain 10, and it might include 990 among the [991, 1000] range.

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