In Ruby, how do I check if at least 80% of elements in an array match a pattern?

Question

I'm using Ruby 2.4. How do I check if at least 80% of my elements in an array match a certain pattern? I want to see if each element matches the regex

/\d\d?\s*-\s*\d\d?/

Filter and compare new array with old array for percentage? – Henrik Andersson Jan 09 '17 at 16:50 — Henrik Andersson, Jan 09 '17 at 16:50

Andrey Deineko · Accepted Answer · 2017-01-09T17:56:24.863

2

You can use Enumerable#grep in conjunction with simple math:

array.grep(/\d\d?\s*-\s*\d\d?/).size / array.size.to_f >= 0.8

To shorten this further you can use Numeric#quo or Numeric#fdiv:

array.grep(/\d\d?\s*-\s*\d\d?/).size.quo(array.size) >= 0.8

edited Jan 09 '17 at 17:56

answered Jan 09 '17 at 16:47

Andrey Deineko

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why not just `>= .8` ? – Scott Weaver Jan 09 '17 at 17:14
1

You can safe the `to_f` if you write `array.grep(...).size >= 0.8 * array.size` – akuhn Jan 09 '17 at 17:46
1

or `.size.fdiv(array.size) >= 0.8` or maybe even `.size.quo(array.size) >= 80/100r` – Stefan Jan 09 '17 at 17:48
@akuhn that's true :) – Andrey Deineko Jan 09 '17 at 17:53
@Stefan nice ones! Did not see `quo` before - TIL (thanks!)! Btw, why not `8/10r` instead of `80/100r` (I suppose intention is more clear with `80/100r`)? :) – Andrey Deineko Jan 09 '17 at 17:54
Yeah, 80% translates literally to `80/100r`, although it's equal to `8/10r` or `4/5r` – Stefan Jan 09 '17 at 18:36

Thomas · Answer 2 · 2017-01-10T10:20:48.363

1

If performance does matter, you don't need to check all the elements to know if at least 80% of them are maching a condition

With ruby 2.3 this implementation is a bit faster than the count solution and twice faster than the grep solution :

def pareto_match(array, proportion: 0.8)
  min_success= (array.count * proportion).round(4)
  max_failures= (array.count * (1 - proportion)).round(4)
  success= 0
  failure= 0
  array.each do |element|
    if yield(element)
      success+= 1
      return true if success >= min_success
    else
      failure+= 1
      return false if failure > max_failures
    end
  end
end

pareto_match(array){|e| e =~ /\d\d?\s*-\s*\d\d?/}

edited Jan 10 '17 at 10:20

answered Jan 09 '17 at 22:40

Thomas

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Beware of floating point arithmetics. For 10 elements, `min_success` will be `8.0`, whereas `max_failures` will be `1.9999999999999996`. – Stefan Jan 10 '17 at 07:55
Indead, I've edited the answer. I use to use Bignum for this kind of problem – Thomas Jan 10 '17 at 10:19
Since you are counting array elements, you could just round the numbers to integers. – Stefan Jan 10 '17 at 10:20
with `to_i` instead of `round(x)` this case is not working : `pareto_match(%w(1 2 3 a)){|e| e =~ /\d/}` – Thomas Jan 10 '17 at 10:22
`round(x)` doesn't make much sense, because an array can't contain a fractional number of elements. I'd use `min_success = (array_count * proportion).round` and `max_failures = array.count - min_success`. You could also use `ceil` or `floor` instead of `round`. – Stefan Jan 10 '17 at 10:30

score 0 · Answer 3 · answered Jan 09 '17 at 19:26

0

I would write:

(array.count{|item| item =~ /\d\d?\s*-\s*\d\d?/} / array.size) >= 0.8

answered Jan 09 '17 at 19:26

hirolau

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In Ruby, how do I check if at least 80% of elements in an array match a pattern?

3 Answers3