More elegant way to do this in Ruby

Question

I've started with Ruby and am finding new, shorter, elegant ways to write code everyday.

In solving Project Euler problems, I've written a lot of code like

if best_score < current_score
  best_score = current_score
end

Is there a more elegant way to write this?

Answer 1

This can be done on a single line:

best_score = current_score if best_score < current_score

Answer 2

Maybe a one-liner?

best_score = current_score if best_score < current_score

Answer 3

Not sure it would qualify as "more elegant", but if you don't want to rewrite the if every time ...

def max(b,c)
 if (b < c)
  c
 else
  b
 end
end

best = 10 
current = 20
best = max(best,current)

Answer 4

This is elegant enough. It's readable and easy to maintain.

If you want shorter, you can go:

best_score = current_score if best_score < current_score

or

best_score = current_score unless best_score >= current_score

... but it's not necessarily an improvement in all cases (keep in mind readability).

Answer 5

Or this way

(current_score > best_score) ? best_score = current_score : best_score

Answer 6

best_score = [best_score, current_score].max

see: Enumerable.max

---

disclaimer: although this is a little more readable (imho), it's less performant:

require 'benchmark'

best_score, current_score, n = 1000, 2000, 100_000

Benchmark.bm do |x|
  x.report { n.times do best_score = [best_score, current_score].max end }
  x.report { n.times do 
    best_score = current_score if best_score < current_score 
  end }
end

will result in (with ruby 1.8.6 (2008-08-11 patchlevel 287)):

    user     system      total        real
0.160000   0.000000   0.160000 (  0.160333)
0.030000   0.000000   0.030000 (  0.030578)

Answer 7

It looks just fine the way you have it already. I would only change the comparison so it reads:

If current score is greater than best score

You may also create a method and call that. That's more OO to me.

 def get_best_score() 
      current_score > best_score ?
          current_score :
          best_score
 end

This is what OOP is all about isn't? Hold the object state.

best_score = get_best_score()

Answer 8

Since I can't see it above, I lean toward this use of the ternary operator:

best_score = current_score > best_score ? current_score : best_score

and there's also this rather less frequently-encountered version:

best_score = (best_score > current_score && best_score) || current_score

...which is harder to read, but shows a (to me) slightly unexpected side-effect of short-circuiting. (See this blog post.)