Ruby - Spaceship operator won't work in a block for .sort

Question

I'm getting an error when attempting to use the spaceship operator with non alpha-numeric characters in the sort function.

word = "out-classed"
letters = word.downcase.split('')
letters.sort! do |x, y|
  if y < 'a'
    next
  else
   value = x <=> y
  end
end

I'm getting ArgumentError: comparison of String with String failed, and I'm almost positive this happens with the spaceship operator and not the < comparison.

The interesting part is that when I do this same comparison in irb outside of the context of a sort block, the comparison works. It also works when the word variable only consists of letters.

Can anybody help me to understand why this doesn't work in this specific context alone?

It's a good question because it's important for you and others to understand why you cannot do what you'd like to to do. — Cary Swoveland
– Cary Swoveland, Commented Jun 30, 2015 at 1:27

Cary Swoveland · Accepted Answer · 2015-06-30 01:23:19Z

3

If you attempt to sort a collection, x<=>y must return 0, 1 or -1 for every pair of elements of the collection. If <=> is defined artificially for some pairs (e.g., 'a'<=>'-' #=> 0 and '-'<=>'a' #=> 0), your sort may return erroneous results.

This is because sort algorithms do not necessarily evaluate all pairs of elements in the collection. If, for example, it finds that:

'a' <=> 'b' #=> 0

and

'b' <=> 'c' #=> 0

it will conclude that:

`a` <=> `c` #=> 0

because the collection being sorted must satisfy transitivity: x <== z if x <= y and y <= z.

For example, if the collection is the array ['z', '-', 'a'] and it finds that 'z' <= '-' and '-' <= 'a', it will conclude that 'z' <= 'a' (and not evaluate 'z' <=> 'a').

That's why:

['z', '-', 'a'].sort { |x,y| p [x,y]; (y < 'a') ? 0 : x<=>y }
  #-> ["z", "-"]
  #-> ["-", "a"]
  #=> ["z", "-", "a"]

doesn't work. You have two choices:

Remove the offending elements before sorting:

['z', '-', 'a'].select { |c| ('a'..'z').cover?(c) }.
                sort { |x,y| (y < 'a') ? 0 : x<=>y }
  #=> ["a", "z"]

or sort all elements of the collection:

['z', '-', 'a'].sort
  #=> ["-", "a", "z"]

If the collection contains non-comparable elements (e.g., [1,2,'cat']), you only choice is to remove elements from the array until all remaining elements are comparable.

edited Jun 30, 2015 at 1:23

answered Jun 29, 2015 at 19:36

Cary Swoveland

111k6 gold badges69 silver badges105 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Christopher Ronning Over a year ago

I'm leaving the selected answer as is because it did very directly answer the question. Upvote on this, however, because it's a very complete discussion of the subject. Incidentally, it helped me to solve the broader issue I was working on. Thank you!!

Kris Over a year ago

Upvote from me too, this is detail I was not aware of.

Kris · Accepted Answer · 2015-06-29 14:29:49Z

2

Instead of next you need to return 0, 1 or -1. Try this:

word = "out-classed"
letters = word.downcase.split('')
letters.sort! do |x, y|
  if y < 'a'
    0
  else
   value = x <=> y
  end
end

answered Jun 29, 2015 at 14:29

Kris

20.1k9 gold badges102 silver badges117 bronze badges

9 Comments

Christopher Ronning Over a year ago

This is working for everything except for closing periods ('.'). So, for example, the sentence "I'm going to do this." comes out as "i'm ggino ot do .hist" Maybe this is outside of the scope of the original question, but do you know why?

tadman Over a year ago

@steenslag You're not supposed to use next in these sorts of functions. You're expected to return -1, 0 or 1 only.

tadman Over a year ago

The value = part is completely useless here. In fact the whole thing could be collapsed to: y < 'a' ? 0 : x <=> y

the Tin Man Over a year ago

There is no reason to set value to the result of the <=> comparison, unless you want only the result of the very last loop, which won't tell you much that is useful. You'd either want all comparisons, or you shouldn't worry about it.

Cary Swoveland Over a year ago

Since you have not defined value prior to sort's block, value's scope is confined to the block. If you reference value outside the block an "undefined local variable or method" exception will be raised.

|

Piotr Kruczek · Accepted Answer · 2015-06-29 14:29:08Z

1

Your problem lies here

if y < 'a'
  next
else
  ...

sort method expects you to return a comparison value between every pair, so when you call next without returning anything, it says that comparison failed.

Try e.g. this:

if y < 'a'
  1
else
  value = x <=> y
end

answered Jun 29, 2015 at 14:29

Piotr Kruczek

2,38813 silver badges18 bronze badges

5 Comments

Hunter McMillen Over a year ago

This example is the default case of sort!, so you could really just say letters.sort! and have the same results

Christopher Ronning Over a year ago

It looks like setting the value to 1 leads to punctuation characters being moved to before the text. What's your reasoning in using 1 as opposed to 0?

Piotr Kruczek Over a year ago

No reasoning at all, simply stating that the value returned should be 1, 0 or -1, depending on how you want to sort this :)

Christopher Ronning Over a year ago

ahh, I see. For practical purposes 0 is what I was looking for, but good thinking on making it widely useful.

Cary Swoveland Over a year ago

['z', '-', '-', 'a'].sort { |x,y| p [x,y]; (y < 'a') ? 0 : x<=>y } #=> ["z", "-", "-", "a"] after printing ["z", "-"], ["-", "a"] and ["-", "-"].

Collectives™ on Stack Overflow

Ruby - Spaceship operator won't work in a block for .sort

3 Answers 3

2 Comments

9 Comments

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

9 Comments

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related