Is Scala idiomatic coding style just a cool trap for writing inefficient code?

Question

I sense that the Scala community has a little big obsession with writing \"concise\", \"cool\", \"scala idiomatic\", \"one-liner\" -if possible- code. This

Answer 1

Let's discuss a fallacy in the question:

So, if 99% of Java developers write more efficient code than 99% of Scala developers, this is a huge obstacle to cross for greater Scala adoption. Is there a way out of this trap?

This is presumed, with absolutely no evidence backing it up. If false, the question is moot.

Is there evidence to the contrary? Well, let's consider the question itself -- it doesn't prove anything, but shows things are not that clear.

Totally non-Scala idiomatic, non-functional, non-concise, but it's very efficient. It traverses the list only once!

Of the four claims in the first sentence, the first three are true, and the fourth, as shown by user unknown, is false! And why it is false? Because, contrary to what the second sentence states, it traverses the list more than once.

The code calls the following methods on it:

res.append(max)
res.append(x)

and

res.toList

Let's consider first append.

1. append takes a vararg parameter. That means max and x are first encapsulated into a sequence of some type (a WrappedArray, in fact), and then passed as parameter. A better method would have been +=.

2. Ok, append calls ++=, which delegates to +=. But, first, it calls ensureSize, which is the second mistake (+= calls that too -- ++= just optimizes that for multiple elements). Because an Array is a fixed size collection, which means that, at each resize, the whole Array must be copied!

So let's consider this. When you resize, Java first clears the memory by storing 0 in each element, then Scala copies each element of the previous array over to the new array. Since size doubles each time, this happens log(n) times, with the number of elements being copied increasing each time it happens.

Take for example n = 16. It does this four times, copying 1, 2, 4 and 8 elements respectively. Since Java has to clear each of these arrays, and each element must be read and written, each element copied represents 4 traversals of an element. Adding all we have (n - 1) * 4, or, roughly, 4 traversals of the complete list. If you count read and write as a single pass, as people often erroneously do, then it's still three traversals.

One can improve on this by initializing the ArrayBuffer with an initial size equal to the list that will be read, minus one, since we'll be discarding one element. To get this size, we need to traverse the list once, though.

Now let's consider toList. To put it simply, it traverses the whole list to create a new list.

So, we have 1 traversal for the algorithm, 3 or 4 traversals for resize, and 1 additional traversal for toList. That's 4 or 5 traversals.

The original algorithm is a bit difficult to analyse, because take, drop and ::: traverse a variable number of elements. Adding all together, however, it does the equivalent of 3 traversals. If splitAt was used, it would be reduced to 2 traversals. With 2 more traversals to get the maximum, we get 5 traversals -- the same number as the non-functional, non-concise algorithm!

So, let's consider improvements.

On the imperative algorithm, if one uses ListBuffer and +=, then all methods are constant-time, which reduces it to a single traversal.

On the functional algorithm, it could be rewritten as:

val max = xs.max
val (before, _ :: after) = xs span (max !=)
before ::: after

That reduces it to a worst case of three traversals. Of course, there are other alternatives presented, based on recursion or fold, that solve it in one traversal.

And, most interesting of all, all of these algorithms are O(n), and the only one which almost incurred (accidentally) in worst complexity was the imperative one (because of array copying). On the other hand, the cache characteristics of the imperative one might well make it faster, because the data is contiguous in memory. That, however, is unrelated to either big-Oh or functional vs imperative, and it is just a matter of the data structures that were chosen.

So, if we actually go to the trouble of benchmarking, analyzing the results, considering performance of methods, and looking into ways of optimizing it, then we can find faster ways to do this in an imperative manner than in a functional manner.

But all this effort is very different from saying the average Java programmer code will be faster than the average Scala programmer code -- if the question is an example, that is simply false. And even discounting the question, we have seen no evidence that the fundamental premise of the question is true.

EDIT

First, let me restate my point, because it seems I wasn't clear. My point is that the code the average Java programmer writes may seem to be more efficient, but actually isn't. Or, put another way, traditional Java style doesn't gain you performance -- only hard work does, be it Java or Scala.

Next, I have a benchmark and results too, including almost all solutions suggested. Two interesting points about it:

1. Depending on list size, the creation of objects can have a bigger impact than multiple traversals of the list. The original functional code by Adrian takes advantage of the fact that lists are persistent data structures by not copying the elements right of the maximum element at all. If a Vector was used instead, both left and right sides would be mostly unchanged, which might lead to even better performance.

2. Even though user unknown and paradigmatic have similar recursive solutions, paradigmatic's is way faster. The reason for that is that he avoids pattern matching. Pattern matching can be really slow.

The benchmark code is here, and the results are here.

Answer 2

The example you gave is not very functional, actually. Here's what you are doing:

// Given a list of Int
def removeMaxCool(xs: List[Int]): List[Int] = {

  // Find the index of the biggest Int
  val maxIndex = xs.indexOf(xs.max);

  // Then take the ints before and after it, and then concatenate then
  xs.take(maxIndex) ::: xs.drop(maxIndex+1)
}

Mind you, it is not bad, but you know when functional code is at its best when it describes what you want, instead of how you want it. As a minor criticism, if you used splitAt instead of take and drop you could improve it slightly.

Another way of doing it is this:

def removeMaxCool(xs: List[Int]): List[Int] = {
  // the result is the folding of the tail over the head 
  // and an empty list
  xs.tail.foldLeft(xs.head -> List[Int]()) {

    // Where the accumulated list is increased by the
    // lesser of the current element and the accumulated
    // element, and the accumulated element is the maximum between them
    case ((max, ys), x) => 
      if (x > max) (x, max :: ys)
      else (max, x :: ys)

  // and of which we return only the accumulated list
  }._2
}

Now, let's discuss the main issue. Is this code slower than the Java one? Most certainly! Is the Java code slower than a C equivalent? You can bet it is, JIT or no JIT. And if you write it directly in assembler, you can make it even faster!

But the cost of that speed is that you get more bugs, you spend more time trying to understand the code to debug it, and you have less visibility of what the overall program is doing as opposed to what a little piece of code is doing -- which might result in performance problems of its own.

So my answer is simple: if you think the speed penalty of programming in Scala is not worth the gains it brings, you should program in assembler. If you think I'm being radical, then I counter that you just chose the familiar as being the "ideal" trade off.

Do I think performance doesn't matter? Not at all! I think one of the main advantages of Scala is leveraging gains often found in dynamically typed languages with the performance of a statically typed language! Performance matters, algorithm complexity matters a lot, and constant costs matters too.

But, whenever there is a choice between performance and readability and maintainability, the latter is preferable. Sure, if performance must be improved, then there isn't a choice: you have to sacrifice something to it. And if there's no lost in readability/maintainability -- such as Scala vs dynamically typed languages -- sure, go for performance.

Lastly, to gain performance out of functional programming you have to know functional algorithms and data structures. Sure, 99% of Java programmers with 5-10 years experience will beat the performance of 99% of Scala programmers with 6 months experience. The same was true for imperative programming vs object oriented programming a couple of decades ago, and history shows it didn't matter.

EDIT

As a side note, your "fast" algorithm suffer from a serious problem: you use ArrayBuffer. That collection does not have constant time append, and has linear time toList. If you use ListBuffer instead, you get constant time append and toList.

Answer 3

First of all, the behavior of the methods you presented is not the same. The first one keeps the element ordering, while the second one doesn't.

Second, among all the possible solution which could be qualified as "idiomatic", some are more efficient than others. Staying very close to your example, you can for instance use tail-recursion to eliminate variables and manual state management:

def removeMax1( xs: List[Int] ) = {
  def rec( max: Int, rest: List[Int], result: List[Int]): List[Int] = {
    if( rest.isEmpty ) result
    else if( rest.head > max ) rec( rest.head, rest.tail, max :: result)
    else rec( max, rest.tail, rest.head :: result )
  }
  rec( xs.head, xs.tail, List() )
}

or fold the list:

def removeMax2( xs: List[Int] ) = {
  val result = xs.tail.foldLeft( xs.head -> List[Int]() ) { 
    (acc,x) =>
      val (max,res) = acc
      if( x > max ) x -> ( max :: res )
      else max -> ( x :: res )
  }
  result._2
}

If you want to keep the original insertion order, you can (at the expense of having two passes, rather than one) without any effort write something like:

def removeMax3( xs: List[Int] ) = {
  val max = xs.max
  xs.filterNot( _ == max )
}

which is more clear than your first example.

Answer 4

Another contender. This uses a ListBuffer, like Daniel's second offering, but shares the post-max tail of the original list, avoiding copying it.

  def shareTail(xs: List[Int]): List[Int] = {
    var res = ListBuffer[Int]()
    var maxTail = xs
    var first = true;
    var x = xs
    while ( x != Nil ) {
      if (x.head > maxTail.head) {
          while (!(maxTail.head == x.head)) {
              res += maxTail.head
              maxTail = maxTail.tail
          }
      }
      x = x.tail
    }
    res.prependToList(maxTail.tail)
  }