scala, guidelines on return type - when prefer seq, iterable, traversable

Scala

Scala Problem Overview


When do you choose to type a given function's return type as Seq vs Iterable vs Traversable (or alternatively even deeper within Seq's hierarchy)?

How do you make that decision? We have a lot of code that returns Seqs by default (usually starting from results of a DB query and successive transformations). I tend to want to make the return types Traversable by default and Seq when specifically expecting a given order. But I don't have a strong justification for doing so.

I am perfectly familiar with the definition of each trait, so please don't answer with defining the terms.

Scala Solutions


Solution 1 - Scala

This is a good question. You have to balance two concerns:

  • (1) try to keep your API general, so you can change the implementation later
  • (2) give the caller some useful operations to perform on the collection

Where (1) asks you to be as little specific about the type (e.g. Iterable over Seq), and (2) asks you the opposite.

Even if the return type is just Iterable, you can still return let's say a Vector, so if the caller wishes to gain extra power, it can just call .toSeq or .toIndexedSeq on it, and that operation is cheap for a Vector.

As a measure of the balance, I would add a third point:

  • (3) use a type that kind of reflects how the data is organised. E.g. when you can assume that the data does have a sequence, give Seq. If you can assume that no two equal objects can occur, give a Set. Etc.

Here are my rules of thumb:

  • try to use only a small set of collections: Set, Map, Seq, IndexedSeq
  • I often violate this previous rule, though, using List in favour of Seq. It allows the caller to do pattern matching with the cons extractors
  • use immutable types only (e.g. collection.immutable.Set, collection.immutable.IndexedSeq)
  • do not use concrete implementations (Vector), but the general type (IndexedSeq) which gives the same API
  • if you are encapsulating a mutable structure, only return Iterator instances, the caller can then easily generate a strict structure, e.g. by calling toList on it
  • if your API is small and clearly tuned towards "big data throughput", use IndexedSeq

Of course, this is my personal choice, but I hope it sounds sane.

Solution 2 - Scala

  • Use Seq by default everywhere.
  • Use IndexedSeq when you need to access by index.
  • Use anything else only in special circumstances.

These are the "common-sense" guidelines. They are simple, practical, and work well in practice while balancing principles and performance. The principles are:

  1. Use a type that reflects how the data is organized (thanks OP and ziggystar).
  2. Use interface types in both method arguments and return types. Both inputs and return types of an API benefit from the flexibility of generality.

Seq satisfies both principles. As described in http://docs.scala-lang.org/overviews/collections/seqs.html:

> A sequence is a kind of iterable that has a [finite] length and whose elements have fixed index positions, starting from 0.

90% of the time, your data is a Seq.

Other notes:

  • List is an implementation type, so you shouldn't use it in an API. A Vector for instance can't be used as a List without going through a conversion.
  • Iterable doesn't define length. Iterable abstracts across finite sequences and potentially infinite streams. Most of the time one is dealing with finite sequences so you "have a length," and Seq reflects that. Frequently you won't actually make use of length. But it's needed often enough, and is easy to provide, so use Seq.

Drawbacks:

There are some slight downsides to these "common-sense" conventions.

Footnotes:

Solution 3 - Scala

Make your method's return type as specific as possible. Then if the caller wants to keep it as a SuperSpecializedHashMap or type it as a GenTraversableOnce, they can. This is why the compiler infers the most specific type by default.

Solution 4 - Scala

A rule of thumb I follow is, depending on implementation, to make the return types as specific as possible and the types of arguments as general as possible. It's an easy to follow rule and it provides you with consistent guarantees on the type properties with maximum freedom.

Say, if you have a function implementation which just traverses a data structure with methods like map, filter or fold - those that are implemented in the trait Traversable, you can expect it to perform equally on any type of input collection - be it a List, Vector, HashSet or even a HashMap, so your input argument should be specified as Traversable[T]. The choice of output type of the function should only depend on its implementation: in this case it should be Traversable too. If however in your function you force this data structure to some more specific type with methods like toList, toSeq or toSet, you should specify the appropriate type. Notice the consistency between the implementation and the return type?

If your function accesses the elements of input by index, the input should be specified as IndexedSeq, as it is the most general type that provides you with guarantees on effective implementation of method apply.

In case of abstract members the same rule applies with the only difference that you should specify the return types based on how you plan to use them instead of implementation, thus most often they will be more general than in implementation. The categorical choices Seq, Set or Map are the most expected.

Following this rule you protect yourself from very common cases of bottleneck when, for instance, items get appended to List or contains gets called on a Seq instead of a Set, yet your program remains a nice degree of freedom and is consistent in sense of choice of types.

Solution 5 - Scala

Quick note: With Scala 2.13.x, Traversable is out of the picture. Iterable is much more general and it was determined that going forward the duality is not justified. Iterable is now at the top of the collections hierarchy and Traversable has been deprecated.

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Solution 3 - ScalaLuigi PlingeView Answer on Stackoverflow
Solution 4 - ScalaNikita VolkovView Answer on Stackoverflow
Solution 5 - ScalaMario RossiView Answer on Stackoverflow