Say, i am looking to better understand what reduceLeft method does when applied on a Array[String]
The scaladoc says:
Ok, i must ask again, what does this method do? And what's more important, if i can't rely on scaladoc to tell me that, where can i find out?
Yeah - that Scaladoc entry could probably be more helpful.
Another useful source of documentation is the Scala Documentation site, which has this to say about reduceLeft:
xs reduceLeft op
Apply binary operation op between successive elements of non-empty collection xs, going left to right.
So what it does is reduce a collection to a single value by successively applying a binary operator. Some examples:
scala> Array(1, 2, 3, 4) reduceLeft (_ + _)
res2: Int = 10
scala> Array("foo", "bar", "baz") reduceLeft (_ + _)
res3: String = foobarbaz
Related
I am little bit confused about +: and :: operators that are available.
It looks like both of them gives the same results.
scala> List(1,2,3)
res0: List[Int] = List(1, 2, 3)
scala> 0 +: res0
res1: List[Int] = List(0, 1, 2, 3)
scala> 0 :: res0
res2: List[Int] = List(0, 1, 2, 3)
For my novice eye source code for both methods looks similar (plus-colon method has additional condition on generics with use of builder factories).
Which one of these methods should be used and when?
+: works with any kind of collection, while :: is specific implementation for List.
If you look at the source for +: closely, you will notice that it actually calls :: when the expected return type is List. That is because :: is implemented more efficiently for the List case: it simply connects the new head to the existing list and returns the result, which is a constant-time operation, as opposed to linear copying the entire collection in the generic case of +:.
+: on the other hand, takes CanBuildFrom, so you can do fancy (albeit, not looking as nicely in this case) things like:
val foo: Array[String] = List("foo").+:("bar")(breakOut)
(It's pretty useless in this particular case, as you could start with the needed type to begin with, but the idea is you can prepend and element to a collection, and change its type in one "go", avoiding an additional copy).
While working trough the exercises and concepts of the book: Functional Programming in Scala by Paul Chiusano and Rúnar Bjarnason, I stumbled upon the exercise of writing my own function to reverse a list.
The "suggestion" the authors gave to motivate readers to learn more, was to see if we can write such a function using a fold.
My "non-fold" version is the following:
def myrev(arr:List[Int]):List[Int] = if (arr.length > 0) { myrev(arr.tail) :+ arr.head } else arr
However, can someone give me some pointers on how to at least start the logic to reverse a list via a fold?
I know that:
List(1,2,3,4).foldLeft(0)({case(x,y)=>x})
gives me the initial "seed" value, which is 0 above, and that:
List(1,2,3,4).foldLeft(0)({case(x,y)=>y})
gives me 4 which is the last element of the list.
So I would need to supply as function to foldLeft a sort of "identity" function that can give me an element in a certain position and maybe use it to reverse the list somehow, but I feel at a loss.
I didn't fully understood some Haskell code I found online and I want to "struggle" and try to get there on my own only with some pointers, instead of simply blindly copy some solution, so any help would be greatly appreciated.
scala> List(1, 2, 3, 4).foldLeft(List.empty[Int])((result, currentElem) => currentElem :: result)
res2: List[Int] = List(4, 3, 2, 1)
An implementation of a function that takes a list and inverse it using foldLeft
def reverse (l: List[String]) =
l.foldLeft(List[String]())( (x: List[String], y: String) => y :: x )
Create a list and call the reverse function
val l = List("1", "2", "3")
reverse(l)
Result
res2: List(3, 2, 1): List[String]
Long time lurker, first time poster.
In Scala, I'm looking for advantages as to why it was preferred to vary operators depending on type. For example, why was this:
Vector(1, 2, 3) :+ 4
determined to be an advantage over:
Vector(1, 2, 3) + 4
Or:
4 +: Vector(1,2,3)
over:
Vector(4) + Vector(1,2,3)
Or:
Vector(1,2,3) ++ Vector(4,5,6)
over:
Vector(1,2,3) + Vector(4,5,6)
So, here we have :+, +:, and ++ when + alone could have sufficed. I'm new at Scala, and I'll succumb. But, this seems unnecessary and obfuscated for a language that tries to be clean with its syntax.
I've done quite a few google and stack overflow searches and have only found questions about specific operators, and operator overloading in general. But, no background on why it was necessary to split +, for example, into multiple variations.
FWIW, I could overload the operators using implicit classes, such as below, but I imagine that would only cause confusion (and tisk tisks) from experienced Scala programmers using/reading my code.
object AddVectorDemo {
implicit class AddVector(vector : Vector[Any]) {
def +(that : Vector[Any]) = vector ++ that
def +(that : Any) = vector :+ that
}
def main(args : Array[String]) : Unit = {
val u = Vector(1,2,3)
val v = Vector(4,5,6)
println(u + v)
println(u + v + 7)
}
}
Outputs:
Vector(1, 2, 3, 4, 5, 6)
Vector(1, 2, 3, 4, 5, 6, 7)
The answer requires a surprisingly long detour through variance. I'll try to make it as short as possible.
First, note that you can add anything to an existing Vector:
scala> Vector(1)
res0: scala.collection.immutable.Vector[Int] = Vector(1)
scala> res0 :+ "fish"
res1: scala.collection.immutable.Vector[Any] = Vector(1, fish)
Why can you do this? Well, if B extends A and we want to be able to use Vector[B] where Vector[A] is called for, we need to allow Vector[B] to add the same sorts of things that Vector[A] can add. But everything extends Any, so we need to allow addition of anything that Vector[Any] can add, which is everything.
Making Vector and most other non-Set collections covariant is a design decision, but it's what most people expect.
Now, let's try adding a vector to a vector.
scala> res0 :+ Vector("fish")
res2: scala.collection.immutable.Vector[Any] = Vector(1, Vector(fish))
scala> res0 ++ Vector("fish")
res3: scala.collection.immutable.Vector[Any] = Vector(1, fish)
If we only had one operation, +, we wouldn't be able to specify which one of these things we meant. And we really might mean to do either. They're both perfectly sensible things to try. We could try to guess based on types, but in practice it's better to just ask the programmer to explicitly say what they mean. And since there are two different things to mean, there need to be two ways to ask.
Does this come up in practice? With collections of collections, yes, all the time. For example, using your +:
scala> Vector(Vector(1), Vector(2))
res4: Vector[Vector[Int]] = Vector(Vector(1), Vector(2))
scala> res4 + Vector(3)
res5: Vector[Any] = Vector(Vector(1), Vector(2), 3)
That's probably not what I wanted.
It's a fair question, and I think it has a lot to do with legacy code and Java compatibility. Scala copied Java's + for String concatenation, which has complicated things.
This + allows us to do:
(new Object) + "foobar" //"java.lang.Object#5bb90b89foobar"
So what should we expect if we had + for List and we did List(1) + "foobar"? One might expect List(1, "foobar") (of type List[Any]), just like we get if we use :+, but the Java-inspired String-concatenation overload would complicate this, since the compiler would fail to resolve the overload.
Odersky even once commented:
One should never have a + method on collections that are covariant in their element type. Sets and maps are non-variant, that's why they can have a + method. It's all rather delicate and messy. We'd be better off if we did not try to duplicate Java's + for String concatenation. But when Scala got designed the idea was to keep essentially all of Java's expression syntax, including String +. And it's too late to change that now.
There is some discussion (although in a different context) on the answers to this similar question.
Alright, Scala has me feeling pretty dense. I'm finding the docs pretty impenetrable -- and worse, you can't Google the term "Scala ++:" because Google drops the operator terms!
I was reading some code and saw this line:
Seq(file) ++: children.flatMap(walkTree(_))
But couldn't figure it out. The docs for Seq show three things:
++
++:
++:
Where the latter two are over loaded to do.. something. The actual explanation in the doc says that they do the same thing as ++. Namely, add one list to another.
So, what exactly is the difference between the operators..?
++ and ++: return different results when the operands are different types of collection. ++ returns the same collection type as the left side, and ++: returns the same collection type as the right side:
scala> List(5) ++ Vector(5)
res2: List[Int] = List(5, 5)
scala> List(5) ++: Vector(5)
res3: scala.collection.immutable.Vector[Int] = Vector(5, 5)
There are two overloaded versions of ++: solely for implementation reasons. ++: needs to be able to take any TraversableOnce, but an overloaded version is provided for Traversable (a subtype of TraversableOnce) for efficiency.
Just to make sure:
A colon (:) in the end of a method name makes the call upside-down.
Let's make two methods and see what's gonna happen:
object Test {
def ~(i: Int) = null
def ~:(i: Int) = null //putting ":" in the tail!
this ~ 1 //compiled
1 ~: this //compiled
this.~(1) //compiled
this.~:(1) //compiled.. lol
this ~: 1 //error
1 ~ this //error
}
So, in seq1 ++: seq2, ++: is actually the seq2's method.
edited: As #okiharaherbst mentions, this is called as right associativity.
Scala function naming will look cryptic unless you learn a few simple rules and their precedence.
In this case, a colon means that the function has right associativity as opposed to the more usual left associativity that you see in imperative languages.
So ++: as in List(10) ++: Vector(10) is not an operator on the list but a function called on the vector even if it appears on its left hand-side, i.e., it is the same as Vector(10).++:(List(10)) and returns a vector.
++ as in List(10) ++ Vector(10) is now function called on the list (left associativity), i.e., it is the same as List(10).++(Vector(10)) and returns a list.
what exactly is the difference between the operators..?
The kind of list a Seq.++ operates on.
def ++[B](that: GenTraversableOnce[B]): Seq[B]
def ++:[B >: A, That](that: Traversable[B])(implicit bf: CanBuildFrom[Seq[A], B, That]): That
def ++:[B](that: TraversableOnce[B]): Seq[B]
As commented in "What is the basic collection type in Scala?"
Traversable extends TraversableOnce (which unites Iterator and Traversable), and
TraversableOnce extends GenTraversableOnce (which units the sequential collections and the parallel.)
According to scaladoc, sliding() returns...
"An iterator producing iterable collections of size size, except the last and the only element will be truncated if there are fewer elements than size."
For me, intuitivelly, sliding(n) would return a sliding window of n elements if available. With the current implementation, I need to perform an extra check to make sure I don't get a list of 1 or 2 elements.
scala> val xs = List(1, 2)
xs: List[Int] = List(1, 2)
scala> xs.sliding(3).toList
res2: List[List[Int]] = List(List(1, 2))
I expected here an empty list instead. Why is sliding() implemented this way instead?
It was a mistake, but wasn't fixed as of 2.9. Everyone occasionally makes design errors, and once one gets into the library it's a nontrivial task to remove it.
Workaround: add a filter.
xs.sliding(3).filter(_.size==3).toList
You can "work around" this by using the GroupedIterator#withPartial modifier.
scala> val xs = List(1, 2)
xs: List[Int] = List(1, 2)
scala> xs.iterator.sliding(3).withPartial(false).toList
res7: List[Seq[Int]] = List()
(I don't know why you need to say xs.iterator but xs.sliding(3).withPartial(false) does not work because you get an Iterator instead of a GroupedIterator.
EDIT:
Check Rex's answer (which is the correct one). I'm leaving this just because (as Rex said on the comments) it was the original (wrong) idea behind that design decision.
I don't know why you would expect an empty list there, returning the full list seems like the best result, consider this example:
def slidingWindowsThing(windows : List[List[Int]]) { // do your thing
For this methods you probably want all these calls to work:
slidingWindowsThing((1 to 10).sliding(3))
slidingWindowsThing((1 to 3).sliding(3))
slidingWindowsThing((1 to 1).sliding(3))
This is why the method defaults to a list of size list.length instead of Nil (empty list).