So I assume my issue has to do with what is going on under the hood, but I don't understand why this doesn't works:
scala> b = b :: "apple";
<console>:8: error: value :: is not a member of java.lang.String
but this does:
scala> b = "apple" :: b;
b: List[java.lang.String] = List(apple, pear)
Thanks.
Method names that end in : are right associative, so b :: "apple" tries to call the :: method on a String, which doesn't exist.
The normal strategy for appending things if you must use a List is to add things to the beginning then reverse the result when you're done. But as Rex says, using a Vector might be better.
:: always joins a new item to the head of a list. Adding an item to the end can be done, but it takes time proportional to the length of the list (since the entire list must not only be traversed but, actually, also rebuilt).
If you really must add a item to the end of a list, use :+:
List("pear","orange") :+ "apple"
Better yet, use Vector when you need to add to the end (it's much faster at double-ended additions):
Vector("grape","peach") :+ "apple"
Another possibility is to explicitly add Nil to the end of chain:
scala> val a = "apple"
a: java.lang.String = apple
scala> val b = "pear"
b: java.lang.String = pear
scala> a::b::Nil
res0: List[java.lang.String] = List(apple, pear)
Related
From the Book programming in Scala I got the following line of code:
val second: List[ Int] => Int = { case x :: y :: _ => y }
//warning: match may not be exhaustive.
It states that this function will return the second element of a list of integers if the list is not empty or nil. Stil this part is a bit awkward to me:
case x :: y :: _
How does this ecxactly work? Does this mathches any list with at least 2 Elements and than return the second? If so can somebody still explain the syntax? I understood that :: is invoked on the right operand. So it could be written as
(_.::(y)).::(X)
Still I than don't get why this would return 2
val second: List[ Int] => Int = { case x :: y :: _ => y }
var x = List(1,2)
second(x) //returns 2
In the REPL, you can type:
scala> val list = "a" :: "b" :: Nil
list: List[String] = List(a, b)
which is to be read from right to left, and means take the end of a List (Nil), prepend String "b" and to this List ("b" :: Nil) prepend String a, a :: ("b" :: Nil) but you don't need the parens, so it can be written "a" :: "b" :: Nil.
In pattern matching you will more often see:
... list match {
case Nil => // ...
case x :: xs => // ...
}
to distinguish between empty list, and nonempty, where xs might be a rest of list, but matches Nil too, if the whole list is ("b" :: Nil) for example, then x="b" and xs=Nil.
But if list= "a" :: "b" :: Nil, then x="a" and xs=(b :: Nil).
In your example, the deconstruction is just one more step, and instead of a name like xs, the joker sign _ is used, indicating, that the name is probably not used and doesn't play a role.
The value second is of function type, it takes List[Int] and returns Int.
If the list has first element ("x"), and a second element ("y"), and whatever comes next (we don't care about it), we simply return the element "y" (which is the second element of the list).
In any other case, the function is not defined. You can check that:
scala> val second: PartialFunction[List[Int], Int] = {
| case x :: y :: _ => y
| }
second: PartialFunction[List[Int],Int] = <function1>
scala> second.isDefinedAt(List(1,2,3))
res18: Boolean = true
scala> second.isDefinedAt(List(1,2))
res19: Boolean = true
scala> second.isDefinedAt(List(0))
res20: Boolean = false
First of all. When you think about pattern matching you should think about matching a structure.
The first part of the case statement describes a structure. This structure may describe one or more things (variables) which are useful to deriving your result.
In your example, you are interested in deriving the second element of a list. A shorthand to build a list in Scala is to use :: method (also called cons). :: can also be used to describe a structure in case statement. At this time, you shouldn't think about evaluation of the :: method in first part of case. May be that's why you are saying about evaluation of _.::(y).::(x). The :: cons operator help us describe the structure of the list in terms of its elements. In this case, the first element (x) , the second element (y) and the rest of it (_ wildcard). We are interested in a structure that is a list with at least 2 elements and the third can be anything - a Nil to indicate end of list or another element - hence the wildcard.
The second part of the case statement, uses the second element to derive the result (y).
More on List and Consing
List in Scala is similar to a LinkedList. You know about the first element called head and start of the rest of the list. When traversing the linked list you stop if the rest of the list is Nil. This :: cons operator helps us visualise the structure of the linked list. Although Scala compile would actually be calling :: methods evaluating from right to left as you described _.::(y).::(x)
As an aside, you might have already noticed that the Scala compiler might be complain that your match isn't exhaustive. This means that this second method would work for list of any size. Because there isn't any case statement to describe list with zero or one element. Also, as mentioned in comments of previous answers, if you aren't interested in first element you can describe it as a wildcard _.
case _ :: y :: _ => y
I hope this helped.
If you see the structure of list in scala its head::tail, first element is treated as head and all remaining ones as tail(Nil will be the last element of tail). whenever you do x::y::_, x will match the head of the list and remaining will be tail and again y will match the head of the next list(tail of first list)
eg:
val l = List(1,2,3,4,5)
you can see this list in differnt ways:
1::2::3::4::5::Nil
1::List(2,3,4,5)
1::2::List(2,3,4,5)
and so on
So try matching the pattern. In your question y will give the second element
I am newbie to scala and just trying out stuff, below is what I am trying
scala> var col = List[String]()
col: List[String] = List()
scala> List("a", "b", "c").foreach(x => x :: col)
scala> println(col)
List()
Actually, I was expecting col to contain a,b,c, what am I missing?
You need an assignment in the foreach
scala> var col = List[String]()
col: List[String] = List()
scala> List("a", "b", "c").foreach(x => {col = x :: col})
scala> col
res0: List[String] = List(c, b, a)
The operation x :: col simply returns a new list consisting of the element x prepended to col, the original col is not changed. You would need to reassign col to this newly generated list.
Note however that this would not typically be considered idiomatic Scala since you are using side-effects.
The :: method on list does not add anything to the list, it creates a new list with the value prepended to it, you are discarding this new list instead of reassigning it to col. x => col = x :: col will add each element of your list to col. Note that col will then be List("c","b","a"), the order is reversed because you are pre-pending the elements to col.
Note that foreach returns nothing and is designed for side-effecting operations. If you simply want to transform a collection or load elements into another collection there are better methods to use.
For your specific operation, the most appropriate method is foldRight which iterates elements in reverse order, right-to-left. We want to iterate in reverse here because when you prepend elements to a list one at a time the order gets reversed.
val col = List("a", "b", "c").foldRight(List[String]())((x, acc) => x :: acc) will produce a List("a", "b", "c"). This has the advantage that we no longer need to use var to declare a mutable variable, and in fact we don'
to need to declare our list ahead of time at all.
note, we could have used some special syntax to save some typing
val col = List("a", "b", "c").foldRight(List[String]())(_ :: _)
The underscores give us a shorter syntax to write function literals, I'll leave up to you to decide if it's more clear or not.
I am currently facing the following issue.
I have code that essentially has the following cases:
val toList = this.toString.match {
case "" => List[MyType]()
case _ => this.val :: this.prev.toList
}
Obviously not exact but its the general gist. It works fine but I want the values appended to the list in the reverse order. Is there any good way to do this? Intellij throws errors if I try to reverse the order and do
this.prev.toList :: this.val
and also if I try to use operations like ++. Is what I'm trying to do impossible based on the structure of my class?
The specific errors I get involve "cannot resolve ::" or whatever symbol I use when I try to put this.prev.toList before this.val.
And yes the "this" aren't necessary- I included it to hopefully make my problem easier to understand.
:: adds an element at the beginning of this list
scala> 1 :: List(2,3)
List(1, 2, 3)
+: is the equivalent of ::
scala> 1 +: List(2,3)
List(1, 2, 3)
:+ append element at the end of the list
scala> List(1,2) :+ 3
List(1, 2, 3)
However the cost of prepending on List is O(1) but the appending one is O(n)!
For "numerous" collections you could consider other datastructure like Vector:
Vector provides very fast append and prepend
http://www.scala-lang.org/api/2.11.7/index.html#scala.collection.immutable.Vector
You can append with this method :+:
this.prev.toList :+ this.val // is `val` really then name?
But keep in mind that appending to a List can be very inefficient for long lists.
I found a piece of code I wrote some time ago using _* to create a flattened set from a list of objects.
The real line of code is a bit more complex and as I didn't remember exactly why was that there, took a bit of experimentation to understand the effect, which is actually very simple as seen in the following REPL session:
scala> val someList = List("a","a","b")
someList: List[java.lang.String] = List(a, a, b)
scala> val x = Set(someList: _*)
x: scala.collection.immutable.Set[java.lang.String] = Set(a, b)
scala> val y = Set(someList).flatten
y: scala.collection.immutable.Set[java.lang.String] = Set(a, b)
scala> x == y
res0: Boolean = true
Just as a reference of what happens without flatten:
scala> val z = Set(someList)
z: scala.collection.immutable.Set[List[java.lang.String]] = Set(List(a, a, b))
As I can't remember where did I get that idiom from I'd like to hear about what is actually happening there and if there is any consequence in going for one way or the other (besides the readability impact)
P.S.: Maybe as an effect of the overuse of underscore in Scala language (IMHO), it is kind of difficult to find documentation about some of its use cases, specially if it comes together with a symbol commonly used as a wildcard in most search engines.
_* is for expand this collection as if it was written here literally, so
val x = Set(Seq(1,2,3,4): _*)
is the same as
val x = Set(1,2,3,4)
Whereas, Set(someList) treats someList as a single argument.
To lookup funky symbols, you could use symbolhound
I have a Map[String, String] and want to concatenate the values to a single string.
I can see how to do this using a List...
scala> val l = List("te", "st", "ing", "123")
l: List[java.lang.String] = List(te, st, ing, 123)
scala> l.reduceLeft[String](_+_)
res8: String = testing123
fold* or reduce* seem to be the right approach I just can't get the syntax right for a Map.
Folds on a map work the same way they would on a list of pairs. You can't use reduce because then the result type would have to be the same as the element type (i.e. a pair), but you want a string. So you use foldLeft with the empty string as the neutral element. You also can't just use _+_ because then you'd try to add a pair to a string. You have to instead use a function that adds the accumulated string, the first value of the pair and the second value of the pair. So you get this:
scala> val m = Map("la" -> "la", "foo" -> "bar")
m: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(la -> la, foo -> bar)
scala> m.foldLeft("")( (acc, kv) => acc + kv._1 + kv._2)
res14: java.lang.String = lalafoobar
Explanation of the first argument to fold:
As you know the function (acc, kv) => acc + kv._1 + kv._2 gets two arguments: the second is the key-value pair currently being processed. The first is the result accumulated so far. However what is the value of acc when the first pair is processed (and no result has been accumulated yet)? When you use reduce the first value of acc will be the first pair in the list (and the first value of kv will be the second pair in the list). However this does not work if you want the type of the result to be different than the element types. So instead of reduce we use fold where we pass the first value of acc as the first argument to foldLeft.
In short: the first argument to foldLeft says what the starting value of acc should be.
As Tom pointed out, you should keep in mind that maps don't necessarily maintain insertion order (Map2 and co. do, but hashmaps do not), so the string may list the elements in a different order than the one in which you inserted them.
The question has been answered already, but I'd like to point out that there are easier ways to produce those strings, if that's all you want. Like this:
scala> val l = List("te", "st", "ing", "123")
l: List[java.lang.String] = List(te, st, ing, 123)
scala> l.mkString
res0: String = testing123
scala> val m = Map(1 -> "abc", 2 -> "def", 3 -> "ghi")
m: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,abc), (2,def), (3,ghi))
scala> m.values.mkString
res1: String = abcdefghi