Why doesn't this work:
val m = Map( 1-> 2, 2-> 4, 3 ->6)
def h(k: Int, v: Int) = if (v > 2) Some(k->v) else None
m.flatMap { case(k,v) => h(k,v) }
m.flatMap { (k,v) => h(k,v) }
The one with the case statement gives me:
res1: scala.collection.immutable.Map[Int,Int] = Map(2 -> 4, 3 -> 6)
but the other one fails and says MIssing Type parameter v, and expected: Int, actual:(Int, Int)
The case keyword signifies pattern matching, so the Tuple2 (a Mapis an Iterable ofTuple2 elements) that you are flatMapping "over" gets decomposed into k and v. (The fact that flatMap works when the h function is producing an Option rather than a Map or Iterable is the Scala collections library being perhaps overly permissive.)
Without the case keyword, you are providing a function that requires two arguments, but flatMap needs a function that accepts a single argument (a Tuple2). So the second version does not typecheck.
For second one you can do this, if you don't want to use case.
m.flatMap { x => h(x._1, x._2) } // x is (key,value) pair here(each element in map), hence accessing the key , value as _1,_2 respectively
Related
Given a string I want to create a map that for each character in a string will give the number of times the character occurs in a string. The following function makes a map from character to a list of Strings.
def wordOccurrences(w: String) = {
val lower = w.toLowerCase.toList
lower.groupBy(t => t)
}
Now I wanted to alter the last line to:
lower.groupBy(t => t) map ( (x,y) => x -> y.length)
But it doesn't work, can someone explain why and how to fix it?
For mapping purposes, a Map[K, V] is an Iterable[(K, V)] (notice the extra pair of parentheses, identifying a tuple type), meaning that when you map over it you have pass a function that goes from (K, V) to your target type.
What you are doing, however, is passing a function that takes two independent arguments, rather then a single tuple argument.
The difference can be seen by inspecting the types of these two functions in the Scala shell:
scala> :t (a: Int, b: Int) => a + b
(Int, Int) => Int
scala> :t (p: (Int, Int)) => p._1 + p._2
((Int, Int)) => Int
Notice how the former takes two arguments while the latter takes a single tuple.
What you can do is pass a function which decomposes the tuple so that you can bind the components of the tuple independently:
lower.groupBy(t => t) map { case (x, y) => x -> y.length }
or alternatively pass a function which uses the tuple without deconstructing it
lower.groupBy(t => t) map (p => p._1 -> p._2.length)
Note
Dotty, which is the current project Scala's original author Martin Odersky is working on and that will probably become Scala 3, supports the syntax you are proposing, calling the feature function arity adaptation. This has been discussed, along with other feature, in Odersky's 2016 Keynote at Scala eXchange, "From DOT to Dotty" (here the video taped at 2017 Voxxed Days CERN).
You can use
lower.groupBy(t => t).mapValues(_.length)
I fould like to understand how foldLeft works for Maps. I do understand how it works if I have a List and call on it foldLeft with a zero-element and a function:
val list1 = List(1,2,3)
list1.foldLeft(0)((a,b) => a + b)
Where I add the zero-element 0 with the first element of list1 and then add the second element of list1 and so on. So output becomes the new input and the first input is the zero-element.
Now I got the code
val map1 = Map(1 -> 2.0, 3 -> 4.0, 5 -> 6.2) withDefaultValue 0.0
val map2 = Map(0 -> 3.0, 3 -> 7.0) withDefaultValue 0.0
def myfct(terms: Map[Int, Double], term: (Int, Double)): Map[Int, Double] = ???
map1.foldLeft(map2)(myfct)
So my first element here is a Tuple2, but since map2 is a Map and not a Tuple2, what is the zero-element?
When we had a List, namely list1, we always "took the next element in list1". What is the "next element in map1? Is it another pair of map1?
In this context, you can think of a Map as a list of tuples. You can create a list like this: List(1 -> 2.0, 3 -> 4.0, 5 -> 6.2), and call foldLeft on it (that is more or less exactly what Map.foldLeft does). If you understand how foldLeft works with lists, then now you know how it works with Maps too :)
To answer your specific questions:
The first parameter of foldLeft can be of any type. You could pass in a Map instead of an Int in your first example too. It does not have to be of the same type as elements of the collection you are processing (although, it could be), as you have in your first example, nor does it need to be the same type as the collection itself, as you have it in the last example.
Consider this for the sake of example:
List(1,2,3,4,5,6).foldLeft(Map.empty[String,Int]) { case(map,elem) =>
map + (elem.toString -> elem)
}
This produces the same result as list.map { x => x.toString -> x }.toMap. As you can see, the first parameter here is a Map, that is neither List nor an Int.
The type you pass to foldLeft is also the type it returns, and the type that the function you pass in returns. It is not "element zero".
foldLeft will pass that parameter to your reducer function, along with the first element of the list. Your function will combine the two elements, and produce a new value of the same type as the first param. That value is passed in again, again with the second element ... etc.
Perhaps, inspecting the signature of foldLeft would be helpful:
foldLeft[B](z: B)(op: (B, A) ⇒ B): B
Here A is the type of your collection elements, and B can be anything, the only requirement is that the four places where it appears have the same type.
Here is another example, that is (almost) equivalent to list.mkString(","):
List(1,2,3,4,5,6).foldLeft("") {
case("", i) => i.toString
case(s,i) => s + "," + i
}
As I explained in the beginning, a map in this context is a kind of a list (a sequence rather). Just like "we always took the next element of the list" when we dealt with lists, we will take the "next element of the map" in this case. You said it yourself, the elements of the map are tuples, so that's what the type of the next element will be:
Map("one" -> 1, "two" -> 2, "three" -> 3)
.foldLeft("") {
case("", (key,value)) => key + "->" + value
case(s, (key,value)) => s + ", " + key + "->" + value
}
As mentioned above, the signature of the foldLeft operation is as shown:
foldLeft[B](z: B)(op: (B, A) ⇒ B): B
The resulting type of the operation is B. So, this answers your first question:
So my first element here is a Tuple2, but since map2 is a Map and not
a Tuple2, what is the zero-element?
The zero-element is whatever you want to output from the foldLeft. It could be an Int or a Map or anything else for that matter.
op in the function signature (i.e the second argument), is how you would like to operate on that for each element of map1, which is a (key,value) pair, or another way of writing it is as key -> value.
Let's try and understand it by building it up with more simpler operations.
val map1 = Map(1 -> 1.0, 4 -> 4.0, 5 -> 5.0) withDefaultValue 0.0
val map2 = Map(0 -> 0.0, 3 -> 3.0) withDefaultValue 0.0
// Here we're just making the output an Int
// So we just add the first element of the key-value pair.
def myOpInt(z: Int, term: (Int, Double)): Int = {
z + term._1
}
// adds all the first elements of the key-value pairs of map1 together
map1.foldLeft(0)(myOpInt) // ... output is 10
// same as above, but for map2 ...
map2.foldLeft(0)(myOpInt) // ... output is 3
Now, taking it to the next step by using a zero element (z) as an existing map...we would essentially add elements to that map we are using as z.
val map1 = Map(1 -> 1.0, 4 -> 4.0, 5 -> 5.0) withDefaultValue 0.0
val map2 = Map(0 -> 0.0, 3 -> 3.0) withDefaultValue 0.0
// Here z is a Map of Int -> Double
// We expect a pair of (Int, Double) as the terms to fold left with z
def myfct(z: Map[Int, Double], term: (Int, Double)): Map[Int, Double] = {
z + term
}
map1.foldLeft(map2)(myfct)
// output is Map(0 -> 0.0, 5 -> 5.0, 1 -> 1.0, 3 -> 3.0, 4 -> 4.0)
// i.e. same elements of both maps concatenated together, albeit out of order, since Maps don't guarantee order ...
When we had a List, namely list1, we always "took the next element in
list1". What is the "next element in map1? Is it another pair of map1?
Yes, it's another key-value pair (k,v) in map1.
This seems not logical for me:
scala> val a = Map((1, "111"), (2, "222"))
a: scala.collection.immutable.Map[Int,String] = Map(1 -> 111, 2 -> 222)
scala> val b = a.map((key, value) => value)
<console>:8: error: wrong number of parameters; expected = 1
val b = a.map((key, value) => value)
^
scala> val c = a.map(x => x._2)
c: scala.collection.immutable.Iterable[String] = List(111, 222)
I know that I can say val d = a.map({ case(key, value) => value })
But why isn't it possible to say a.map((key, value) => value) ? There is only one argument there of type Tuple2[Int, String] or Pair of Int, String. What's the difference between a.map((key, value) => value) and a.map(x => x._2) ?
UPDATE:
val myTuple2 = (1, 2) -- this is one variable, correct?
for ( (k, v) <- a ) yield v -- (k, v) is also only one variable, correct?
map((key, value) => value) -- 2 variables. weird.
So how do I specify a variable of type Tuple2 (or any other type) in map without using case?
UPDATE2:
What's wrong with that?
Map((1, "111"), (2, "222")).map( ((x,y):Tuple2[Int, String]) => y) -- wrong
Map((1, "111"), (2, "222")).map( ((x):Tuple2[Int, String]) => x._2) -- ok
Okay, you still not convinced. In cases like this it is pretty reasonable to fallback to the source of the truth (well, kinda): The Holy Specification (aka, Scala Language Specification).
So, in anonymous function parameters are treated on individual basis, not as a whole tuple band (and it is pretty smart, otherwise, how would you call the anonymous function with 2, ... n parameters?).
At the same time
val x = (1, 2)
is a single item of type Tiple2[Int,Int] (if you're interested you may find corresponding section of spec as well).
for ( (k, v) <- a ) yield v
In this case you have one variable unpacked to two variables. It is similar to
val x = (1, 2) // one variable -- tuple
val (y,z) = x // two integer variables unpacked from one
Some call this destructuring assignment and this is a particular case of pattern matching. And you've already provided another example of pattern matching in action:
a.map({ case(key, value) => value })
Which we can read as map accepts a function produced by a partial function literal, which enables use of pattern matching.
You're basically asking this same questions:
Scala - can a lambda parameter match a tuple?
You've already listed most of the options they listed there, including the accepted answer of using a PartialFunction.
However, since you're using your lambda in a map function, you could use a for comprehension instead:
for ( (k, v) <- a ) yield v
Alternatively, you can use the Function2.tupled method to fix your lambda's type:
scala> val a = Map((1, "111"), (2, "222"))
a: scala.collection.immutable.Map[Int,String] = Map(1 -> 111, 2 -> 222)
scala> a.map( ((k:Int,v:String) => v).tupled )
res1: scala.collection.immutable.Iterable[String] = List(111, 222)
To answer your question in your thread with om-nom-nom above, look at this output:
scala> ( (x:Int,y:String) => y ).getClass.getSuperclass
res0: Class[?0] forSome { type ?0 >: ?0; type ?0 <: (Int, String) => String } = class scala.runtime.AbstractFunction2
Notice that the superclass of the anonymous function (x:Int,y:String) => y is Function2[Int, String, String], not Function1[(Int, String), String].
You can use pattern matching (or partial function, in this instance this is the same), notice angular brackets:
val b = a.map{ case (key, value) => value }
To create a new class that can be used in a Scala for comprehension, it seems that all you have to do is define a map function:
scala> class C[T](items: T*) {
| def map[U](f: (T) => U) = this.items.map(f)
| }
defined class C
scala> for (x <- new C(1 -> 2, 3 -> 4)) yield x
res0: Seq[(Int, Int)] = ArrayBuffer((1,2), (3,4))
But that only works for simple for loops where there is no pattern matching on the left hand side of <-. If you try to pattern match there, you get a complaint that the filter method is not defined:
scala> for ((k, v) <- new C(1 -> 2, 3 -> 4)) yield k -> v
<console>:7: error: value filter is not a member of C[(Int, Int)]
for ((k, v) <- new C(1 -> 2, 3 -> 4)) yield k -> v
Why is filter required to implement the pattern matching here? I would have thought Scala would just translate the above loop into the equivalent map call:
scala> new C(1 -> 2, 3 -> 4).map{case (k, v) => k -> v}
res2: Seq[(Int, Int)] = ArrayBuffer((1,2), (3,4))
But that seems to work fine, so the for loop must be translated into something else. What is it translated into that needs the filter method?
The short answer: according to the Scala specs, you shouldn't need to define a 'filter' method for the example you gave, but there is an open bug that means it is currently required.
The long answer: the desugaring algorithm applied to for comprehensions is described in the Scala language specification. Let's start with section 6.19 "For Comprehensions and For Loops" (I'm looking at version 2.9 of the specification):
In a first step, every generator p <- e, where p is not irrefutable (§8.1) for the type of e is replaced by p <- e.withFilter { case p => true; case _ => false }
The important point for your question is whether the pattern in the comprehension is "irrefutable" for the given expression or not. (The pattern is the bit before the '<-'; the expression is the bit afterwards.) If it is "irrefutable" then the withFilter will not be added, otherwise it will be needed.
Fine, but what does "irrefutable" mean? Skip ahead to section 8.1.14 of the spec ("Irrefutable Patterns"). Roughly speaking, if the compiler can prove that the pattern cannot fail when matching the expression then the pattern is irrefutable and the withFilter call will not be added.
Now your example that works as expected is the first type of irrefutable pattern from section 8.1.14, a variable pattern. So the first example is easy for the compiler to determine that withFilter is not required.
Your second example is potentially the third type of irrefutable pattern, a constructor pattern. Trying to match (k,v) which is Tuple2[Any,Any] against a Tuple2[Int,Int] (see section 8.1.6 and 8.1.7 from the specification)) succeeds since Int is irrefutable for Any. Therefore the second pattern is also irrefutable and doesn't (shouldn't) need a withFilter method.
In Daniel's example, Tuple2[Any,Any] isn't irrefutable against Any, so the withFilter calls gets added.
By the way, the error message talks about a filter method but the spec talks about withFilter - it was changed with Scala 2.8, see this question and answer for the gory details.
See the difference:
scala> for ((k, v) <- List(1 -> 2, 3 -> 4, 5)) yield k -> v
res22: List[(Any, Any)] = List((1,2), (3,4))
scala> List(1 -> 2, 3 -> 4, 5).map{case (k, v) => k -> v}
scala.MatchError: 5
I had a List of Scala tuples like the following:
val l = List((1,2),(2,3),(3,4))
and I wanted to map it in a list of Int where each item is the sum of the Ints in a the corresponding tuple. I also didn't want to use to use the x._1 notation so I solved the problem with a pattern matching like this
def addTuple(t: (Int, Int)) : Int = t match {
case (first, second) => first + second
}
var r = l map addTuple
Doing that I obtained the list r: List[Int] = List(3, 5, 7) as expected. At this point, almost by accident, I discovered that I can achieve the same result with an abbreviated form like the following:
val r = l map {case(first, second) => first + second}
I cannot find any reference to this syntax in the documentation I have. Is that normal? Am I missing something trivial?
See Section 8.5 of the language reference, "Pattern Matching Anonymous Functions".
An anonymous function can be defined by a sequence of cases
{case p1 =>b1 ... case pn => bn }
which appear as an expression without a prior match. The expected type of such an expression must in part be defined. It must be either scala.Functionk[S1, ..., Sk, R] for some k > 0, or scala.PartialFunction[S1, R], where the argument type(s) S1, ..., Sk must be fully determined, but the result type R may be undetermined.
The expected type deternines whether this is translated to a FunctionN or PartialFunction.
scala> {case x => x}
<console>:6: error: missing parameter type for expanded function ((x0$1) => x0$1 match {
case (x # _) => x
})
{case x => x}
^
scala> {case x => x}: (Int => Int)
res1: (Int) => Int = <function1>
scala> {case x => x}: PartialFunction[Int, Int]
res2: PartialFunction[Int,Int] = <function1>
{case(first, second) => first + second} is treated as a PartialFunction literal. See examples in "Partial Functions" section here: http://programming-scala.labs.oreilly.com/ch08.html or section 15.7 of Programming in Scala.
Method map accepts a function. In your first example you create a function, assign it to a variable, and pass it to the map method. In the second example you pass your created function directly, omitting assigning it to a variable. You are doing just the same thing.