I have a List[Either[Error, Satisfaction]], and I need to take the first element in that list that isRight. If there are no elements that satisfy isRight, then I should return the last Error.
For now I've thought on a takeWhile(_.isLeft), find the size of this last list, and then get the nth element in the original list, with n being the size of the first-lefts list, but I think there's a better solution for this.
Any help will be strongly appreciated.
You might reduce the list like so:
lst.reduceLeft((a,b) => if (a.isRight) a else b)
The result will be the first Right element or, if there is none, the final Left element.
You can do this, which has the dual problem to #jwvh's answer: if rest is empty, it has to go over the list again.
val (lefts, rest) = list.span(_.isLeft)
rest.headOption.getOrElse(lefts.last)
// or orElse(lefts.lastOption) if list can be empty
And the recursive solution without either problem for completeness (assuming non-empty list):
def foo(list: List[Either[Error, Satisfaction]]) = list match {
case x :: tail => if (x.isRight || tail.isEmpty) x else foo(tail)
}
Not that much more code.
Related
I have a input file like this:
The Works of Shakespeare, by William Shakespeare
Language: English
and I want to use flatMap with the combinations method to get the K-V pairs per line.
This is what I do:
var pairs = input.flatMap{line =>
line.split("[\\s*$&#/\"'\\,.:;?!\\[\\(){}<>~\\-_]+")
.filter(_.matches("[A-Za-z]+"))
.combinations(2)
.toSeq
.map{ case array => array(0) -> array(1)}
}
I got 17 pairs after this, but missed 2 of them: (by,shakespeare) and (william,shakespeare). I think there might be something wrong with the last word of the first sentence, but I don't know how to solve it, can anyone tell me?
The combinations method will not give duplicates even if the values are in the opposite order. So the values you are missing already appear in the solution in the other order.
This code will create all ordered pairs of words in the text.
for {
line <- input
t <- line.split("""\W+""").tails if t.length > 1
a = t.head
b <- t.tail
} yield a -> b
Here is the description of the tails method:
Iterates over the tails of this traversable collection. The first value will be this traversable collection and the final one will be an empty traversable collection, with the intervening values the results of successive applications of tail.
I have a template sequence like this:
val template = Seq(0,0,0,0,0,0,0,0,0,0)
I have another sequence which contains the index should be modified, like this:
val indices = Seq(1,3,7)
I want to modify the template sequence, in the way that, if the element's index is in the indices sequence, then replace the element to 1.
So the output should look like:
(0,1,0,1,0,0,0,1,0,0)
What is the simplest way to implement this function?
Figured it by myself:
indices.foldLeft(template)((b, a) => b.updated(a, 1))
The foldLeft in your solution repeatedly rebuilds the sequence. This could be inefficient if the sequence is for example a List and all indices point to the end of the list.
You could also achieve it without repeatedly rebuilding the collection:
val indexSet = indices.toSet
val result = template.zipWithIndex.map{ case (v, i) =>
if (indexSet contains i) 1 else v
}
For a List this solution would be amortized O(n), because checking whether an integer is in the set of indices is amortized O(1).
Maybe there are also use-cases where your solution is preferable, it's not so clear for general Seqs.
So i'm trying to make a basic hitori solver, but i am not sure where i should start. I'm still new to Scala.
My first issue is that i'm trying to have an array of some ints (1,2,3,4,2)
and making the program output them like this: (1,2,3,4,B)
notice that the duplicate has become a char B.
Where do i start? Here is what i already did, but didn't do what i excatly need.
val s = lines.split(" ").toSet;
var jetSet = s
for(i<-jetSet){
print(i);
}
One way is to fold over the numbers, left to right, building the Set[Int], for the uniqueness test, and the list of output, as you go along.
val arr = Array(1,2,3,4,2)
arr.foldLeft((Set[Int](),List[String]())){case ((s,l),n) =>
if (s(n)) (s,"B" :: l)
else (s + n, n.toString :: l)
}._2.reverse // res0: List[String] = List(1, 2, 3, 4, B)
From here you can use mkString() to format the output as desired.
What I'd suggest is to break your program into a number of steps and try to solve those.
As a first step you could transform the list into tuples of the numbers and the number of times they have appeared so far ...
(1,2,3,4,2) becomes ((1,1),(2,1),(3,1),(4,1),(2,2)
Next step it's easy to map over this list returning the number if the count is 1 or the letter if it is greater.
That first step is a little bit tricky because as you walk through the list you need to keep track of how many you've seen so far of each letter.
When want to process a sequence and maintain some changing state as you do, you should use a fold. If you're not familiar with fold it has the following signature:
def foldLeft[B](z: B)(op: (B, A) => B): B
Note that the type of z (the initial value) has to match the type of the return value from the fold (B).
So one way to do this would be for type B to be a tuple of (outputList, seensofarCounts)
outputList would accumulate in each step by taking the next number and updating the map of how many of each numbers you've seen so far. "seensofarCounts" would be a map of the numbers and the current count.
So what you get out of the foldLeft is a tuple of (((1,1),(2,1),(3,1),(4,1),(2,2), Map(1 -> 1, 2, 2 ETC ... ))
Now you can map over that first element of the tuple as described above.
Once it's working you could avoid the last step by updating the numbers to letters as you work through the fold.
Usually this technique of breaking things into steps makes it simple to reason about, then when it's working you may see that some steps trivially collapse into each other.
Hope this helps.
I have a tuple like the following:
(Age, List(19,17,11,3,2))
and I would like to get the position of the first element where their position in the list is greater than their value. To do this I tried to use .indexOf() and .indexWhere() but I probably can't find exactly the right syntax and so I keep getting:
value indexWhere is not a member of org.apache.spark.rdd.RDD[(String,
Iterable[Int])]
My code so far is:
val test =("Age", List(19,17,11,3,2))
test.indexWhere(_.2(_)<=_.2(_).indexOf(_.2(_)) )
I also searched the documentation here with no result: http://www.scala-lang.org/api/current/index.html#scala.collection.immutable.List
If you want to perform this for each element in an RDD, you can use RDD's mapValues (which would only map the right-hand-side of the tuple) and pass a function that uses indexWhere:
rdd.mapValues(_.zipWithIndex.indexWhere { case (v, i) => i+1 > v} + 1)
Notes:
Your example seems wrong, if you want the last matching item it should be 5 (position of 2) and not 4
You did not define what should be done when no item matches your condition, e.g. for List(0,0,0) - in this case the result would be 0 but not sure that's what you need
One of my exercises requires me to write a recursive method in which a list is given, and it returns the same list with only every other element on it.
for example : List {"a", "b", "c"} would return
List{"a","c"}
I am writing in scala, and I understand that it has built in library but I am not supposed to use those. I can only use if/else, helper methods,and patterns.
How could I parse thru a list using head and tail only?
so far I have this:
def removeLetter(list:List[String]):List[String]=list match{
case Nil => Nil
case n::rest=>
if (n == rest){ // I understand that this doesn't quite work.
tail
}
else
head::removeLetter(tail)
}
}
I am looking for the logic and not code.
Using pattern matching, you can also deconstruct a list on it's first two elements in the same way you're doing with your n::rest construction. Just remember to also take lists with uneven length into account.
You correctly stated one base-case to the recursion: In case of an empty list, the result is again the empty list. case Nil => Nil
There is a second base-case: A list containing a single element is again the list itself. case x :: Nil => x :: Nil
You can formulate the recursive step as follows: Given a list with at least two elements, the result is a list containing the first element, followed by every other element of the list after the second element. case x :: y :: z => x :: removeLetter(z) (Note that here, x and y are both of type String while z is of type List[String])
Remark: If you wanted to, you could also combine both base-cases, as in both cases, the input to the function is its output.