I currently have a value of result that is a string which represents cycles in a graph
> scala result
String =
0:0->52->22;
5:5->70->77;
8:8->66->24;8->42->32;
. //
. // trimmed to get by point across
. //
71:71->40->45;
77:77->34->28;77->5->70;
84:84->22->29
However, I want to have the output have the numbers in between be included and up to a certain value included. The example code would have value = 90
0:0->52->22;
1:
2:
3:
4:
5:5->70->77;
6:
7:
8:8->66->24;8->42->32;
. //
. // trimmed
. //
83:
84:84->22->29;
85:
86:
87:
88:
89:
90:
If it helps or makes any difference, this value is changed to a list for later purposes, such like
list_result = result.split("\n").toList
List[String] = List(0:0->52->22;, 5:5->70->77;, 8:8->66->24;8->42->32;, 11:11->26->66;11->17->66;
My initial thought was to insert the missing numbers into the list and then sort it, but I had trouble with the sorting so I instead look here for a better method.
Turn your list_result into a Map with default values. Then walk through the desired number range, exchanging each for its Map value.
val map_result: Map[String,List[String]] =
list_result.groupBy("\\d+:".r.findFirstIn(_).getOrElse("bad"))
.withDefault(List(_))
val full_result: String =
(0 to 90).flatMap(n => map_result(s"$n:")).mkString("\n")
Here's a Scastie session to see it in action.
One option would be to use a Map as an intermediate data structure:
val l: List[String] = List("0:0->52->22;", "5:5->70->77;", "8:8->66->24;8->42->32;", "11:11->26->66;11->17->66;")
val byKey: List[Array[String]] = l.map(_.split(":"))
val stop = 90
val mapOfValues = (1 to stop).map(_->"").toMap
val output = byKey.foldLeft(mapOfValues)((acc, nxt) => acc + (nxt.head.toInt -> nxt.tail.head))
output.toList.sorted.map {case (key, value) => println(s"$key, $value")}
This will give you the output you are after. It breaks your input strings into pseudo key-value pairs, creates a map to hold the results, inserts the elements of byKey into the map, then returns a sorted list of the results.
Note: If you are using this in anything like production code you'd need to properly check that each Array in byKey does have two elements to prevent any nullPointerExceptions with the later calls to head and tail.head.
The provided solutions are fine, but I would like to suggest one that can process the data lazily and doesn't need to keep all data in memory at once.
It uses a nice function called unfold, which allows to "unfold" a collection from a starting state, up to a point where you deem the collection to be over (docs).
It's not perfectly polished but I hope it may help:
def readLines(s: String): Iterator[String] =
util.Using.resource(io.Source.fromString(s))(_.getLines)
def emptyLines(from: Int, until: Int): Iterator[(String)] =
Iterator.range(from, until).map(n => s"$n:")
def indexOf(line: String): Int =
Integer.parseInt(line.substring(0, line.indexOf(':')))
def withDefaults(from: Int, to: Int, it: Iterator[String]): Iterator[String] = {
Iterator.unfold((from, it)) { case (n, lines) =>
if (lines.hasNext) {
val next = lines.next()
val i = indexOf(next)
Some((emptyLines(n, i) ++ Iterator.single(next), (i + 1, lines)))
} else if (n < to) {
Some((emptyLines(n, to + 1), (to, lines)))
} else {
None
}
}.flatten
}
You can see this in action here on Scastie.
What unfold does is start from a state (in this case, the line number from and the iterator with the lines) and at every iteration:
if there are still elements in the iterator it gets the next item, identifies its index and returns:
as the next item an Iterator with empty lines up to the latest line number followed by the actual line
e.g. when 5 is reached the empty lines between 1 and 4 are emitted, terminated by the line starting with 5
as the next state, the index of the line after the last in the emitted item and the iterator itself (which, being stateful, is consumed by the repeated calls to unfold at each iteration)
e.g. after processing 5, the next state is 6 and the iterator
if there are no elements in the iterator anymore but the to index has not been reached, it emits another Iterator with the remaining items to be printed (in your example, those after 84)
if both conditions are false we don't need to emit anything anymore and we can close the "unfolding" collection, signalling this by returning a None instead of Some[(Item, State)]
This returns an Iterator[Iterator[String]] where every nested iterator is a range of values from one line to the next, with the default empty lines "sandwiched" in between. The call to flatten turns it into the desired result.
I used an Iterator to make sure that only the essential state is kept in memory at any time and only when it's actually used.
Related
I have tried to get multiple user inputs to print them in Scala IDE.
I have tried the this piece of code
println(scala.io.StdIn.readLine())
which works, as the IDE takes my input and then print it in the line but this works only for a single input.
I want the code to take multiple inputs till only newline is entered. example,
1
2
3
so i decided we needed an iterator for the input, which led me to try the following 2 lines of code seperately
var in = Iterator.continually{ scala.io.StdIn.readLine() }.takeWhile { x => x != null}
and
var in = io.Source.stdin.getLines().takeWhile { x => x != null}
Unfortunately none of them worked as the IDE is not taking my input at all.
You're really close.
val in = Iterator.continually(io.StdIn.readLine).takeWhile(_.nonEmpty).toList
This will read input until an empty string is entered and saves the input in a List[String]. The reason for toList is because an Iterator element doesn't become real until next is called on it, so readLine won't be called until the next element is required. The transition to List creates all the elements of the Iterator.
update
As #vossad01 has pointed out, this can be made safer for unexpected input.
val in = Iterator.continually(io.StdIn.readLine)
.takeWhile(Option(_).fold(false)(_.nonEmpty))
.toList
3 tuples in a list
val l = List(("a","b"),("c","d"),("e","f"))
choice one element from each tuple then return this 3 letters word every time
for example: fca or afd or cbf ...
how to realize it
the same as:
echo {a,b}{c,d}{e,f}|xargs -n1|shuf -n1|sed 's/\B/\n/g'|shuf|paste -sd ''
Working with tuples can be a bit of a pain. You can't easily index them and tuples of different sizes are considered different types in the type system.
val ts = List(("a","b"),("c","d"),("e","f"))
val str = ts.map{t =>
t.productElement(util.Random.nextInt(t.productArity))
}.mkString("")
Every time I run this I get a different result: bde, acf, bdf, etc.
I am trying to build a list in scala that given input (length,and a function) the output would be a list from 0 up to that length-1.
for example:
listMaker(3,f) = List(0,1,2)
so far I have created a helper class that takes 2 int and returns a list in that range.
the listMaker function is as follows:
def listMaker[A](length:Int, f:Int =>A):List[A] = length match{
case 0 => List()
case _ => listMaker(length,f)
}
my f function just takes a variable x and returns that:
def f(x:Int)=x
the comment below makes sense, but it still gets me errors. I think the edited code is an easier way to get where I would like to
However, now I get an infinite loop. What part of the logic am I missing?
A recursive function typically has to gradually "bite off" pieces of the input data until there is nothing left - otherwise it can never terminate.
What this means in your particular case is that length must decrease on each recursive call until it reaches zero.
def listMaker[A](length:Int, f:Int =>A):List[A] = length match{
case 0 => List()
case _ => listMaker(length,f)
}
But you are not reducing length - you are passing it unchanged to the next recursive call, so, your function cannot terminate.
(There are other problems too - you need to build up your result list as you recurse, but your current code simply returns an empty list. I assume this is a learning exercise, so I'm not supplying working code...).
I'm trying to create a map which goes through all the ngrams in a document and counts how often they appear. Ngrams are sets of n consecutive words in a sentence (so in the last sentence, (Ngrams, are) is a 2-gram, (are, sets) is the next 2-gram, and so on). I already have code that creates a document from a file and parses it into sentences. I also have a function to count the ngrams in a sentence, ngramsInSentence, which returns Seq[Ngram].
I'm getting stuck syntactically on how to create my counts map. I am iterating through all the ngrams in the document in the for loop, but don't know how to map the ngrams to the count of how often they occur. I'm fairly new to Scala and the syntax is evading me, although I'm clear conceptually on what I need!
def getNGramCounts(document: Document, n: Int): Counts = {
for (sentence <- document.sentences; ngram <- nGramsInSentence(sentence,n))
//I need code here to map ngram -> count how many times ngram appears in document
}
The type Counts above, as well as Ngram, are defined as:
type Counts = Map[NGram, Double]
type NGram = Seq[String]
Does anyone know the syntax to map the ngrams from the for loop to a count of how often they occur? Please let me know if you'd like more details on the problem.
If I'm correctly interpreting your code, this is a fairly common task.
def getNGramCounts(document: Document, n: Int): Counts = {
val allNGrams: Seq[NGram] = for {
sentence <- document.sentences
ngram <- nGramsInSentence(sentence, n)
} yield ngram
allNgrams.groupBy(identity).mapValues(_.size.toDouble)
}
The allNGrams variable collects a list of all the NGrams appearing in the document.
You should eventually turn to Streams if the document is big and you can't hold the whole sequence in memory.
The following groupBycreates a Map[NGram, List[NGram]] which groups your values by its identity (the argument to the method defines the criteria for "aggregate identification") and groups the corresponding values in a list.
You then only need to map the values (the List[NGram]) to its size to get how many recurring values there were of each NGram.
I took for granted that:
NGram has the expected correct implementation of equals + hashcode
document.sentences returns a Seq[...]. If not you should expect allNGrams to be of the corresponding collection type.
UPDATED based on the comments
I wrongly assumed that the groupBy(_) would shortcut the input value. Use the identity function instead.
I converted the count to a Double
Appreciate the help - I have the correct code now using the suggestions above. The following returns the desired result:
def getNGramCounts(document: Document, n: Int): Counts = {
val allNGrams: Seq[NGram] = (for(sentence <- document.sentences;
ngram <- ngramsInSentence(sentence,n))
yield ngram)
allNGrams.groupBy(l => l).map(t => (t._1, t._2.length.toDouble))
}
I have worked on python
In python there is a function .pop() which delete the last value in a list and return that
deleted value
ex. x=[1,2,3,4]
x.pop() will return 4
I was wondering is there is a scala equivalent for this function?
If you just wish to retrieve the last value, you can call x.last. This won't remove the last element from the list, however, which is immutable. Instead, you can call x.init to obtain a list consisting of all elements in x except the last one - again, without actually changing x. So:
val lastEl = x.last
val rest = x.init
will give you the last element (lastEl), the list of all bar the last element (rest), and you still also have the original list (x).
There are a lot of different collection types in Scala, each with its own set of supported and/or well performing operations.
In Scala, a List is an immutable cons-cell sequence like in Lisp. Getting the last element is not a well optimised solution (the head element is fast). Similarly Queue and Stack are optimised for retrieving an element and the rest of the structure from one end particularly. You could use either of them if your order is reversed.
Otherwise, Vector is a good performing general structure which is fast both for head and last calls:
val v = Vector(1, 2, 3, 4)
val init :+ last = v // uses pattern matching extractor `:+` to get both init and last
Where last would be the equivalent of your pop operation, and init is the sequence with the last element removed (you can also use dropRight(1) as suggested in the other answers). To just retrieve the last element, use v.last.
I tend to use
val popped :: newList = list
which assigns the first element of the list to popped and the remaining list to newList
The first answer is correct but you can achieve the same doing:
val last = x.last
val rest = x.dropRight(1)
If you're willing to relax your need for immutable structures, there's always Stack and Queue:
val poppable = scala.collection.mutable.Stack[String]("hi", "ho")
val popped = poppable.pop
Similar to Python's ability to pop multiple elements, Queue handles that:
val multiPoppable = scala.collection.mutable.Queue[String]("hi", "ho")
val allPopped = poppable.dequeueAll(_ => true)
If it is mutable.Queue, use dequeue function
/** Returns the first element in the queue, and removes this element
* from the queue.
*
* #throws java.util.NoSuchElementException
* #return the first element of the queue.
*/
def dequeue(): A =
if (isEmpty)
throw new NoSuchElementException("queue empty")
else {
val res = first0.elem
first0 = first0.next
decrementLength()
res
}