Related
Sorry if this is a stupid question as I am a total beginner. I have a function factors which looks like this:
def factors (n:Int):List[Int] = {
var xs = List[Int]()
for(i <- 2 to (n-1)) {
if(n%i==0) {xs :+ i}
}
return xs
}
However if I do println(factors(10)) I always get List().
What am I doing wrong?
The :+ operation returns a new List, you never assign it to xs.
def factors (n:Int):List[Int] = {
var xs = List[Int]()
for (i <- 2 to (n - 1)) {
if(n%i==0) {xs = xs :+ i}
}
return xs
}
But, you really shouldn't be using var. We don't like them very much in Scala.
Also don't don't don't use return in Scala. It is a much more loaded keyword than you might think. Read about it here
Here is a better way of doing this.
def factors (n:Int): List[Int] =
for {
i <- (2 to (n - 1)).toList
if (n % i) == 0
} yield i
factors(10)
You don't need .toList either but didn't want to mess with your return types. You are welcome to adjust
Working link: https://scastie.scala-lang.org/haGESfhKRxqDdDIpaHXfpw
You can think of this problem as a filtering operation. You start with all the possible factors and you keep the ones where the remainder when dividing the input by that number is 0. The operation that does this in Scala is filter, which keeps values where a particular test is true and removes the others:
def factors(n: Int): List[Int] =
(2 until n).filter(n % _ == 0).toList
To keep the code short I have also used the short form of a function where _ stands for the argument to the function, so n % _ means n divided by the current number that is being tested.
I'm trying to define a Gen[Stream[A]] for an infinite (lazily evaluated) stream of As where each element A can depend on previous elements.
As a minimal case, we can take Gen[Stream[Int]] where the next element is either +1 or +2 of the previous element. For reference here is a haskell implementation:
increasingInts :: Gen [Int]
increasingInts = arbitrary >>= go
where
go seed = do
inc <- choose (1,2)
let next = seed + inc
rest <- go next
return (next : rest)
I have tried Gen.sequence on a Stream[Gen[A]] but got a stackoverflow. I also tried defining the Gen from scratch, but the constructor gen for Gen is private and works with private methods/types.
This attempt also gives a stackoverflow.
def go(seed: Int): Gen[Stream[Int]] =
for {
inc <- Gen.choose(1, 2)
next = seed + inc
rest <- Gen.lzy(go(next))
} yield next #:: rest
val increasingInts: Gen[Stream[Int]] = go(0)
increasingInts(Gen.Parameters.default, Seed.random()).get foreach println
So I'm stuck. Any ideas?
What you want can be achieved with this:
val increasingInts = {
val increments = Gen.choose(1, 2)
val initialSeed = 0
for {
stream <- Gen.infiniteStream(increments)
} yield stream.scanLeft(initialSeed)(_ + _)
}
.scanLeft is like a .foldLeft but retains the intermediate values, thus giving you another Stream.
I've written about scanLeft here: https://www.scalawilliam.com/most-important-streaming-abstraction/
I am using a for comprehension on a stream and I would like to know how many iterations took to get o the final results.
In code:
var count = 0
for {
xs <- xs_generator
x <- xs
count = count + 1 //doesn't work!!
if (x prop)
yield x
}
Is there a way to achieve this?
Edit: If you don't want to return only the first item, but the entire stream of solutions, take a look at the second part.
Edit-2: Shorter version with zipWithIndex appended.
It's not entirely clear what you are attempting to do. To me it seems as if you are trying to find something in a stream of lists, and additionaly save the number of checked elements.
If this is what you want, consider doing something like this:
/** Returns `x` that satisfies predicate `prop`
* as well the the total number of tested `x`s
*/
def findTheX(): (Int, Int) = {
val xs_generator = Stream.from(1).map(a => (1 to a).toList).take(1000)
var count = 0
def prop(x: Int): Boolean = x % 317 == 0
for (xs <- xs_generator; x <- xs) {
count += 1
if (prop(x)) {
return (x, count)
}
}
throw new Exception("No solution exists")
}
println(findTheX())
// prints:
// (317,50403)
Several important points:
Scala's for-comprehension have nothing to do with Python's "yield". Just in case you thought they did: re-read the documentation on for-comprehensions.
There is no built-in syntax for breaking out of for-comprehensions. It's better to wrap it into a function, and then call return. There is also breakable though, but it works with Exceptions.
The function returns the found item and the total count of checked items, therefore the return type is (Int, Int).
The error in the end after the for-comprehension is to ensure that the return type is Nothing <: (Int, Int) instead of Unit, which is not a subtype of (Int, Int).
Think twice when you want to use Stream for such purposes in this way: after generating the first few elements, the Stream holds them in memory. This might lead to "GC-overhead limit exceeded"-errors if the Stream isn't used properly.
Just to emphasize it again: the yield in Scala for-comprehensions is unrelated to Python's yield. Scala has no built-in support for coroutines and generators. You don't need them as often as you might think, but it requires some readjustment.
EDIT
I've re-read your question again. In case that you want an entire stream of solutions together with a counter of how many different xs have been checked, you might use something like that instead:
val xs_generator = Stream.from(1).map(a => (1 to a).toList)
var count = 0
def prop(x: Int): Boolean = x % 317 == 0
val xsWithCounter = for {
xs <- xs_generator;
x <- xs
_ = { count = count + 1 }
if (prop(x))
} yield (x, count)
println(xsWithCounter.take(10).toList)
// prints:
// List(
// (317,50403), (317,50721), (317,51040), (317,51360), (317,51681),
// (317,52003), (317,52326), (317,52650), (317,52975), (317,53301)
// )
Note the _ = { ... } part. There is a limited number of things that can occur in a for-comprehension:
generators (the x <- things)
filters/guards (if-s)
value definitions
Here, we sort-of abuse the value-definition syntax to update the counter. We use the block { counter += 1 } as the right hand side of the assignment. It returns Unit. Since we don't need the result of the block, we use _ as the left hand side of the assignment. In this way, this block is executed once for every x.
EDIT-2
If mutating the counter is not your main goal, you can of course use the zipWithIndex directly:
val xsWithCounter =
xs_generator.flatten.zipWithIndex.filter{x => prop(x._1)}
It gives almost the same result as the previous version, but the indices are shifted by -1 (it's the indices, not the number of tried x-s).
I want to implement a lazy iterator that yields the next element in each call, in a 3-level nested loop.
Is there something similar in scala to this snippet of c#:
foreach (int i in ...)
{
foreach (int j in ...)
{
foreach (int k in ...)
{
yield return do(i,j,k);
}
}
}
Thanks, Dudu
Scala sequence types all have a .view method which produces a lazy equivalent of the collection. You can play around with the following in the REPL (after issuing :silent to stop it from forcing the collection to print command results):
def log[A](a: A) = { println(a); a }
for (i <- 1 to 10) yield log(i)
for (i <- (1 to 10) view) yield log(i)
The first will print out the numbers 1 to 10, the second will not until you actually try to access those elements of the result.
There is nothing in Scala directly equivalent to C#'s yield statement, which pauses the execution of a loop. You can achieve similar effects with the delimited continuations which were added for scala 2.8.
If you join iterators together with ++, you get a single iterator that runs over both. And the reduceLeft method helpfully joins together an entire collection. Thus,
def doIt(i: Int, j: Int, k: Int) = i+j+k
(1 to 2).map(i => {
(1 to 2).map(j => {
(1 to 2).iterator.map(k => doIt(i,j,k))
}).reduceLeft(_ ++ _)
}).reduceLeft(_ ++ _)
will produce the iterator you want. If you want it to be even more lazy than that, you can add .iterator after the first two (1 to 2) also. (Replace each (1 to 2) with your own more interesting collection or range, of course.)
You can use a Sequence Comprehension over Iterators to get what you want:
for {
i <- (1 to 10).iterator
j <- (1 to 10).iterator
k <- (1 to 10).iterator
} yield doFunc(i, j, k)
If you want to create a lazy Iterable (instead of a lazy Iterator) use Views instead:
for {
i <- (1 to 10).view
j <- (1 to 10).view
k <- (1 to 10).view
} yield doFunc(i, j, k)
Depending on how lazy you want to be, you may not need all of the calls to iterator / view.
If your 3 iterators are generally small (i.e., you can fully iterate them without concern for memory or CPU) and the expensive part is computing the result given i, j, and k, you can use Scala's Stream class.
val tuples = for (i <- 1 to 3; j <- 1 to 3; k <- 1 to 3) yield (i, j, k)
val stream = Stream(tuples: _*) map { case (i, j, k) => i + j + k }
stream take 10 foreach println
If your iterators are too large for this approach, you could extend this idea and create a Stream of tuples that calculates the next value lazily by keeping state for each iterator. For example (although hopefully someone has a nicer way of defining the product method):
def product[A, B, C](a: Iterable[A], b: Iterable[B], c: Iterable[C]): Iterator[(A, B, C)] = {
if (a.isEmpty || b.isEmpty || c.isEmpty) Iterator.empty
else new Iterator[(A, B, C)] {
private val aItr = a.iterator
private var bItr = b.iterator
private var cItr = c.iterator
private var aValue: Option[A] = if (aItr.hasNext) Some(aItr.next) else None
private var bValue: Option[B] = if (bItr.hasNext) Some(bItr.next) else None
override def hasNext = cItr.hasNext || bItr.hasNext || aItr.hasNext
override def next = {
if (cItr.hasNext)
(aValue get, bValue get, cItr.next)
else {
cItr = c.iterator
if (bItr.hasNext) {
bValue = Some(bItr.next)
(aValue get, bValue get, cItr.next)
} else {
aValue = Some(aItr.next)
bItr = b.iterator
(aValue get, bValue get, cItr.next)
}
}
}
}
}
val stream = product(1 to 3, 1 to 3, 1 to 3).toStream map { case (i, j, k) => i + j + k }
stream take 10 foreach println
This approach fully supports infinitely sized inputs.
I think the below code is what you're actually looking for... I think the compiler ends up translating it into the equivalent of the map code Rex gave, but is closer to the syntax of your original example:
scala> def doIt(i:Int, j:Int) = { println(i + ","+j); (i,j); }
doIt: (i: Int, j: Int)(Int, Int)
scala> def x = for( i <- (1 to 5).iterator;
j <- (1 to 5).iterator ) yield doIt(i,j)
x: Iterator[(Int, Int)]
scala> x.foreach(print)
1,1
(1,1)1,2
(1,2)1,3
(1,3)1,4
(1,4)1,5
(1,5)2,1
(2,1)2,2
(2,2)2,3
(2,3)2,4
(2,4)2,5
(2,5)3,1
(3,1)3,2
(3,2)3,3
(3,3)3,4
(3,4)3,5
(3,5)4,1
(4,1)4,2
(4,2)4,3
(4,3)4,4
(4,4)4,5
(4,5)5,1
(5,1)5,2
(5,2)5,3
(5,3)5,4
(5,4)5,5
(5,5)
scala>
You can see from the output that the print in "doIt" isn't called until the next value of x is iterated over, and this style of for generator is a bit simpler to read/write than a bunch of nested maps.
Turn the problem upside down. Pass "do" in as a closure. That's the entire point of using a functional language
Iterator.zip will do it:
iterator1.zip(iterator2).zip(iterator3).map(tuple => doSomething(tuple))
Just read the 20 or so first related links that are show on the side (and, indeed, where shown to you when you first wrote the title of your question).
How do I break out a loop?
var largest=0
for(i<-999 to 1 by -1) {
for (j<-i to 1 by -1) {
val product=i*j
if (largest>product)
// I want to break out here
else
if(product.toString.equals(product.toString.reverse))
largest=largest max product
}
}
How do I turn nested for loops into tail recursion?
From Scala Talk at FOSDEM 2009 http://www.slideshare.net/Odersky/fosdem-2009-1013261
on the 22nd page:
Break and continue
Scala does not have them. Why?
They are a bit imperative; better use many smaller functions
Issue how to interact with closures.
They are not needed!
What is the explanation?
You have three (or so) options to break out of loops.
Suppose you want to sum numbers until the total is greater than 1000. You try
var sum = 0
for (i <- 0 to 1000) sum += i
except you want to stop when (sum > 1000).
What to do? There are several options.
(1a) Use some construct that includes a conditional that you test.
var sum = 0
(0 to 1000).iterator.takeWhile(_ => sum < 1000).foreach(i => sum+=i)
(warning--this depends on details of how the takeWhile test and the foreach are interleaved during evaluation, and probably shouldn't be used in practice!).
(1b) Use tail recursion instead of a for loop, taking advantage of how easy it is to write a new method in Scala:
var sum = 0
def addTo(i: Int, max: Int) {
sum += i; if (sum < max) addTo(i+1,max)
}
addTo(0,1000)
(1c) Fall back to using a while loop
var sum = 0
var i = 0
while (i <= 1000 && sum <= 1000) { sum += 1; i += 1 }
(2) Throw an exception.
object AllDone extends Exception { }
var sum = 0
try {
for (i <- 0 to 1000) { sum += i; if (sum>=1000) throw AllDone }
} catch {
case AllDone =>
}
(2a) In Scala 2.8+ this is already pre-packaged in scala.util.control.Breaks using syntax that looks a lot like your familiar old break from C/Java:
import scala.util.control.Breaks._
var sum = 0
breakable { for (i <- 0 to 1000) {
sum += i
if (sum >= 1000) break
} }
(3) Put the code into a method and use return.
var sum = 0
def findSum { for (i <- 0 to 1000) { sum += i; if (sum>=1000) return } }
findSum
This is intentionally made not-too-easy for at least three reasons I can think of. First, in large code blocks, it's easy to overlook "continue" and "break" statements, or to think you're breaking out of more or less than you really are, or to need to break two loops which you can't do easily anyway--so the standard usage, while handy, has its problems, and thus you should try to structure your code a different way. Second, Scala has all sorts of nestings that you probably don't even notice, so if you could break out of things, you'd probably be surprised by where the code flow ended up (especially with closures). Third, most of Scala's "loops" aren't actually normal loops--they're method calls that have their own loop, or they are recursion which may or may not actually be a loop--and although they act looplike, it's hard to come up with a consistent way to know what "break" and the like should do. So, to be consistent, the wiser thing to do is not to have a "break" at all.
Note: There are functional equivalents of all of these where you return the value of sum rather than mutate it in place. These are more idiomatic Scala. However, the logic remains the same. (return becomes return x, etc.).
This has changed in Scala 2.8 which has a mechanism for using breaks. You can now do the following:
import scala.util.control.Breaks._
var largest = 0
// pass a function to the breakable method
breakable {
for (i<-999 to 1 by -1; j <- i to 1 by -1) {
val product = i * j
if (largest > product) {
break // BREAK!!
}
else if (product.toString.equals(product.toString.reverse)) {
largest = largest max product
}
}
}
It is never a good idea to break out of a for-loop. If you are using a for-loop it means that you know how many times you want to iterate. Use a while-loop with 2 conditions.
for example
var done = false
while (i <= length && !done) {
if (sum > 1000) {
done = true
}
}
To add Rex Kerr answer another way:
(1c) You can also use a guard in your loop:
var sum = 0
for (i <- 0 to 1000 ; if sum<1000) sum += i
Simply We can do in scala is
scala> import util.control.Breaks._
scala> object TestBreak {
def main(args : Array[String]) {
breakable {
for (i <- 1 to 10) {
println(i)
if (i == 5)
break;
} } } }
output :
scala> TestBreak.main(Array())
1
2
3
4
5
Since there is no break in Scala yet, you could try to solve this problem with using a return-statement. Therefore you need to put your inner loop into a function, otherwise the return would skip the whole loop.
Scala 2.8 however includes a way to break
http://www.scala-lang.org/api/rc/scala/util/control/Breaks.html
An approach that generates the values over a range as we iterate, up to a breaking condition, instead of generating first a whole range and then iterating over it, using Iterator, (inspired in #RexKerr use of Stream)
var sum = 0
for ( i <- Iterator.from(1).takeWhile( _ => sum < 1000) ) sum += i
// import following package
import scala.util.control._
// create a Breaks object as follows
val loop = new Breaks;
// Keep the loop inside breakable as follows
loop.breakable{
// Loop will go here
for(...){
....
// Break will go here
loop.break;
}
}
use Break module
http://www.tutorialspoint.com/scala/scala_break_statement.htm
Just use a while loop:
var (i, sum) = (0, 0)
while (sum < 1000) {
sum += i
i += 1
}
Here is a tail recursive version. Compared to the for-comprehensions it is a bit cryptic, admittedly, but I'd say its functional :)
def run(start:Int) = {
#tailrec
def tr(i:Int, largest:Int):Int = tr1(i, i, largest) match {
case x if i > 1 => tr(i-1, x)
case _ => largest
}
#tailrec
def tr1(i:Int,j:Int, largest:Int):Int = i*j match {
case x if x < largest || j < 2 => largest
case x if x.toString.equals(x.toString.reverse) => tr1(i, j-1, x)
case _ => tr1(i, j-1, largest)
}
tr(start, 0)
}
As you can see, the tr function is the counterpart of the outer for-comprehensions, and tr1 of the inner one. You're welcome if you know a way to optimize my version.
Close to your solution would be this:
var largest = 0
for (i <- 999 to 1 by -1;
j <- i to 1 by -1;
product = i * j;
if (largest <= product && product.toString.reverse.equals (product.toString.reverse.reverse)))
largest = product
println (largest)
The j-iteration is made without a new scope, and the product-generation as well as the condition are done in the for-statement (not a good expression - I don't find a better one). The condition is reversed which is pretty fast for that problem size - maybe you gain something with a break for larger loops.
String.reverse implicitly converts to RichString, which is why I do 2 extra reverses. :) A more mathematical approach might be more elegant.
I am new to Scala, but how about this to avoid throwing exceptions and repeating methods:
object awhile {
def apply(condition: () => Boolean, action: () => breakwhen): Unit = {
while (condition()) {
action() match {
case breakwhen(true) => return ;
case _ => { };
}
}
}
case class breakwhen(break:Boolean);
use it like this:
var i = 0
awhile(() => i < 20, () => {
i = i + 1
breakwhen(i == 5)
});
println(i)
if you don’t want to break:
awhile(() => i < 20, () => {
i = i + 1
breakwhen(false)
});
The third-party breakable package is one possible alternative
https://github.com/erikerlandson/breakable
Example code:
scala> import com.manyangled.breakable._
import com.manyangled.breakable._
scala> val bkb2 = for {
| (x, xLab) <- Stream.from(0).breakable // create breakable sequence with a method
| (y, yLab) <- breakable(Stream.from(0)) // create with a function
| if (x % 2 == 1) continue(xLab) // continue to next in outer "x" loop
| if (y % 2 == 0) continue(yLab) // continue to next in inner "y" loop
| if (x > 10) break(xLab) // break the outer "x" loop
| if (y > x) break(yLab) // break the inner "y" loop
| } yield (x, y)
bkb2: com.manyangled.breakable.Breakable[(Int, Int)] = com.manyangled.breakable.Breakable#34dc53d2
scala> bkb2.toVector
res0: Vector[(Int, Int)] = Vector((2,1), (4,1), (4,3), (6,1), (6,3), (6,5), (8,1), (8,3), (8,5), (8,7), (10,1), (10,3), (10,5), (10,7), (10,9))
import scala.util.control._
object demo_brk_963
{
def main(args: Array[String])
{
var a = 0;
var b = 0;
val numList1 = List(1,2,3,4,5,6,7,8,9,10);
val numList2 = List(11,12,13);
val outer = new Breaks; //object for break
val inner = new Breaks; //object for break
outer.breakable // Outer Block
{
for( a <- numList1)
{
println( "Value of a: " + a);
inner.breakable // Inner Block
{
for( b <- numList2)
{
println( "Value of b: " + b);
if( b == 12 )
{
println( "break-INNER;");
inner.break;
}
}
} // inner breakable
if( a == 6 )
{
println( "break-OUTER;");
outer.break;
}
}
} // outer breakable.
}
}
Basic method to break the loop, using Breaks class.
By declaring the loop as breakable.
Ironically the Scala break in scala.util.control.Breaks is an exception:
def break(): Nothing = { throw breakException }
The best advice is: DO NOT use break, continue and goto! IMO they are the same, bad practice and an evil source of all kind of problems (and hot discussions) and finally "considered be harmful". Code block structured, also in this example breaks are superfluous.
Our Edsger W. Dijkstra† wrote:
The quality of programmers is a decreasing function of the density of go to statements in the programs they produce.
I got a situation like the code below
for(id<-0 to 99) {
try {
var symbol = ctx.read("$.stocks[" + id + "].symbol").toString
var name = ctx.read("$.stocks[" + id + "].name").toString
stocklist(symbol) = name
}catch {
case ex: com.jayway.jsonpath.PathNotFoundException=>{break}
}
}
I am using a java lib and the mechanism is that ctx.read throw a Exception when it can find nothing.
I was trapped in the situation that :I have to break the loop when a Exception was thrown, but scala.util.control.Breaks.break using Exception to break the loop ,and it was in the catch block thus it was caught.
I got ugly way to solve this: do the loop for the first time and get the count of the real length.
and use it for the second loop.
take out break from Scala is not that good,when you are using some java libs.
Clever use of find method for collection will do the trick for you.
var largest = 0
lazy val ij =
for (i <- 999 to 1 by -1; j <- i to 1 by -1) yield (i, j)
val largest_ij = ij.find { case(i,j) =>
val product = i * j
if (product.toString == product.toString.reverse)
largest = largest max product
largest > product
}
println(largest_ij.get)
println(largest)
Below is code to break a loop in a simple way
import scala.util.control.Breaks.break
object RecurringCharacter {
def main(args: Array[String]) {
val str = "nileshshinde";
for (i <- 0 to str.length() - 1) {
for (j <- i + 1 to str.length() - 1) {
if (str(i) == str(j)) {
println("First Repeted Character " + str(i))
break() //break method will exit the loop with an Exception "Exception in thread "main" scala.util.control.BreakControl"
}
}
}
}
}
I don't know how much Scala style has changed in the past 9 years, but I found it interesting that most of the existing answers use vars, or hard to read recursion. The key to exiting early is to use a lazy collection to generate your possible candidates, then check for the condition separately. To generate the products:
val products = for {
i <- (999 to 1 by -1).view
j <- (i to 1 by -1).view
} yield (i*j)
Then to find the first palindrome from that view without generating every combination:
val palindromes = products filter {p => p.toString == p.toString.reverse}
palindromes.head
To find the largest palindrome (although the laziness doesn't buy you much because you have to check the entire list anyway):
palindromes.max
Your original code is actually checking for the first palindrome that is larger than a subsequent product, which is the same as checking for the first palindrome except in a weird boundary condition which I don't think you intended. The products are not strictly monotonically decreasing. For example, 998*998 is greater than 999*997, but appears much later in the loops.
Anyway, the advantage of the separated lazy generation and condition check is you write it pretty much like it is using the entire list, but it only generates as much as you need. You sort of get the best of both worlds.