I'm using Mockito as a part of Specs in scala code and I've stumbled upon the following task:
Given an ArrayBuffer that emulates a chess board (8x8 = 64 cells). If we querying ArrayBuffer for cell that doesn't exist (has number more than 63 or less than 0) we should receive None. Otherwise we returning Some(0) (in almost all cases) or Some(1) (just in few specified cells).
Right now I'm thinking about spies and something that starts like:
val spiedArray = spy(new ArrayBuffer[Int])
for (x <- 1 to 8; y <- 1 to 8) {
doReturn(Some(0)).when(spiedArray).apply(x * y-1)
}
And then explicitly respecify cells with Some(1).
But how about out-of-bound cells that should return None?
Is there a simplest and natural way to achieve that mocking?
The main issue here is that the specification is wrong: an ArrayBuffer cannot work as expected in the spec. Thus you must either:
Change the expected behavior
Change ArrayBuffer for an homemade trait
Related
I have a comparator like this:
lazy val seq = mapping.toSeq.sortWith { case ((_, set1), (_, set2)) =>
// Just propose all the most connected nodes first to the users
// But also allow less connected nodes to pop out sometimes
val popOutChance = random.nextDouble <= 0.1D && set2.size > 5
if (popOutChance) set1.size < set2.size else set1.size > set2.size
}
It is my intention to compare sets sizes such that smaller sets may appear higher in a sorted list with 10% chance.
But compiler does not let me do that and throws an Exception: java.lang.IllegalArgumentException: Comparison method violates its general contract! once I try to use it in runtime. How can I override it?
I think the problem here is that, every time two elements are compared, the outcome is random, thus violating the transitive property required of a comparator function in any sorting algorithm.
For example, let's say that some instance a compares as less than b, and then b compares as less than c. These results should imply that a compares as less than c. However, since your comparisons are stochastic, you can't guarantee that outcome. In fact, you can't even guarantee that a will be less than b next time they're compared.
So don't do that. No sort algorithm can handle it. (Such an approach also violates the referential transparency principle of functional programming and will make your program much harder to reason about.)
Instead, what you need to do is to decorate your map's members with a randomly assigned weighting - before attempting to sort them - so that they can be sorted consistently. However, since this happens at the start of a sort operation, the result of the sort will be different each time, which I think is what you're looking for.
It's not clear what type mapping has in your example, but it appears to be something like: Map[Any, Set[_]]. (You can replace the types as required - it's not that important to this approach. For example, say mapping actually has the type Map[String, Set[SomeClass]], then you would replace references below to Any with String and Set[_] to Set[SomeClass].)
First, we'll create a case class that we'll use to score and compare the map elements. Then we'll map the contents of mapping to a sequence of elements of this case class. Next, we sort those elements. Finally, we extract the tuple from the decorated class. The result should look something like this:
final case class Decorated(x: (Any, Set[_]), rand: Double = random.nextDouble)
extends Ordered[Decorated] {
// Calculate a rank for this element. You'll need to change this to suit your precise
// requirements. Here, if rand is less than 0.1 (a 10% chance), I'm adding 5 to the size;
// otherwise, I'll report the actual size. This allows transitive comparisons, since
// rand doesn't change once defined. Values are negated so bigger sets come to the fore
// when sorted.
private def rank: Int = {
if(rand < 0.1) -(x._2.size + 5)
else -x._2.size
}
// Compare this element with another, by their ranks.
override def compare(that: Decorated): Int = rank.compare(that.rank)
}
// Now sort your mapping elements as follows and convert back to tuples.
lazy val seq = mapping.map(x => Decorated(x)).toSeq.sorted.map(_.x)
This should put the elements with larger sets towards the front, but there's 10% chance that sets appear 5 bigger and so move up the list. The result will be different each time the last line is re-executed, since map will create new random values for each element. However, during sorting, the ranks will be fixed and will not change.
(Note that I'm setting the rank to a negative value. The Ordered[T] trait sorts elements in ascending order, so that - if we sorted purely by set size - smaller sets would come before larger sets. By negating the rank value, sorting will put larger sets before smaller sets. If you don't want this behavior, remove the negations.)
New to Scala. I'm iterating a for loop 100 times. 10 times I want condition 'a' to be met and 90 times condition 'b'. However I want the 10 a's to occur at random.
The best way I can think is to create a val of 10 random integers, then loop through 1 to 100 ints.
For example:
val z = List.fill(10)(100).map(scala.util.Random.nextInt)
z: List[Int] = List(71, 5, 2, 9, 26, 96, 69, 26, 92, 4)
Then something like:
for (i <- 1 to 100) {
whenever i == to a number in z: 'Condition a met: do something'
else {
'condition b met: do something else'
}
}
I tried using contains and == and =! but nothing seemed to work. How else can I do this?
Your generation of random numbers could yield duplicates... is that OK? Here's how you can easily generate 10 unique numbers 1-100 (by generating a randomly shuffled sequence of 1-100 and taking first ten):
val r = scala.util.Random.shuffle(1 to 100).toList.take(10)
Now you can simply partition a range 1-100 into those who are contained in your randomly generated list and those who are not:
val (listOfA, listOfB) = (1 to 100).partition(r.contains(_))
Now do whatever you want with those two lists, e.g.:
println(listOfA.mkString(","))
println(listOfB.mkString(","))
Of course, you can always simply go through the list one by one:
(1 to 100).map {
case i if (r.contains(i)) => println("yes: " + i) // or whatever
case i => println("no: " + i)
}
What you consider to be a simple for-loop actually isn't one. It's a for-comprehension and it's a syntax sugar that de-sugares into chained calls of maps, flatMaps and filters. Yes, it can be used in the same way as you would use the classical for-loop, but this is only because List is in fact a monad. Without going into too much details, if you want to do things the idiomatic Scala way (the "functional" way), you should avoid trying to write classical iterative for loops and prefer getting a collection of your data and then mapping over its elements to perform whatever it is that you need. Note that collections have a really rich library behind them which allows you to invoke cool methods such as partition.
EDIT (for completeness):
Also, you should avoid side-effects, or at least push them as far down the road as possible. I'm talking about the second example from my answer. Let's say you really need to log that stuff (you would be using a logger, but println is good enough for this example). Doing it like this is bad. Btw note that you could use foreach instead of map in that case, because you're not collecting results, just performing the side effects.
Good way would be to compute the needed stuff by modifying each element into an appropriate string. So, calculate the needed strings and accumulate them into results:
val results = (1 to 100).map {
case i if (r.contains(i)) => ("yes: " + i) // or whatever
case i => ("no: " + i)
}
// do whatever with results, e.g. print them
Now results contains a list of a hundred "yes x" and "no x" strings, but you didn't do the ugly thing and perform logging as a side effect in the mapping process. Instead, you mapped each element of the collection into a corresponding string (note that original collection remains intact, so if (1 to 100) was stored in some value, it's still there; mapping creates a new collection) and now you can do whatever you want with it, e.g. pass it on to the logger. Yes, at some point you need to do "the ugly side effect thing" and log the stuff, but at least you will have a special part of code for doing that and you will not be mixing it into your mapping logic which checks if number is contained in the random sequence.
(1 to 100).foreach { x =>
if(z.contains(x)) {
// do something
} else {
// do something else
}
}
or you can use a partial function, like so:
(1 to 100).foreach {
case x if(z.contains(x)) => // do something
case _ => // do something else
}
If I have a custom generator then the shrinker will remember my suchThat clause and not shrink with invalid values:
val myGen = Gen.identifier.suchThat { _.length > 3 }
// all shrinks have > 3 characters
property("failing case") = forAll (myGen) { (a: String) =>
println(s"Gen suchThat Value: $a")
a == "Impossible"
}
If I do something further to the generated value (ie map it) then the shrinker "forgets" my suchThat clause:
// the shrinker will shrink all the way down to ""
property("failing case") = forAll (myGen.map{_ + "bbb"}) { (a: String) =>
println(s"Gen with map Value: $a")
a == "Impossible"
}
Is it possible to have suchThat values propagate through generators. In my real project I am doing more than a simple map but that seems to be the simplest example of the limitation I am hitting.
I'm fairly certain the answer is no (at least at this point in time).
This is quite annoying although perhaps not as trivial as it seems. The generator result does attempt to keep track of the sieve although it gets lost in map and flatMap. Apart from applying the sieve to the result of the shrink there isn't any other connection back to the generator. Even if there were all the intermediate results would need to be retained and applied to each sieve at the correct points. That then raises the question of: What exactly is being shrunk? The generated result or the original generator(s)?
The only solution that I have found so far is to either:
Disable shrinking, or
Implement a custom Shrink, or
Add a whenever clause that rechecks the generated value.
This can be quite challenging, especially when composing multiple generators.
Full disclosure: I am (was?) taking Coursera's Scala course but was stumped by the second assignment on Sets. I'm not looking for just the answers (which are easily obtainable) and would receive marginal credit anyway. But I would really like to understand what is happening.
Okay, so here is the first question: "Define a function which creates a singleton set from one integer value: the set represents the set of the one given element." So my first attempt was this:
def singletonSet(elem: Int): Set = Set(elem)
So this function, singletonSet, just returns a newly created Set. It could be invoked thusly:
val why = singletonSet(3)
// now why is a singleton set with a single integer, 3
This implementation seemed trivial, so I Googled for the answer, which seems to be this:
def singletonSet(elem: Int): Set = (x => x == elem)
Now my understanding is that (x => x == elem) is an anonymous function which takes an integer x and returns a boolean. But... what? So as a JavaScript developer, I decided to translate it:
function singletonSet(elem) {
return function(x) {
return x === elem;
};
};
So then I can write (am I currying?):
singletonSet(3)(4)
// singletonSet(3) => returns an anonymous function, function(x) { x === 3; };
// function(4) { return 4 === 3; }
// false
If this is even close to what is happening in Scala, it seems like I am not creating a singleton set. Rather, I am just checking if two numbers are the same.
What am I missing here? I feel like it must be something very basic.
Thanks in advance.
Remember this implementation of a set is a function. In particular its a boolean function, so the function can just be seen as asking the question: "Is this number in the set? - true or false." The function can be called as many times as you want, in effect asking the question multiple times:
"is this number in the set? Is that number in the set?" etc, etc.
As the set is a singleton set, there is only one number in the set. So you use the set by calling the function, asking the question, in effect, "is this number the one and only number that is in the set?" So you are correct this set, the singleton set is just asking are these two numbers the same.
It should be emphasised that this example is from the course Functional Programming Principles in Scala. The course is not meant as an easy introduction to Scala. In fact the course is deliberately making things difficult, in order to enable a deep understanding of functional programming. Normally one would just use the in scope immutable Set class.
If you wanted to work with say the even numbers between -1000 and 1000, you'd probably use an iterator like:
(-1000 to 1000).withFilter(_ %2 == 0)
or:
(-1000 to 1000 by 2)
So, while working my way through "Scala for the Impatient" I found myself wondering: Can you use a Scala for loop without a sequence?
For example, there is an exercise in the book that asks you to build a counter object that cannot be incremented past Integer.MAX_VALUE. In order to test my solution, I wrote the following code:
var c = new Counter
for( i <- 0 to Integer.MAX_VALUE ) c.increment()
This throws an error: sequences cannot contain more than Int.MaxValue elements.
It seems to me that means that Scala is first allocating and populating a sequence object, with the values 0 through Integer.MaxValue, and then doing a foreach loop on that sequence object.
I realize that I could do this instead:
var c = new Counter
while(c.value < Integer.MAX_VALUE ) c.increment()
But is there any way to do a traditional C-style for loop with the for statement?
In fact, 0 to N does not actually populate anything with integers from 0 to N. It instead creates an instance of scala.collection.immutable.Range, which applies its methods to all the integers generated on the fly.
The error you ran into is only because you have to be able to fit the number of elements (whether they actually exist or not) into the positive part of an Int in order to maintain the contract for the length method. 1 to Int.MaxValue works fine, as does 0 until Int.MaxValue. And the latter is what your while loop is doing anyway (to includes the right endpoint, until omits it).
Anyway, since the Scala for is a very different (much more generic) creature than the C for, the short answer is no, you can't do exactly the same thing. But you can probably do what you want with for (though maybe not as fast as you want, since there is some performance penalty).
Wow, some nice technical answers for a simple question (which is good!) But in case anyone is just looking for a simple answer:
//start from 0, stop at 9 inclusive
for (i <- 0 until 10){
println("Hi " + i)
}
//or start from 0, stop at 9 inclusive
for (i <- 0 to 9){
println("Hi " + i)
}
As Rex pointed out, "to" includes the right endpoint, "until" omits it.
Yes and no, it depends what you are asking for. If you're asking whether you can iterate over a sequence of integers without having to build that sequence first, then yes you can, for instance using streams:
def fromTo(from : Int, to : Int) : Stream[Int] =
if(from > to) {
Stream.empty
} else {
// println("one more.") // uncomment to see when it is called
Stream.cons(from, fromTo(from + 1, to))
}
Then:
for(i <- fromTo(0, 5)) println(i)
Writing your own iterator by defining hasNext and next is another option.
If you're asking whether you can use the 'for' syntax to write a "native" loop, i.e. a loop that works by incrementing some native integer rather than iterating over values produced by an instance of an object, then the answer is, as far as I know, no. As you may know, 'for' comprehensions are syntactic sugar for a combination of calls to flatMap, filter, map and/or foreach (all defined in the FilterMonadic trait), depending on the nesting of generators and their types. You can try to compile some loop and print its compiler intermediate representation with
scalac -Xprint:refchecks
to see how they are expanded.
There's a bunch of these out there, but I can't be bothered googling them at the moment. The following is pretty canonical:
#scala.annotation.tailrec
def loop(from: Int, until: Int)(f: Int => Unit): Unit = {
if (from < until) {
f(from)
loop(from + 1, until)(f)
}
}
loop(0, 10) { i =>
println("Hi " + i)
}