What is the difference between methods ## and hashCode?
They seem to be outputting the same values no matter which class or hashCode overloading I use. Google doesn't help, either, as it cannot find symbol ##.
"Subclasses" of AnyVal do not behave properly from a hashing perspective:
scala> 1.0.hashCode
res14: Int = 1072693248
Of course this is boxed to a call to:
scala> new java.lang.Double(1.0).hashCode
res16: Int = 1072693248
We might prefer it to be:
scala> new java.lang.Double(1.0).##
res17: Int = 1
scala> 1.0.##
res15: Int = 1
We should expect this given that the int 1 is also the double 1. Of course this issue does not arise in Java. Without it, we'd have this problem:
Set(1.0) contains 1 //compiles but is false
Luckily:
scala> Set(1.0) contains 1
res21: Boolean = true
## was introduced because hashCode is not consistent with the == operator in Scala. If a == b then a.## == b.## regardless of the type of a and b (if custom hashCode implementations are correct). The same is not true for hashCode as can be seen in the examples given by other posters.
Just want to add to the answers of other posters that although the ## method strives to keep the contract between equality and hash codes, it is apparently not good enough in some cases, like when you are comparing doubles and longs (scala 2.10.2):
> import java.lang._
import java.lang._
> val lng = Integer.MAX_VALUE.toLong + 1
lng: Long = 2147483648
> val dbl = Integer.MAX_VALUE.toDouble + 1
dbl: Double = 2.147483648E9
> lng == dbl
res65: Boolean = true
> lng.## == dbl.##
res66: Boolean = false
> (lng.##, lng.hashCode)
res67: (Int, Int) = (-2147483647,-2147483648)
> (dbl.##, dbl.hashCode)
res68: (Int, Int) = (-2147483648,1105199104)
In addition to what everyone else said, I'd like to say that ## is null-safe, because null.## returns 0 whereas null.hashCode throws NullPointerException.
From scaladoc:
Equivalent to x.hashCode except for boxed numeric types and null. For numerics, it returns a hash value which is consistent with value equality: if two value type instances compare as true, then ## will produce the same hash value for each of them. For null returns a hashcode where null.hashCode throws a NullPointerException.
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What is the difference between methods ## and hashCode?
They seem to be outputting the same values no matter which class or hashCode overloading I use. Google doesn't help, either, as it cannot find symbol ##.
"Subclasses" of AnyVal do not behave properly from a hashing perspective:
scala> 1.0.hashCode
res14: Int = 1072693248
Of course this is boxed to a call to:
scala> new java.lang.Double(1.0).hashCode
res16: Int = 1072693248
We might prefer it to be:
scala> new java.lang.Double(1.0).##
res17: Int = 1
scala> 1.0.##
res15: Int = 1
We should expect this given that the int 1 is also the double 1. Of course this issue does not arise in Java. Without it, we'd have this problem:
Set(1.0) contains 1 //compiles but is false
Luckily:
scala> Set(1.0) contains 1
res21: Boolean = true
## was introduced because hashCode is not consistent with the == operator in Scala. If a == b then a.## == b.## regardless of the type of a and b (if custom hashCode implementations are correct). The same is not true for hashCode as can be seen in the examples given by other posters.
Just want to add to the answers of other posters that although the ## method strives to keep the contract between equality and hash codes, it is apparently not good enough in some cases, like when you are comparing doubles and longs (scala 2.10.2):
> import java.lang._
import java.lang._
> val lng = Integer.MAX_VALUE.toLong + 1
lng: Long = 2147483648
> val dbl = Integer.MAX_VALUE.toDouble + 1
dbl: Double = 2.147483648E9
> lng == dbl
res65: Boolean = true
> lng.## == dbl.##
res66: Boolean = false
> (lng.##, lng.hashCode)
res67: (Int, Int) = (-2147483647,-2147483648)
> (dbl.##, dbl.hashCode)
res68: (Int, Int) = (-2147483648,1105199104)
In addition to what everyone else said, I'd like to say that ## is null-safe, because null.## returns 0 whereas null.hashCode throws NullPointerException.
From scaladoc:
Equivalent to x.hashCode except for boxed numeric types and null. For numerics, it returns a hash value which is consistent with value equality: if two value type instances compare as true, then ## will produce the same hash value for each of them. For null returns a hashcode where null.hashCode throws a NullPointerException.
This is the code snippet:
OS.config flatMap {_.Allocation } flatMap {_.memory}
The value obtained from the memory parameter is Option[Long].
How can I compare and check if it is greater than zero?
I tried using filter, but the answer obtained is Option[Boolean]. I need to check for many objects and if the value is greater than zero increment the counter.
You could use the exists method. This method, when run on an option, takes a function that returns a boolean. If your Option is a None, it will return false.
eg.
scala> var x = Some(123)
x: Some[Int] = Some(123)
scala> x.exists(_ > 0)
res0: Boolean = true
scala> x.exists(_ < 0)
res1: Boolean = false
You can fold over it to either get false if it hasn't been specified or check if it's greater than 0:
(OS.config flatMap {_.Allocation } flatMap {_.memory}).fold(false)(_ > 0)
I want to write really nice looking idiomatic Scala code list indexOf foo getOrElse Int.MaxValue but now I have to settle for idiotic Java looking code val result = list indexOf foo; if (result < 0) Int.MaxValue else result. Is there a good reason that indexOf in Scala returns Int instead of Option[Int]
It's for compatibility with Java and for speed. You'd have to double-box the answer (first in java.lang.Integer then Option) instead of just returning a negative number. This can take on the order of ten times longer.
You can always write something that will convert negative numbers to None and non-negative to Some(n) if it bothers you:
implicit class OptionOutNegatives(val underlying: Int) extends AnyVal {
def asIndex = if (underlying < 0) None else Some(underlying)
}
It doesn't have to be like that.
scala> "abcde" index 'c'
res0: psp.std.Index = 2
scala> "abcde" index 'z'
res1: psp.std.Index = -1
scala> "abcde" index 'z' match { case Index(n) => n ; case _ => MaxInt }
res2: Int = 2147483647
// Emphasizing that at the bytecode level we still return an Int - no boxing.
scala> :javap psp.std.SeqLikeExtensionOps
[...]
public abstract int index(A);
descriptor: (Ljava/lang/Object;)I
That's from psp-std, you can run "sbt console" and then the above.
To address the secondary question, squeezed between the primary and tertiary ones, there are other methods for doing things like processing indices and finding or collecting elements that satisfy a predicate.
scala> ('a' to 'z').zipWithIndex find (_._1 == 'k') map (_._2)
res6: Option[Int] = Some(10)
Usually you're doing something interesting with the element you find.
I am desperately trying to find a way to sort a List of strings, where the strings are predefined identifiers of following form: a1.1, a1.2,..., a1.100, a2.1, a2.2,....,a2.100,...,b1.1, b1.2,.. and so on, which is alread the correct ordering. So each identifier is first ordered by its first character (descending alphabetic order) and within this ordering descending ordered by consecutive numbers. I have tried sortWith by providing a sorting function specifying the above rule for all two consecutive list members.
scala> List("a1.102", "b2.2", "b2.1", "a1.1").sortWith((a: String, b: String) => a.take(1) < b.take(1) && a.drop(1).toDouble < b.drop(1).toDouble)
res2: List[java.lang.String] = List(a1.102, a1.1, b2.2, b2.1)
This is not the ordering I expected. However, by swapping the ordering of the expressions, as
scala> List("a1.102", "b2.2", "b2.1", "a1.1").sortWith((a: String, b: String) => (a.drop(1).toDouble < b.drop(1).toDouble && a.take(1) < b.take(2)))
res3: List[java.lang.String] = List(a1.1, a1.102, b2.1, b2.2)
this indeed gives me (at least for this example) the desired ordering, which I do not understand neither.
I would be so thankful, if somebody could give me a hint what exactly is going on there and how I can sort lists as I wish (with a more complex boolean expression than only comparing < or >). A further question: The strings I am sorting (in my example) are actually keys from a HashMap m. Will any solution effect sorting m by its keys within
m.toSeq.sortWith((a: (String, String), b: (String, String)) => a._1.drop(1).toDouble < b._1.drop(1).toDouble && a._1.take(1) < b._1.take(1))
Many thanks in advanced!
Update: I misread your example—you want a1.2 to precede a1.102, which the toDouble versions below won't get right. I'd suggest the following instead:
items.sortBy { s =>
val Array(x, y) = s.tail.split('.')
(s.head, x.toInt, y.toInt)
}
Here we use Scala's Ordering instance for Tuple3[Char, Int, Int].
It looks like you have a typo in your second ("correct") version: b.take(2) should doesn't make sense, and should be b.take(1) to match the first. Once you fix that, you get the same (incorrect) ordering.
The real problem is that you only need the second condition in the case where the numbers match. So the following works as desired:
val items = List("a1.102", "b2.2", "b2.1", "a1.1")
items.sortWith((a, b) =>
a.head < b.head || (a.head == b.head && a.tail.toDouble < b.tail.toDouble)
)
I'd actually suggest the following, though:
items.sortBy(s => s.head -> s.tail.toDouble)
Here we take advantage of the fact that Scala provides an appropriate Ordering instance for Tuple2[Char, Double], so we can just provide a transformation function that turns your items into that type.
And to answer your last question: yes, either of these approaches should work just fine with your Map example.
Create a tuple containing the string before the "." and then the integer after the ".". This will use a lexicographic order for the first part and an order on the integer for the second part.
scala> val order = Ordering.by((s:String) => (s.split("\\.")(0),s.split("\\.")(1).toInt))
order: scala.math.Ordering[String] = scala.math.Ordering$$anon$7#384eb259
scala> res2
res8: List[java.lang.String] = List(a1.5, a2.2, b1.11, b1.8, a1.10)
scala> res2.sorted(order)
res7: List[java.lang.String] = List(a1.5, a1.10, a2.2, b1.8, b1.11)
So consider what happens when your sorting function is passed a="a1.1" and b="a1.102".
What you'd like is for the function to return true. However, a.take(1) < b.take(1) returns false, so the function returns false.
Think about your cases a bit more carefully
if the prefix is equal, and the tails are ordered properly, then the arguments are ordered properly
if the prefixes are not equal, then the arguments are ordered properly only if the prefixes are.
So try this instead:
(a: String, b: String) => if (a.take(1) == b.take(1)) a.drop(1).toDouble < b.drop(1).toDouble else a.take(1) < b.take(1)
And that returns the proper ordering:
scala> List("a1.102", "b2.2", "b2.1", "a1.1").sortWith((a: String, b: String) => if (a.take(1) == b.take(1)) a.drop(1).toDouble < b.drop(1).toDouble else a.take(1) < b.take(1))
res8: List[java.lang.String] = List(a1.1, a1.102, b2.1, b2.2)
The reason it worked for you with the reversed ordering was luck. Consider the extra input "c0" to see what was happening:
scala> List("c0", "a1.102", "b2.2", "b2.1", "a1.1").sortWith((a: String, b: String) => (a.drop(1).toDouble < b.drop(1).toDouble && a.take(1) < b.take(2)))
res1: List[java.lang.String] = List(c0, a1.1, a1.102, b2.1, b2.2)
The reversed function sorts on the numeric part of the string first, then on the prefix. It just so happens that your numeric ordering you gave also preserved the prefix ordering, but that won't always be the case.
I wanted to compare the cardinality of two sets in Scala. Since stuff sometimes "just work" in Scala, I tried using < between the sets. It seems to go through, but I can't make any sense out of the result.
Example:
scala> Set(1,2,3) < Set(1,4)
res20: Boolean = true
What does it return?
Where can I read about this method in the API?
Why isn't it listed anywhere under scala.collection.immutable.Set?
Update: Even the order(??) of the elements in the sets seem to matter:
scala> Set(2,3,1) < Set(1,3)
res24: Boolean = false
scala> Set(1,2,3) < Set(1,3)
res25: Boolean = true
This doesn't work with 2.8. On Scala 2.7, what happens is this:
scala.Predef.iterable2ordered(Set(1, 2, 3): Iterable[Int]) < (Set(1, 3, 2): Iterable[Int])
In other words, there's an implicit conversion defined on scala.Predef, which is "imported" for all Scala code, from an Iterable[A] to an Ordered[Iterable[A]], provided there's an implicit A => Ordered[A] available.
Given that the order of an iterable for sets is undefined, you can't really predict much about it. If you add elements to make the set size bigger than four, for instance, you'll get entirely different results.
If you want to compare the cardinality, just do so directly:
scala> Set(1, 2, 3).size < Set(2, 3, 4, 5).size
res0: Boolean = true
My knowledge of Scala is not extensive, but doing some test, I get the following:
scala> Set(1,2) <
<console>:5: error: missing arguments for method < in trait Ordered;
follow this method with `_' if you want to treat it as a partially applied function
Set(1,2) <
^
That tells me that < comes from the trait Ordered. More hints:
scala> Set(1,2) < _
res4: (Iterable[Int]) => Boolean = <function>
That is, the Set is evaluated into an Iterable, because maybe there is some implicit conversion from Iterable[A] to Ordered[Iterable[A]], but I'm not sure anymore... Tests are not consistent. For example, these two might suggest a kind of lexicographical compare:
scala> Set(1,2,3) < Set(1,2,4)
res5: Boolean = true
1 is equal, 2 is equal, 3 is less than 4.
scala> Set(1,2,4) < Set(1,2,3)
res6: Boolean = false
But these ones don't:
scala> Set(2,1) < Set(2,4)
res11: Boolean = true
scala> Set(2,1) < Set(2,2)
res12: Boolean = false
I think the correct answer is that found in the Ordered trait proper: There is no implementation for < between sets more than comparing their hashCode:
It is important that the hashCode method for an instance of Ordered[A] be consistent with the compare method. However, it is not possible to provide a sensible default implementation. Therefore, if you need to be able compute the hash of an instance of Ordered[A] you must provide it yourself either when inheiriting or instantiating.