Reputation: 26550
Consider the simple problem of using a mutable map to keep track of occurrences/counts, i.e. with:
val counts = collection.mutable.Map[SomeKeyType, Int]()
My current approach to incrementing a count is:
counts(key) = counts.getOrElse(key, 0) + 1
// or equivalently
counts.update(key, counts.getOrElse(key, 0) + 1)
This somehow feels a bit clumsy, because I have to specify the key twice. In terms of performance, I would also expect that key
has to be located twice in the map, which I would like to avoid. Interestingly, this access and update problem would not occur if Int
would provide some mechanism to modify itself. Changing from Int
to a Counter
class that provides an increment
function would for instance allow:
// not possible with Int
counts.getOrElseUpdate(key, 0) += 1
// but with a modifiable counter
counts.getOrElseUpdate(key, new Counter).increment
Somehow I'm always expecting to have the following functionality with a mutable map (somewhat similar to transform
but without returning a new collection and on a specific key with a default value):
// fictitious use
counts.updateOrElse(key, 0, _ + 1)
// or alternatively
counts.getOrElseUpdate(key, 0).modify(_ + 1)
However as far as I can see, such a functionality does not exist. Wouldn't it make sense in general (performance and syntax wise) to have such a f: A => A
in-place modification possibility? Probably I'm just missing something here... I guess there must be some better solution to this problem making such a functionality unnecessary?
Update:
I should have clarified that I'm aware of withDefaultValue
but the problem remains the same: performing two lookups is still twice as slow than one, no matter if it is a O(1) operation or not. Frankly, in many situations I would be more than happy to achieve a speed-up of factor 2. And obviously the construction of the modification closure can often be moved outside of the loop, so imho this is not a big issue compared to running an operation unnecessarily twice.
Upvotes: 27
Views: 18699
Reputation: 61666
Starting Scala 2.13
, Map#updateWith
serves this exact purpose:
map.updateWith("a")({
case Some(count) => Some(count + 1)
case None => Some(1)
})
def updateWith(key: K)(remappingFunction: (Option[V]) => Option[V]): Option[V]
For instance, if the key doesn't exist:
val map = collection.mutable.Map[String, Int]()
// map: collection.mutable.Map[String, Int] = HashMap()
map.updateWith("a")({ case Some(count) => Some(count + 1) case None => Some(1) })
// Option[Int] = Some(1)
map
// collection.mutable.Map[String, Int] = HashMap("a" -> 1)
and if the key exists:
map.updateWith("a")({ case Some(count) => Some(count + 1) case None => Some(1) })
// Option[Int] = Some(2)
map
// collection.mutable.Map[String, Int] = HashMap("a" -> 2)
Upvotes: 6
Reputation: 3343
I wanted to lazy-initialise my mutable map instead of doing a fold (for memory efficiency). The collection.mutable.Map.getOrElseUpdate() method suited my purposes. My map contained a mutable object for summing values (again, for efficiency).
val accum = accums.getOrElseUpdate(key, new Accum)
accum.add(elem.getHours, elem.getCount)
collection.mutable.Map.withDefaultValue() does not keep the default value for a subsequent requested key.
Upvotes: 1
Reputation: 1480
You could create the map with a default value, which would allow you to do the following:
scala> val m = collection.mutable.Map[String, Int]().withDefaultValue(0)
m: scala.collection.mutable.Map[String,Int] = Map()
scala> m.update("a", m("a") + 1)
scala> m
res6: scala.collection.mutable.Map[String,Int] = Map(a -> 1)
As Impredicative mentioned, map lookups are fast so I wouldn't worry about 2 lookups.
Update:
As Debilski pointed out you can do this even more simply by doing the following:
scala> val m = collection.mutable.Map[String, Int]().withDefaultValue(0)
scala> m("a") += 1
scala> m
res6: scala.collection.mutable.Map[String,Int] = Map(a -> 1)
Upvotes: 29