I have 3 variables that have already been initialized, and I want to assign a new value to all three of them. e.g.
var a = 1
var b = 2
var c = 3
and I want to reassign them to the same value e.g.
a = b = c = 4
But the above expression is invalid. Is there a right way to do this in Scala?
It is possible to slightly shorten the var definition code as follows
var (a, b, c) = (1, 2, 3)
This works because of extractor objects in scala. A tuple of 3 is extracted into 3 components it was created with.
But following does not work becase the extraction is applied on val or var definitions.
(a, b, c) = (4, 4, 4)
You can do this:
var Seq(a, b, c) = Seq.fill(3)(4)
As with Ivan's solution, this only works when declaring vars or vals, not when reassigning. Since the second parameter is computed for each element, it even works well for mutable objects:
import scala.collection.mutable.ListBuffer
var Seq(a, b, c) = Seq.fill(3)(ListBuffer[Int]())
a += 1 // Only modifies a, not b or c.
By contrast, something like a = b = c = [] will only create one list in most programming languages (e.g. Python, JavaScript, Java). If you don't want to create your object each time (perhaps because it is immutable and creation is expensive), declare it as val first to prevent this behavior:
val largeObject = ???
var Seq(a, b, c) = Seq.fill(3)(largeObject)
Given the following tuple assignment:
let (a, b, c, d) = (Item::new(1), Item::new(10), Item::new(100), Item::new(1000));
Can this be simplified so that the constructor could be removed and made into a macro. e.g.:
let (a, b, c, d) = item_tuple!(1, 10, 100, 1000);
From looking into recursive macros, it seems each macro instantiation needs to create a valid tuple, so the macro would create tuple pairs, e.g.: let (a, (b, (c, d))) = ...; (see this example).
Is it possible to write a macro that expands into a flat-tuple of constructors?
You can accept a variadic argument in your macro and expand it with a call to Item::new() around like so:
macro_rules! item_tuple {
($($arg:expr),*) => {
(
$(Item::new($arg),)*
)
}
}
With this macro this call will work and behave as expected:
let (a, b, c, d) = item_tuple!(1, 10, 100, 1000);
Playground link
I have been looking and I cannot find an example or discussion of the aggregate function in Scala that I can understand. It seems pretty powerful.
Can this function be used to reduce the values of tuples to make a multimap-type collection? For example:
val list = Seq(("one", "i"), ("two", "2"), ("two", "ii"), ("one", "1"), ("four", "iv"))
After applying aggregate:
Seq(("one" -> Seq("i","1")), ("two" -> Seq("2", "ii")), ("four" -> Seq("iv"))
Also, can you give example of parameters z, segop, and combop? I'm unclear on what these parameters do.
Let's see if some ascii art doesn't help. Consider the type signature of aggregate:
def aggregate [B] (z: B)(seqop: (B, A) ⇒ B, combop: (B, B) ⇒ B): B
Also, note that A refers to the type of the collection. So, let's say we have 4 elements in this collection, then aggregate might work like this:
z A z A z A z A
\ / \ /seqop\ / \ /
B B B B
\ / combop \ /
B _ _ B
\ combop /
B
Let's see a practical example of that. Say I have a GenSeq("This", "is", "an", "example"), and I want to know how many characters there are in it. I can write the following:
Note the use of par in the below snippet of code. The second function passed to aggregate is what is called after the individual sequences are computed. Scala is only able to do this for sets that can be parallelized.
import scala.collection.GenSeq
val seq = GenSeq("This", "is", "an", "example")
val chars = seq.par.aggregate(0)(_ + _.length, _ + _)
So, first it would compute this:
0 + "This".length // 4
0 + "is".length // 2
0 + "an".length // 2
0 + "example".length // 7
What it does next cannot be predicted (there are more than one way of combining the results), but it might do this (like in the ascii art above):
4 + 2 // 6
2 + 7 // 9
At which point it concludes with
6 + 9 // 15
which gives the final result. Now, this is a bit similar in structure to foldLeft, but it has an additional function (B, B) => B, which fold doesn't have. This function, however, enables it to work in parallel!
Consider, for example, that each of the four computations initial computations are independent of each other, and can be done in parallel. The next two (resulting in 6 and 9) can be started once their computations on which they depend are finished, but these two can also run in parallel.
The 7 computations, parallelized as above, could take as little as the same time 3 serial computations.
Actually, with such a small collection the cost in synchronizing computation would be big enough to wipe out any gains. Furthermore, if you folded this, it would only take 4 computations total. Once your collections get larger, however, you start to see some real gains.
Consider, on the other hand, foldLeft. Because it doesn't have the additional function, it cannot parallelize any computation:
(((0 + "This".length) + "is".length) + "an".length) + "example".length
Each of the inner parenthesis must be computed before the outer one can proceed.
The aggregate function does not do that (except that it is a very general function, and it could be used to do that). You want groupBy. Close to at least. As you start with a Seq[(String, String)], and you group by taking the first item in the tuple (which is (String, String) => String), it would return a Map[String, Seq[(String, String)]). You then have to discard the first parameter in the Seq[String, String)] values.
So
list.groupBy(_._1).mapValues(_.map(_._2))
There you get a Map[String, Seq[(String, String)]. If you want a Seq instead of Map, call toSeq on the result. I don't think you have a guarantee on the order in the resulting Seq though
Aggregate is a more difficult function.
Consider first reduceLeft and reduceRight.
Let as be a non empty sequence as = Seq(a1, ... an) of elements of type A, and f: (A,A) => A be some way to combine two elements of type A into one. I will note it as a binary operator #, a1 # a2 rather than f(a1, a2). as.reduceLeft(#) will compute (((a1 # a2) # a3)... # an). reduceRight will put the parentheses the other way, (a1 # (a2 #... # an)))). If # happens to be associative, one does not care about the parentheses. One could compute it as (a1 #... # ap) # (ap+1 #...#an) (there would be parantheses inside the 2 big parantheses too, but let's not care about that). Then one could do the two parts in parallel, while the nested bracketing in reduceLeft or reduceRight force a fully sequential computation. But parallel computation is only possible when # is known to be associative, and the reduceLeft method cannot know that.
Still, there could be method reduce, whose caller would be responsible for ensuring that the operation is associative. Then reduce would order the calls as it sees fit, possibly doing them in parallel. Indeed, there is such a method.
There is a limitation with the various reduce methods however. The elements of the Seq can only be combined to a result of the same type: # has to be (A,A) => A. But one could have the more general problem of combining them into a B. One starts with a value b of type B, and combine it with every elements of the sequence. The operator # is (B,A) => B, and one computes (((b # a1) # a2) ... # an). foldLeft does that. foldRight does the same thing but starting with an. There, the # operation has no chance to be associative. When one writes b # a1 # a2, it must mean (b # a1) # a2, as (a1 # a2) would be ill-typed. So foldLeft and foldRight have to be sequential.
Suppose however, that each A can be turned into a B, let's write it with !, a! is of type B. Suppose moreover that there is a + operation (B,B) => B, and that # is such that b # a is in fact b + a!. Rather than combining elements with #, one could first transform all of them to B with !, then combine them with +. That would be as.map(!).reduceLeft(+). And if + is associative, then that can be done with reduce, and not be sequential: as.map(!).reduce(+). There could be an hypothetical method as.associativeFold(b, !, +).
Aggregate is very close to that. It may be however, that there is a more efficient way to implement b#a than b+a! For instance, if type B is List[A], and b#a is a::b, then a! will be a::Nil, and b1 + b2 will be b2 ::: b1. a::b is way better than (a::Nil):::b. To benefit from associativity, but still use #, one first splits b + a1! + ... + an!, into (b + a1! + ap!) + (ap+1! + ..+ an!), then go back to using # with (b # a1 # an) + (ap+1! # # an). One still needs the ! on ap+1, because one must start with some b. And the + is still necessary too, appearing between the parantheses. To do that, as.associativeFold(!, +) could be changed to as.optimizedAssociativeFold(b, !, #, +).
Back to +. + is associative, or equivalently, (B, +) is a semigroup. In practice, most of the semigroups used in programming happen to be monoids too, i.e they contain a neutral element z (for zero) in B, so that for each b, z + b = b + z = b. In that case, the ! operation that make sense is likely to be be a! = z # a. Moreover, as z is a neutral element b # a1 ..# an = (b + z) # a1 # an which is b + (z + a1 # an). So is is always possible to start the aggregation with z. If b is wanted instead, you do b + result at the end. With all those hypotheses, we can do as.aggregate(z, #, +). That is what aggregate does. # is the seqop argument (applied in a sequence z # a1 # a2 # ap), and + is combop (applied to already partially combined results, as in (z + a1#...#ap) + (z + ap+1#...#an)).
To sum it up, as.aggregate(z)(seqop, combop) computes the same thing as as.foldLeft(z)( seqop) provided that
(B, combop, z) is a monoid
seqop(b,a) = combop(b, seqop(z,a))
aggregate implementation may use the associativity of combop to group the computations as it likes (not swapping elements however, + has not to be commutative, ::: is not). It may run them in parallel.
Finally, solving the initial problem using aggregate is left as an exercise to the reader. A hint: implement using foldLeft, then find z and combo that will satisfy the conditions stated above.
The signature for a collection with elements of type A is:
def aggregate [B] (z: B)(seqop: (B, A) ⇒ B, combop: (B, B) ⇒ B): B
z is an object of type B acting as a neutral element. If you want to count something, you can use 0, if you want to build a list, start with an empty list, etc.
segop is analoguous to the function you pass to fold methods. It takes two argument, the first one is the same type as the neutral element you passed and represent the stuff which was already aggregated on previous iteration, the second one is the next element of your collection. The result must also by of type B.
combop: is a function combining two results in one.
In most collections, aggregate is implemented in TraversableOnce as:
def aggregate[B](z: B)(seqop: (B, A) => B, combop: (B, B) => B): B
= foldLeft(z)(seqop)
Thus combop is ignored. However, it makes sense for parallel collections, becauseseqop will first be applied locally in parallel, and then combopis called to finish the aggregation.
So for your example, you can try with a fold first:
val seqOp =
(map:Map[String,Set[String]],tuple: (String,String)) =>
map + ( tuple._1 -> ( map.getOrElse( tuple._1, Set[String]() ) + tuple._2 ) )
list.foldLeft( Map[String,Set[String]]() )( seqOp )
// returns: Map(one -> Set(i, 1), two -> Set(2, ii), four -> Set(iv))
Then you have to find a way of collapsing two multimaps:
val combOp = (map1: Map[String,Set[String]], map2: Map[String,Set[String]]) =>
(map1.keySet ++ map2.keySet).foldLeft( Map[String,Set[String]]() ) {
(result,k) =>
result + ( k -> ( map1.getOrElse(k,Set[String]() ) ++ map2.getOrElse(k,Set[String]() ) ) )
}
Now, you can use aggregate in parallel:
list.par.aggregate( Map[String,Set[String]]() )( seqOp, combOp )
//Returns: Map(one -> Set(i, 1), two -> Set(2, ii), four -> Set(iv))
Applying the method "par" to list, thus using the parallel collection(scala.collection.parallel.immutable.ParSeq) of the list to really take advantage of the multi core processors. Without "par", there won't be any performance gain since the aggregate is not done on the parallel collection.
aggregate is like foldLeft but may executed in parallel.
As missingfactor says, the linear version of aggregate(z)(seqop, combop) is equivalent to foldleft(z)(seqop). This is however impractical in the parallel case, where we would need to combine not only the next element with the previous result (as in a normal fold) but we want to split the iterable into sub-iterables on which we call aggregate and need to combine those again. (In left-to-right order but not associative as we might have combined the last parts before the fist parts of the iterable.) This re-combining in in general non-trivial, and therefore, one needs a method (S, S) => S to accomplish that.
The definition in ParIterableLike is:
def aggregate[S](z: S)(seqop: (S, T) => S, combop: (S, S) => S): S = {
executeAndWaitResult(new Aggregate(z, seqop, combop, splitter))
}
which indeed uses combop.
For reference, Aggregate is defined as:
protected[this] class Aggregate[S](z: S, seqop: (S, T) => S, combop: (S, S) => S, protected[this] val pit: IterableSplitter[T])
extends Accessor[S, Aggregate[S]] {
#volatile var result: S = null.asInstanceOf[S]
def leaf(prevr: Option[S]) = result = pit.foldLeft(z)(seqop)
protected[this] def newSubtask(p: IterableSplitter[T]) = new Aggregate(z, seqop, combop, p)
override def merge(that: Aggregate[S]) = result = combop(result, that.result)
}
The important part is merge where combop is applied with two sub-results.
Here is the blog on how aggregate enable performance on the multi cores processor with bench mark.
http://markusjais.com/scalas-parallel-collections-and-the-aggregate-method/
Here is video on "Scala parallel collections" talk from "Scala Days 2011".
http://days2011.scala-lang.org/node/138/272
The description on the video
Scala Parallel Collections
Aleksandar Prokopec
Parallel programming abstractions become increasingly important as the number of processor cores grows. A high-level programming model enables the programmer to focus more on the program and less on low-level details such as synchronization and load-balancing. Scala parallel collections extend the programming model of the Scala collection framework, providing parallel operations on datasets.
The talk will describe the architecture of the parallel collection framework, explaining their implementation and design decisions. Concrete collection implementations such as parallel hash maps and parallel hash tries will be described. Finally, several example applications will be shown, demonstrating the programming model in practice.
The definition of aggregate in TraversableOnce source is:
def aggregate[B](z: B)(seqop: (B, A) => B, combop: (B, B) => B): B =
foldLeft(z)(seqop)
which is no different than a simple foldLeft. combop doesn't seem to be used anywhere. I am myself confused as to what the purpose of this method is.
Just to clarify explanations of those before me, in theory the idea is that
aggregate should work like this, (I have changed the names of the parameters to make them clearer):
Seq(1,2,3,4).aggragate(0)(
addToPrev = (prev,curr) => prev + curr,
combineSums = (sumA,sumB) => sumA + sumB)
Should logically translate to
Seq(1,2,3,4)
.grouped(2) // split into groups of 2 members each
.map(prevAndCurrList => prevAndCurrList(0) + prevAndCurrList(1))
.foldLeft(0)(sumA,sumB => sumA + sumB)
Because the aggregation and mapping are separate, the original list could theoretically be split into different groups of different sizes and run in parallel or even on different machine.
In practice scala current implementation does not support this feature by default but you can do this in your own code.
In Python
def cross(A, B):
"Cross product of elements in A and elements in B."
return [a+b for a in A for b in B]
returns an one-dimensional array if you call it with two arrays (or strings).
But in CoffeeScript
cross = (A, B) -> (a+b for a in A for b in B)
returns a two-dimensional array.
Do you think it's by design in CoffeeScript or is it a bug?
How do I flatten arrays in CoffeScript?
First I would say say that 2 array comprehensions in line is not a very maintainable pattern. So lets break it down a little.
cross = (A, B) ->
for a in A
for b in B
a+b
alert JSON.stringify(cross [1,2], [3,4])
What's happening here is that the inner creates a closure, which has its own comprehension collector. So it runs all the b's, then returns the results as an array which gets pushed onto the parent comprehension result collector. You are sort of expecting a return value from an inner loop, which is a bit funky.
Instead I would simply collect the results myself.
cross = (A, B) ->
results = []
for a in A
for b in B
results.push a + b
results
alert JSON.stringify(cross [1,2], [3,4])
Or if you still wanted to do some crazy comprehension magic:
cross = (A, B) ->
results = []
results = results.concat a+b for b in B for a in A
results
alert JSON.stringify(cross [1,2], [3,4])
Whether this is a bug in CS or not is a bit debatable, I suppose. But I would argue it's good practice to do more explicit comprehension result handling when dealing with nested iterators.
https://github.com/jashkenas/coffee-script/issues/1191