What is the equivalent of Scala groupBy in D? - d

Question

Whenever you want to create a frequency Map in Scala, you can easily call .groupBy on a collection.
val seq = Seq("a", "a", "b", "c", "b", "a")
seq.groupBy(a => a) // Map(b -> List(b, b), a -> List(a, a, a), c -> List(c))
Same is easily done with nested collections.
val nseq = Seq(Seq("a", 1), Seq("a", -1), Seq("b", -5), Seq("c", 100), Seq("b", 5), Seq("a", 0))
nseq.groupBy(a => a(0)) // Map(b -> List(List(b, -5), List(b, 5)), a -> List(List(a, 1), List(a, -1), List(a, 0)), c -> List(List(c, 100)))
Notice how values get aggregated together for each key.
I tried to find analogous function in D and found group. It works somewhat differently since it returns tuple pairs.
int[] arr = [1, 1, 2, 2, 3, 2, 2, 5];
arr.sort.group; // [Tuple!(int, uint)(1, 2), Tuple!(int, uint)(2, 4), Tuple!(int, uint)(3, 1), Tuple!(int, uint)(5, 1)]
arr.sort.group.assocArray; // [5:1, 3:1, 2:4, 1:2]
However, when it comes to nested collections.
Tuple!(string, int)[] arr = [tuple("a", 1), tuple("a", -1), tuple("b", 2), tuple("b", 25), tuple("c", 100), tuple("b", 21)];
arr.sort!("a[0] > b[0]").group!((a, b) => a[0] == b[0]); //(Tuple!(string, int)("c", 100), 1), (Tuple!(string, int)("b", 25), 3), (Tuple!(string, int)("a", 1), 2)]
The value aggregation does not happen and only the first value is taken. But what's the use of taking just one first value? One can of course circumvent this via
int[][string] res;
arr.each!(s => res[s[0]] ~= [s[1]]);
writeln(res) // ["c":[100], "a":[1, -1], "b":[25, 2, 21]]
but is it possible to this in one line without predefining res array?

Answer 1

Tuple!(string, int)[] arr = [tuple("a", 1), tuple("a", -1), tuple("b", 2), tuple("b", 25), tuple("c", 100), tuple("b", 21)];
arr.sort!("a[0] > b[0]").group!((a, b) => a[0] == b[0]); // [Tuple!(string, int)("b", 25):3, Tuple!(string, int)("c", 100):1, Tuple!(string, int)("a", 1):2]
That's not the result I'm getting:
https://run.dlang.io/is/9lVPkv
It results in [tuple(tuple("c", 100), 1), tuple(tuple("b", 25), 3), tuple(tuple("a", 1), 2)], which looks correct.
but is it possible to this in one line without predefining res array?
I think you're looking for chunkBy:
arr.sort!("a[0] > b[0]").chunkBy!((a, b) => a[0] == b[0]).each!writeln;
This results in:
[Tuple!(string, int)("c", 100)]
[Tuple!(string, int)("b", 25), Tuple!(string, int)("b", 2), Tuple!(string, int)("b", 21)]
[Tuple!(string, int)("a", 1), Tuple!(string, int)("a", -1)]