Re: iterating through a range, operating on last few elements at a time

[Timon Gehr via Digitalmars-d-learn]

On 08/14/2015 05:12 AM, H. S. Teoh via Digitalmars-d-learn wrote:

On Fri, Aug 14, 2015 at 02:42:26AM +0000, Laeeth Isharc via Digitalmars-d-learn 
wrote:

I have a range that is an array of structs.  I would like to iterate
through the range, calling a function with the prior k items in the
range up to that point and storing the result of the function in a new
range/array.

what's the best way to do this?  for low fixed k I could use zip with
staggered slices (s[3..$],s[2..$-1],s[1..$-2],s[0..$-3]) and then map.
I can't think of how to do it elegantly.

any thoughts?

Maybe something like this?

        import std.algorithm;
        import std.stdio;
        import std.range;
        
        auto slidingWindow(R)(R range, int k) {
                return iota(k).map!(i => range.save.drop(i))
                              .array
                              .transposed;
        }
        
        void main() {
                auto data = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ];
                writeln(data.slidingWindow(3));
        }

To apply the function to each slice, just write:

        data.slidingWindow(k).map!myFunc ...

I didn't figure out how to eliminate the short slices toward the end,

import std.algorithm;
import std.stdio;
import std.range;

auto transp(RoR)(RoR ror){
    static struct Transp{
        typeof(transposed(ror)) orig;
        alias orig this;
        @property bool empty(){
            return orig.tupleof[0].any!(a=>a.empty);
        }
    }
    return Transp(transposed(ror));
}

auto slidingWindow(R)(R range, int k) {
    return iota(k).map!(i => range.save.drop(i)).array.transp;
}

void main() {
    auto data = iota(1,11).array;
    writeln(data.slidingWindow(3));
}

:o)

but all you need to do is to somehow drop the last (k-1) elements from
the range returned by slidingWindow.