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.
>