Re: Is there a smart way to process a range of range by front ?
On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear wrote: On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote: I was thinking to a general *interleave()* algorithm for any compatible Range of Range but I can't find any smart way to process each sub range by front Can you show a sample input and output to clarify what you mean by interleave? It's possible that what you want is std.range.frontTransversal, std.range.transversal, or std.range.transposed. --- auto r0 = [[0,2],[1,3]]; auto r1 = interleave(r0); assert(r1 = [0,1,2,3]); auto r2 = [[0,3],[1,4],[2,5]]; auto r3 = interleave(r2); assert(r3 = [0,1,2,3,4,5]); --- the fact that the numbers are ordered is just an helper.
Re: Is there a smart way to process a range of range by front ?
On Wed, 23 Sep 2015 21:17:27 +, BBasile wrote: > On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear wrote: >> On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote: >> >>> I was thinking to a general *interleave()* algorithm for any >>> compatible Range of Range but I can't find any smart way to process >>> each sub range by front >> >> Can you show a sample input and output to clarify what you mean by >> interleave? It's possible that what you want is >> std.range.frontTransversal, std.range.transversal, or >> std.range.transposed. > > --- > auto r0 = [[0,2],[1,3]]; > auto r1 = interleave(r0); > assert(r1 = [0,1,2,3]); > auto r2 = [[0,3],[1,4],[2,5]]; > auto r3 = interleave(r2); > assert(r3 = [0,1,2,3,4,5]); > --- > > the fact that the numbers are ordered is just an helper. OK, I think what you're after is std.range.roundRobin.
Re: Is there a smart way to process a range of range by front ?
On Wednesday, 23 September 2015 at 21:17:29 UTC, BBasile wrote: On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear wrote: On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote: I was thinking to a general *interleave()* algorithm for any compatible Range of Range but I can't find any smart way to process each sub range by front Can you show a sample input and output to clarify what you mean by interleave? It's possible that what you want is std.range.frontTransversal, std.range.transversal, or std.range.transposed. --- auto r0 = [[0,2],[1,3]]; auto r1 = interleave(r0); assert(r1 = [0,1,2,3]); auto r2 = [[0,3],[1,4],[2,5]]; auto r3 = interleave(r2); assert(r3 = [0,1,2,3,4,5]); --- the fact that the numbers are ordered is just an helper. just imagine that there are double equal symbols in the assertions...
Re: Is there a smart way to process a range of range by front ?
On Wednesday, 23 September 2015 at 21:24:22 UTC, Justin Whear
wrote:
On Wed, 23 Sep 2015 21:17:27 +, BBasile wrote:
On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear
wrote:
On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote:
I was thinking to a general *interleave()* algorithm for any
compatible Range of Range but I can't find any smart way to
process each sub range by front
Can you show a sample input and output to clarify what you
mean by interleave? It's possible that what you want is
std.range.frontTransversal, std.range.transversal, or
std.range.transposed.
---
auto r0 = [[0,2],[1,3]];
auto r1 = interleave(r0);
assert(r1 = [0,1,2,3]);
auto r2 = [[0,3],[1,4],[2,5]];
auto r3 = interleave(r2);
assert(r3 = [0,1,2,3,4,5]);
---
the fact that the numbers are ordered is just an helper.
OK, I think what you're after is std.range.roundRobin.
---
import std.range;
auto interleave(RoR)(RoR r)
{
return r.transposed.join;
}
void main()
{
auto r0 = [[0,2],[1,3]];
auto r1 = interleave(r0);
assert(r1 == [0,1,2,3]);
auto r2 = [[0,3],[1,4],[2,5]];
auto r3 = interleave(r2);
assert(r3 == [0,1,2,3,4,5]);
}
--
thx, but as you was suposing initially 'transposed' works.
didn't know this function before. works fine.
Re: Is there a smart way to process a range of range by front ?
On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote: > I was thinking to a general *interleave()* algorithm for any compatible > Range of Range but I can't find any smart way to process each sub range > by front Can you show a sample input and output to clarify what you mean by interleave? It's possible that what you want is std.range.frontTransversal, std.range.transversal, or std.range.transposed.
Re: Is there a smart way to process a range of range by front ?
On Wednesday, 23 September 2015 at 21:30:37 UTC, BBasile wrote:
auto interleave(RoR)(RoR r)
{
return r.transposed.join;
If you use joiner it will even be lazy and avoid the allocation.
