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.
Is there a smart way to process a range of range by front ?
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, eg:
---
void interleave(RoR)(RoR r)
{
r.each!(a => a.writeln);
}
void main()
{
auto r = [[0,2],[1,3]];
interleave(r);
}
---
will print:
[0,2]
[1,3]
while to interleave i need to take the front of each sub range
before poping each ror element.
Currently I'm here (don't run this ;)) :
---
auto interleave(RoR)(RoR r)
{
alias T = ElementType!r[0];
T[] result;
while (!empty(r[0]))
r.each!(a => (result ~= a.front, a.popFront));
return result;
}
void main()
{
auto r = [[0,2],[1,3]];
interleave(r);
}
---
but it doesn't work because 'a' is not consumed. It looks like
it's saved from the input parameter at each iteration of the
while loop hence it never returns.
Is it possible ?
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.
