Re: Zip with range of ranges

[ARaspiK via Digitalmars-d-learn]

On Sunday, 25 February 2018 at 20:18:27 UTC, Paul Backus wrote:

On Sunday, 25 February 2018 at 16:22:19 UTC, ARaspiK wrote:
Instead of passing std.range.zip a set of ranges as different 
arguments, is it possible to hand the m a range of ranges, and 
get them to zip together each element of every subrange?


`std.range.transposed` does this, but it requires that the 
range of ranges has assignable elements, so it may not work in 
all cases. For example:


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

auto rr1 = [[1, 2, 3], [4, 5, 6]];
rr1.transposed.each!writeln; // Works

auto rr2 = only(only(1, 2, 3), only(4, 5, 6));
rr2.transposed.each!writeln; // Doesn't work


Thank you so much. It works now. I was already receiving a 
forward range, so copying was easy.

Re: Zip with range of ranges

[Paul Backus via Digitalmars-d-learn]

On Sunday, 25 February 2018 at 16:22:19 UTC, ARaspiK wrote:
Instead of passing std.range.zip a set of ranges as different 
arguments, is it possible to hand the m a range of ranges, and 
get them to zip together each element of every subrange?


`std.range.transposed` does this, but it requires that the range 
of ranges has assignable elements, so it may not work in all 
cases. For example:


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

auto rr1 = [[1, 2, 3], [4, 5, 6]];
rr1.transposed.each!writeln; // Works

auto rr2 = only(only(1, 2, 3), only(4, 5, 6));
rr2.transposed.each!writeln; // Doesn't work

Zip with range of ranges

[ARaspiK via Digitalmars-d-learn]

Instead of passing std.range.zip a set of ranges as different 
arguments, is it possible to hand the m a range of ranges, and 
get them to zip together each element of every subrange?