Title says it all. Is there a trivial way to do this?

On Wednesday, 27 June 2018 at 13:27:46 UTC, Uknown wrote:
> Title says it all. Is there a trivial way to do this?

There are
https://dlang.org/library/std/algorithm/mutation/reverse.html
and
https://dlang.org/library/std/range/retro.html

both require a bidirectional range, which Indexed, luckily is.

On Wednesday, 27 June 2018 at 14:21:39 UTC, Alex wrote:
> On Wednesday, 27 June 2018 at 13:27:46 UTC, Uknown wrote:
>> Title says it all. Is there a trivial way to do this?
>
> There are
> https://dlang.org/library/std/algorithm/mutation/reverse.html
> and
> https://dlang.org/library/std/range/retro.html
>
> both require a bidirectional range, which Indexed, luckily is.

I wasn't clear enough. I meant getting back the underlying `Source` range with _its_ elements in the order that the indices specify. This wouldn't be possible in the generic case, but the special case when indices.length == source.length, it should be possible. So indexed(myRange, [2, 3, 5, 1, 4]).sourceWithSwappedElements should return a typeof(myRange) with the elements swapped in that order.

On Wednesday, 27 June 2018 at 14:29:33 UTC, Uknown wrote:
> On Wednesday, 27 June 2018 at 14:21:39 UTC, Alex wrote:
>> On Wednesday, 27 June 2018 at 13:27:46 UTC, Uknown wrote:
>>> Title says it all. Is there a trivial way to do this?
>>
>> There are
>> https://dlang.org/library/std/algorithm/mutation/reverse.html
>> and
>> https://dlang.org/library/std/range/retro.html
>>
>> both require a bidirectional range, which Indexed, luckily is.
>
> I wasn't clear enough. I meant getting back the underlying `Source` range with _its_ elements in the order that the indices specify. This wouldn't be possible in the generic case, but the special case when indices.length == source.length, it should be possible. So indexed(myRange, [2, 3, 5, 1, 4]).sourceWithSwappedElements should return a typeof(myRange) with the elements swapped in that order.

I see. Ok, one possibility is

source = indexed(source, indices).array;

but I assume you want something without extra allocation, right?

On Wednesday, 27 June 2018 at 14:50:25 UTC, Alex wrote:
> On Wednesday, 27 June 2018 at 14:29:33 UTC, Uknown wrote:
>> On Wednesday, 27 June 2018 at 14:21:39 UTC, Alex wrote:
>>> On Wednesday, 27 June 2018 at 13:27:46 UTC, Uknown wrote:
>>>> [...]
>
> I see. Ok, one possibility is
>
> source = indexed(source, indices).array;
>
> but I assume you want something without extra allocation, right?

This doesn't work for some reason.

"123".byCodeUnit.permutations.array.writeln //[123, 123, 123, 123, 123, 123]

On Wednesday, 27 June 2018 at 15:07:57 UTC, Uknown wrote:
> On Wednesday, 27 June 2018 at 14:50:25 UTC, Alex wrote:
>> On Wednesday, 27 June 2018 at 14:29:33 UTC, Uknown wrote:
>>> On Wednesday, 27 June 2018 at 14:21:39 UTC, Alex wrote:
>>>> On Wednesday, 27 June 2018 at 13:27:46 UTC, Uknown wrote:
>>>>> [...]
>>
>> I see. Ok, one possibility is
>>
>> source = indexed(source, indices).array;
>>
>> but I assume you want something without extra allocation, right?
>
> This doesn't work for some reason.
>
> "123".byCodeUnit.permutations.array.writeln //[123, 123, 123, 123, 123, 123]

This would:

["123".byCodeUnit.permutations].joiner.writeln;

On Wednesday, 27 June 2018 at 15:18:05 UTC, Alex wrote:
> On Wednesday, 27 June 2018 at 15:07:57 UTC, Uknown wrote:
>> On Wednesday, 27 June 2018 at 14:50:25 UTC, Alex wrote:
>>> On Wednesday, 27 June 2018 at 14:29:33 UTC, Uknown wrote:
>>>> On Wednesday, 27 June 2018 at 14:21:39 UTC, Alex wrote:
>>>>> On Wednesday, 27 June 2018 at 13:27:46 UTC, Uknown wrote:
>>>>>> [...]
> This would:
>
> ["123".byCodeUnit.permutations].joiner.writeln;

Thanks. This works, but it still seems silly that permutations doesn't just return a range with front returning the original type