If you have an array of structs, such as...

     struct Foo
     {
         int x;
         int y;
     }

     Foo[] foos;

...and you wanted to sort the foos then you'd do something like...

     foos.sort!(a.x < b.x),

..and, of course, both of the fields x and y get sorted together.
If you have a so-called struct of arrays, or an equivalent
situation, such as...

     int[] fooX;
     int[] fooY;

...is there a simple way to sort fooX and fooY
"together"/coherently (keyed on, say, fooX), using the standard
lib?

On Friday, 9 May 2014 at 14:23:41 UTC, Luís Marques wrote:
> If you have an array of structs, such as...
>
>      struct Foo
>      {
>          int x;
>          int y;
>      }
>
>      Foo[] foos;
>
> ...and you wanted to sort the foos then you'd do something like...
>
>      foos.sort!(a.x < b.x),
>
> ..and, of course, both of the fields x and y get sorted together.
> If you have a so-called struct of arrays, or an equivalent
> situation, such as...
>
>      int[] fooX;
>      int[] fooY;
>
> ...is there a simple way to sort fooX and fooY
> "together"/coherently (keyed on, say, fooX), using the standard
> lib?

std.range.zip(fooX, fooY).sort!((a, b) => a[0] < b[0]);

I wasn't sure if that's supposed to work. Turns out the
documentation on zip [1] has this exact use case as an example.

[1] http://dlang.org/phobos/std_range.html#zip

On Friday, 9 May 2014 at 14:48:50 UTC, anonymous wrote:
> std.range.zip(fooX, fooY).sort!((a, b) => a[0] < b[0]);
>
> I wasn't sure if that's supposed to work. Turns out the
> documentation on zip [1] has this exact use case as an example.
>
> [1] http://dlang.org/phobos/std_range.html#zip

Ha! Awesome! Sorry that I missed that example.

On Friday, 9 May 2014 at 14:23:41 UTC, Luís Marques wrote:
> If you have an array of structs, such as...
>
>      struct Foo
>      {
>          int x;
>          int y;
>      }
>
>      Foo[] foos;
>
> ...and you wanted to sort the foos then you'd do something like...
>
>      foos.sort!(a.x < b.x),
>
> ..and, of course, both of the fields x and y get sorted together.
> If you have a so-called struct of arrays, or an equivalent
> situation, such as...
>
>      int[] fooX;
>      int[] fooY;
>
> ...is there a simple way to sort fooX and fooY
> "together"/coherently (keyed on, say, fooX), using the standard
> lib?

For some situations (expensive/impossible to move/copy elements of fooY), you would be best with this:

auto indices = zip(iota(fooX.length).array, fooX).sort!"a[1] < b[1]".map!"a[0]";
auto sortedFooY = fooY.indexed(indices);

bearing in mind that this causes an allocation for the index, but if you really can't move the elements of fooY (and fooX isn't already indices of fooY) then you don't have much of a choice.

On Friday, 9 May 2014 at 15:52:51 UTC, John Colvin wrote:
> On Friday, 9 May 2014 at 14:23:41 UTC, Luís Marques wrote:
>> If you have an array of structs, such as...
>>
>>     struct Foo
>>     {
>>         int x;
>>         int y;
>>     }
>>
>>     Foo[] foos;
>>
>> ...and you wanted to sort the foos then you'd do something like...
>>
>>     foos.sort!(a.x < b.x),
>>
>> ..and, of course, both of the fields x and y get sorted together.
>> If you have a so-called struct of arrays, or an equivalent
>> situation, such as...
>>
>>     int[] fooX;
>>     int[] fooY;
>>
>> ...is there a simple way to sort fooX and fooY
>> "together"/coherently (keyed on, say, fooX), using the standard
>> lib?
>
> For some situations (expensive/impossible to move/copy elements of fooY), you would be best with this:
>
> auto indices = zip(iota(fooX.length).array, fooX).sort!"a[1] < b[1]".map!"a[0]";
> auto sortedFooY = fooY.indexed(indices);
>
> bearing in mind that this causes an allocation for the index, but if you really can't move the elements of fooY (and fooX isn't already indices of fooY) then you don't have much of a choice.

It's probably better to use makeIndex:
http://dlang.org/phobos/std_algorithm.html#makeIndex

On Friday, 9 May 2014 at 16:26:22 UTC, Rene Zwanenburg wrote:
> On Friday, 9 May 2014 at 15:52:51 UTC, John Colvin wrote:
>> On Friday, 9 May 2014 at 14:23:41 UTC, Luís Marques wrote:
>>> If you have an array of structs, such as...
>>>
>>>    struct Foo
>>>    {
>>>        int x;
>>>        int y;
>>>    }
>>>
>>>    Foo[] foos;
>>>
>>> ...and you wanted to sort the foos then you'd do something like...
>>>
>>>    foos.sort!(a.x < b.x),
>>>
>>> ..and, of course, both of the fields x and y get sorted together.
>>> If you have a so-called struct of arrays, or an equivalent
>>> situation, such as...
>>>
>>>    int[] fooX;
>>>    int[] fooY;
>>>
>>> ...is there a simple way to sort fooX and fooY
>>> "together"/coherently (keyed on, say, fooX), using the standard
>>> lib?
>>
>> For some situations (expensive/impossible to move/copy elements of fooY), you would be best with this:
>>
>> auto indices = zip(iota(fooX.length).array, fooX).sort!"a[1] < b[1]".map!"a[0]";
>> auto sortedFooY = fooY.indexed(indices);
>>
>> bearing in mind that this causes an allocation for the index, but if you really can't move the elements of fooY (and fooX isn't already indices of fooY) then you don't have much of a choice.
>
> It's probably better to use makeIndex:
> http://dlang.org/phobos/std_algorithm.html#makeIndex

good call, I didn't realise that existed.