Re: sort struct of arrays
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.
Re: sort struct of arrays
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
Re: sort struct of arrays
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.
Re: sort struct of arrays
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.
Re: sort struct of arrays
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
sort struct of arrays
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?