Front page | perl.perl6.language.data |
Postings from September 2000
Buddha Buck wrote: > At 12:00 AM 9/12/00 +1100, Jeremy Howard wrote: > >c.soeller@auckland.ac.nz wrote: > >> Reading through the examples left me wondering about some > >> technicalities: > >> > >> > @t[|i;|j] = @a[|j;|i]; # transpose 2-d @a > >> > >> Written like this it would require that @a is exact 2-dim, i.e. it would > >> not just swap the first two dims of any n-dim array? I suppose if I'd > >> want that I'd write > >> > >> @t[|i;|j;] = @a[|j;|i;]; # trailing ';' implies there might be > >> trailing dims > >> > >Not necessary. Since arrays support all the syntax of a plain old list of > >lists, and the |i syntax just creates an implicit loop, the example quoted > >from the RFC will work with other sized arrays. In fact, if it was only 2d, > >it would more properly be: > > > > $t[|i;|j] = $a[|j;|i]; # transpose 2-d @a > <...> > > While something like "$t[|i;|j] = $a[|j;|i];" looks clearly like > scalar-to-scaler assignment of 2-D array elements, something like "@p = > $a[$k;|i]*$b[|i;$k];" isn't so clear. Multiply two scalers to get an > array?!? In truth, it's assigning an array to an array, so there is no > problem except appearances. > I don't think: @p = $a[$k;|i]*$b[|i;$k]; is meaningful. I would have thought that the iterator indexes must appear on _both_ sides of an expression: $p[|i] = $a[$k;|i]*$b[|i;$k]; unless using a reduction: $sum = reduce ^_ + ^_, $a[$k;|i]*$b[|i;$k]; That way it is easy to conceptualise the implicit loop that is being created. However, there is an issue with the width of the implicit loop. In the above expression I want reduce() to be sent the array of values $a[$k;|i]*$b[|i;$k]. However, in: $sum[|i] = reduce ^_ + ^_, $a[$k;|i]*$b[|i;$k]; I want reduce to be sent the list at each iteration of $a[$k;|i]*$b[|i;$k], where @a and @b and both 3d arrays. I have no idea how to do this...Thread Previous | Thread Next