HI,
thanks for all the answers, and the insights into J and sorting.
The version applicable to my sorting needs is indeed the classic primary
key, secondary key sort.
And the R.E. Boss version is going into my toolbox.
Jimmy
On Mon, Jan 21, 2019 at 12:10 PM Roger Hui <rogerhui.can...@gmail.com>
wrote:
> It's is possible that we have a different interpretation of what is the
> problem to be solved. My interpretation is the classic definition of
> lexicographic ordering. First, you order by column 0, resulting in groups
> of rows with identical values in column 0. Then, within each group, you
> order by column 1, getting subgroups with identical values in columns 0 and
> 1. Then, within each subgroup, you order by column 1+k, getting
> subsubgroups with identical values in columns i.k. The "twist" is that
> each ordering in the process can be ascending or descending, to be
> specified separately for each column. I have seen system sort routines
> which deliver exactly this (ahem) sort of functionality.
>
> Therefore, for x=. ,/^:2 >{3#<i.3, the transposed result
>
> |: 1 1 _1 (] /: *"1 ) x
> 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
> 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2
> 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0
>
> is exactly what I intended and consistent with my interpretation of the
> problem.
>
> It's possible that your interpretation is what was asked in the original
> question posed by Jimmy Gauvin. No matter, I find my version an
> interesting problem and that's the one I am solving.
>
>
> On Mon, Jan 21, 2019 at 4:36 AM R.E. Boss <r.e.b...@outlook.com> wrote:
>
> > >
> > > On Sat, Jan 19, 2019 at 8:40 AM Roger Hui <rogerhui.can...@gmail.com>
> > > wrote:
> > >
> > > > If the keys are numeric you can multiply the ascending column by 1
> and
> > > > the descending one by _1, and then apply /: .
> > > >
> >
> > It is not enough to multiply with 1 or _1 for ascending or descending
> > columns. You might have to reorder the columns as well.
> >
> > Suppose the 3-column matrix x with
> > |:x=. ,/^:2 >{3#<i.3
> > 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
> > 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2
> > 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2
> > has to be sorted ascending by column 0, descending by column 2 and
> > ascending by column 1, in that order.
> >
> > Multiplying columns by 1 1 _1 and sorting would deliver
> > |:1 1 _1( ]/: *"1 ) x
> > 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
> > 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2
> > 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0
> >
> > Whereas this is (the transposed of) what we want
> > |: 1 1 _1( ]/: *"1)&.:(0 2 1&C."1) x
> > 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
> > 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2
> > 2 2 2 1 1 1 0 0 0 2 2 2 1 1 1 0 0 0 2 2 2 1 1 1 0 0 0
> >
> >
> > R.E. Boss
> > ----------------------------------------------------------------------
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> ----------------------------------------------------------------------
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