If it's a general approach and (more importantly :-) it solves your
problem, then take the general solution and simplify it for your specific
problem.



On Tue, Sep 11, 2012 at 10:49 PM, June Kim (김창준) <junea...@gmail.com> wrote:

> Hi Roger,
>
> Thanks, Roger, as always. I am aware of that wiki page, but I think it's a
> more general approach. How could I use that to make removen simpler and
> more elegant?
>
> Best regards
>
> June
>
> On Tue, Sep 11, 2012 at 3:42 PM, Roger Hui <rogerhui.can...@gmail.com
> >wrote:
>
> > You may want to look at
> > http://www.jsoftware.com/jwiki/Essays/Progressive%20Index-Of
> >
> >
> >
> > On Mon, Sep 10, 2012 at 11:37 PM, June Kim (김창준) <junea...@gmail.com>
> > wrote:
> >
> > > Hello
> > >
> > > If you want to remove all occurences of x in y, it's simple in J:
> > >
> > > y-.x
> > >
> > > For example,
> > >
> > >    1 2 3 1 9 10 8 1 -. 1
> > > 2 3 9 10 8
> > >
> > > Now, if you want to remove n occurences of x in y, how would you do
> that?
> > >
> > > Following is my quick and dirty solution:
> > >
> > >    removen=.13 : 'y#~-.((x>:]) *. 0&~:) (*+/\) (={.)y'
> > >    3 removen 'fasdffgh  flkjffif'
> > > asdgh  flkjffif
> > >    3 removen 1 1 2 3 1 9 10 8 1
> > > 2 3 9 10 8 1
> > >
> > > I used the first item in y as the "item to remove" -- smiliar to the
> cut
> > > verb.
> > >
> > > I believe there are simpler solutions as always(you may relax the
> > constrain
> > > for keeping the order of y). Any suggestions?
> > >
> > > June
> > > ----------------------------------------------------------------------
> > > For information about J forums see http://www.jsoftware.com/forums.htm
> > >
> > ----------------------------------------------------------------------
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> ----------------------------------------------------------------------
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