Given F=: +./@E."1 S:0 'ab cd ef' F ;: 'abcd defg zabc' 0 0 0 It's just a matter of how you form the arguments.
-- Raul On Fri, Nov 11, 2011 at 3:50 PM, Kip Murray <k...@math.uh.edu> wrote: > However, > > 'ab cd ef' f ;: 'abcd defg zabc' > 0 0 0 > > is the desired result. > > That is, none of the three strings 'abcd' 'defg' 'zabc' contains the > string 'ab cd ef' > > ;:'abcd defg zabc' > +----+----+----+ > |abcd|defg|zabc| > +----+----+----+ > > > On 11/11/2011 1:36 PM, Raul Miller wrote: >> I think I would use >> F=: +./@E."1 S:0 >> >> 'ab' F 'abcd' >> 1 >> 'ab' F ;:'abcd defg zabc' >> 1 0 1 >> 'ab cd ef' F&;: 'abcd defg zabc' >> 1 0 0 >> > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm