The intent of my question is not for some arbitrary v that breaks the equivalence, but for a v with some practical use that breaks the equivalence.
The thing with $ $ v@, is, here you have an argument with structure. v@, discards the structure before applying v , but after that is restructured back to the same shape. (Is the structure necessary or not?) It feels peculiar. ----- Original Message ----- From: Viktor Cerovski <[email protected]> Date: Monday, October 4, 2010 8:41 Subject: Re: [Jprogramming] Another Abbreviation To: [email protected] > > > Roger Hui wrote: > > > > I suspect v"0 would serve. If it does not, > > I'd like to see what such a v is. > > > > > > > > ----- Original Message ----- > > From: Justin Paston-Cooper <[email protected]> > > Date: Friday, October 1, 2010 9:12 > > Subject: [Jprogramming] Another Abbreviation > > To: Programming forum <[email protected]> > > > >> Hello, > >> > >> Is there an abbreviation or better way of doing <<($ $ > >> v@,)>>, where v > >> is a monad? > > --------------------------------------------------------------- > ------- > > For information about J forums see > http://www.jsoftware.com/forums.htm> > > > I think this much is clear: If v returns scalar, then (monadic) > v"0 is > equivalent to ($ $ v@,) and if result of v has rank > 0, then it > is not. > > For instance: > > a =: 1 : '$ $ u@,' > v =: 5&, > > v"0 i. 2 2 > 5 0 > 5 1 > > 5 2 > 5 3 > > v a i. 2 2 > 5 0 > 1 2 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
