On 9/3/07, Terrence Brannon <[EMAIL PROTECTED]> wrote: > I am studying this: http://www.jsoftware.com/jwiki/Essays/Rank > > The thing that confuses me is the expression < @ $ > > First things first - @ is traditional function composition, right?
Right. > Ok, but what I dont get about that expression, is that there is no > Noun for $ to take the shape of. Which means that <@$ is a verb (just like $ is a verb). When you give it a noun, the verb will give you back a new noun. > Also please confirm that my Anglicization as means of Elucidation of > the boxr verb is indeed accurate: > > ; Left =: [ > ; Right =: ] > ;; boxr =: ]`(<@$ , [ $: */@[}.])@.([EMAIL PROTECTED]@]) > ; boxr =: Right ` BigVerb @. GerundIndexer > ; GerundIndexer =: Signum (Tally Right) > ; BigVerb =: V1 Append V2 > ; V1 =: Box ShapeOf NB. here is where I am lost > ; V2 =: Left SelfCall (LeftProd Take Right) > ; LeftProd = (Insert *) Left No. If I make your two names distinct: ;; boxra =: ]`(<@$ , [ $: */@[}.])@.([EMAIL PROTECTED]@]) ; boxrb =: Right ` BigVerb @. GerundIndexer and define Tally=: 3 Signum=: * then boxra 1 2 +-+ |2| +-+ boxrb 1 2 |length error: Signum | boxrb 1 2 That said, it would be good to understand the domain, and or purpose of these verbs: boxra 0 |stack error: boxra | boxra 0 -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
