I get different results in jhs where none of the lines have an empty line. NB. Here 5#'' (5#'')*/i.5 ''*/i.5 '' NB. To here
I actually always expected to have an empty line for each expression. Linda -----Original Message----- From: programming-boun...@forums.jsoftware.com [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Don Kelly Sent: Wednesday, September 10, 2014 7:44 PM To: programm...@jsoftware.com Subject: Re: [Jprogramming] Replace one item of a list I get 5#'' (5#'')*/i.5 ''*/i.5 NOTE-. blank line using J802 Don Kelly On 10/09/2014 4:16 PM, Linda Alvord wrote: > Dan, after your 4 series, I tried this. Also odd: > > NB. Begin here > 5#'' > > (5#'')*/i.5 > ''*/i.5 > '' > > NB. Note the spacing between lines above > > Linda > > -----Original Message----- > From: programming-boun...@forums.jsoftware.com > [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Dan Bron > Sent: Wednesday, September 10, 2014 11:56 AM > To: programm...@jsoftware.com > Subject: Re: [Jprogramming] Replace one item of a list > > Linda wrote: >> *i.5 >> 0 1 1 1 1 >> */"0 i.5 >> 0 1 2 3 4 > Joe wrote: >> It looks like /"0 yields the right side without evaluating the left side >> >> asdfasdf/"0 i.5 >> 0 1 2 3 4 >> >> I don't totally get it myself > This boils down to foo/ scalar_value . > > Note that because of the "0 (meaning "apply to each scalar value > individually") > > asdfasdf/"0 i.5 > 0 1 2 3 4 > > is equivalent to > > (asdfasdf/ 0),(asdfasdf/ 1),(asdfasdf/ 2),(asdfasdf/ > 3),(asdfasdf/ 4) > 0 1 2 3 4 > > In Linda's example, asdfasdf is *, so her example is > > (*/ 0),(*/ 1),(*/ 2),(*/ 3),(*/ 4) > 0 1 2 3 4 > > And this is quite different from the other expression she contrasts it > with, which is > > *i.5 > 0 1 1 1 1 > > which, because monad * (signum) is a scalar function, is equivalent to: > > *"0 i.5 > 0 1 1 1 1 > > which, after expanding in the same manner as above, is equivalent to: > > (* 0),(* 1),(* 2),(* 3),(* 4) > 0 1 1 1 1 > > Note the lack of any / which is the key issue here. That is, the > difference Linda spotted boils down to the difference between these two > expressions: > > */ 4 > 4 > > * 4 > 1 > > > Now, *4 is 1 because 4>0 ; no mystery there. But why does */4 produce > 4 ? Well, */ is product, and > > */ 4 4 4 4 NB. Product of four 4s is 4^4, 256 > 256 > */ 4 4 4 NB. Product of three 4s is 4^3, 64 > 64 > */ 4 4 NB. Product of two 4s is 4^2, 16 > 16 > */ 4 NB. Product of one 4 is 4^1, 4 > 4 > > And in general, foo/ scalar_value (or asdfasdf/ scalar_value) is simply > scalar_value . Because foo/ noun says "insert foo between all _pairs > of items_ in noun", but when noun is scalar_value, there _are no pairs_, > so no insertion is done. > > -Dan > > PS: and, of course, > > */ '' NB. Product of zero "4s" is 4^0, 1 > 1 > > There are no pairs to insert * between here, either, but empty arguments > gets into identity functions. > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
