Please do not ignore rank when reading the vocabulary. For example: http://www.jsoftware.com/help/dictionary/d630v.htm says that the rank of the derived verb is mv mv mv
This means that the monad rank of verb v will be used for all verb ranks of a verb derived using & So, for the example #&> the verb v will be > and the monad rank of this verb is 0 so when considering equivalent definitions you must retain that rank. You can either use an explicit declaration "0 0 0 or "0 or you can constrain your examples so that you only present rank 0 nouns to your example code. In any event, you cannot take a part of the definition in isolation unless you ensure that the constraints imposed by the rest of the definition have been handled properly. Thanks, -- Raul On Sat, Jan 12, 2013 at 2:37 AM, Linda Alvord <lindaalv...@verizon.net> wrote: > My issue is with the vocabulary. Here is compose: > > u&v y ↔ u v y . Thus +:&- 7 is _14 (double the negation). Moreover, the > monads u&v and u@v are equivalent. > > f=: 13 :'#&>y' > f 'one';'two';'three' > 3 3 5 > > g=: 13 :'#>y' > g 'one';'two';'three' > 3 > > I play the part of a perpetual beginner. When a statement is in the > vocabulary I should be able to trust that it will work. As I see it, it does > not work. > > My question is why not? > > Linda > > -----Original Message----- > From: programming-boun...@forums.jsoftware.com > [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Linda Alvord > Sent: Saturday, January 12, 2013 2:19 AM > To: programm...@jsoftware.com > Subject: Re: [Jprogramming] joining to an empty list > > Fortunately it shows up in my sent messages in small print. > > Linda > > -----Original Message----- > From: programming-boun...@forums.jsoftware.com > [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Raul Miller > Sent: Friday, January 11, 2013 10:14 PM > To: programm...@jsoftware.com > Subject: Re: [Jprogramming] joining to an empty list > > On Fri, Jan 11, 2013 at 10:07 PM, Linda Alvord <lindaalv...@verizon.net> > wrote: >> I don't understand why f and g do not agree? > > Because they are doing something different. > >> #&>'one';'two';'three' >> 3 3 5 >> f=: 13 :'#&>y' >> f 'one';'two';'three' >> 3 3 5 > > f counts the number of items in each box. > >> NB. Compose: u&v y ↔ u v y >> >> g=: 13 :'#>y' >> g=: 13 :'#>y' >> g 'one';'two';'three' >> 3 > > g counts the number of items after unboxing. > > Let's try replacing # with < > > <@> 'one';'two';'three' > +---+---+-----+ > |one|two|three| > +---+---+-----+ > <> 'one';'two';'three' > +-----+ > |one | > |two | > |three| > +-----+ > > Or, let's change the data and use $ > > $@> i.each 2 2; 3 3; 4 4 > 2 2 > 3 3 > 4 4 > $> i. each 2 2; 3 3; 4 4 > 3 4 4 > > Remember that > has rank 0 so F@> is going to run an independent instance of > F for each box that you are unpacking. > > Does this help? > > -- > Raul > ---------------------------------------------------------------------- > 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
