There's some very subtle things going on here. Thanks. That's just the exposition I needed. I think I now see how the behaviour of 'x $ y' with x rank of 0 (actual rank 0 this time! see my response to Raul) extends to explain (1), (2), and (3).
To achieve a satisfactory understanding of a language, J or otherwise, I feel a strong need to predict its behaviour from a minimum set of assumptions. The 'x $ y' verb looks a lot simpler now. Thanks again. On Fri, Feb 5, 2016 at 10:45 PM, bill lam <[email protected]> wrote: > Your x is not rank 0, specifically rank of your x > (which is a noun) is #$x > but x is empty since 0 e. $x > > I think it works like this, > > for sentence like > (1 0 1$0) $ 1 2 3 > since the left argument of ... $ 1 2 3 is shape 1 0 1 > so that the shape of frame is 1 0 > thus the shape of result is 1 0 , ($ (1$0) $ 1 2 3) > which is 1 0 0 > > for (1 0 0$0) $ 1 2 3 > the shape of result is 1 0 , ($ (0$0) $ 1 2 3) > which is 1 0 > > for (1 0$0) $ 1 2 3 > the shape of result is 1 , ($ (0$0) $ 1 2 3) > which is 1 > since this shape contains no 0 so the array must be non-empty > Perhaps interpreting this way > result of '' $ 1 2 3 is 1 > but in case (2), the shape of frame is (1 0) so the the > result 1 cannot be put into the frame so that the assembled result > is empty. > > When the rank of argument is less than the rank of the verb, it > is up to the verb how to handle it. > > OTOH, when the rank of argument is larger, then the verb will > never see the original argument because J interpreter has sliced > it into k-cells which are what to be received by the verb. > (actual implementation is more efficient, eg atomic verbs such as +) > > Пт, 05 фев 2016, Matthew Baulch написал(а): > > The 'x $ y' dyad, like you say, has rank 1 for x and _ for y. Examples > > (1)-(3) have rank 0 arguments for x. The x arguments are arrays with > > non-empty shape but no atoms in these cases. Your x arguments all have > > atoms so, I'm sorry, I can't see their significance. My x arguments have > > shapes containing only non-negative integers (as you mentioned), although > > some are zero. I'm not sure what you mean by: > > > > "So your first example is not even going to get to the reshape part of > the > > operation. Your frame, which will hold your results, is empty." > > > > I understand and agree that the frame of the left argument is empty. That > > doesn't surprise or concern me. What I don't understand are the reasons > for > > the result shapes in (1)-(3), and why the result for (3) has an atom. > > > > All verbs (that I can think of) with an argument of rank >= 1 are defined > > also for arguments of rank 0. For instance, 'i. y', 'x # y' (x has rank > > 1), and so on. Of course, there are very many verbs having arguments of > > infinite rank too. Therefore, I don't see it as remarkable to apply a > rank > > 0 array as an argument to a verb with rank > 0. > > > > Your comment > > > > "the dictionary definitions for verbs mostly only deal with arguments > which > > fit within the dictionary defined rank for the verb" > > > > was enlightening and made me think. Is it simply that the case that most > > verb definitions don't cover the case of rank 0 arguments with no atoms? > > Perhaps my question relates to an undefined part of J? > > > > On Fri, Feb 5, 2016 at 7:27 PM, Raul Miller <[email protected]> > wrote: > > > > > I'm going to go with the original vocab definition, because that's how > > > I understand reshape. > > > > > > But also keep in mind that the left rank of reshape is 1. A shape is a > > > (usually quite short) list of non-negative integers. This is something > > > that we all have to go through - remembering that the dictionary > > > definitions for verbs mostly only deal with arguments which fit within > > > the dictionary defined rank for the verb. That's good enough for most > > > purposes, when you need to concern yourself with rank it's the same > > > rules for all verbs (except for the actual numeric values being > > > different on some verbs). > > > > > > Perhaps http://www.jsoftware.com/help/dictionary/dictb.htm would be > > > worth reviewing? > > > > > > Anyways, to see what that means for your reshape examples, consider > this: > > > > > > <"1 x=: 1 0 1 $ 0 > > > > > > (that's an empty result - no boxes.) > > > > > > $<"1 x=: 1 0 1 $ 0 > > > 1 0 > > > > > > (that's what your frame looks like, for this example.) > > > > > > So your first example is not even going to get to the reshape part of > > > the operation. Your frame, which will hold your results, is empty. > > > > > > In your first example, x is empty, so you'll be using an array of > > > fills (which are zeros) for the test run of reshape. So you actually > > > do get a final dimension here - the last dimension is 1, and although > > > x is empty, the interpreter uses that 1 to decide how many zeros it's > > > going to use. > > > > > > Your second and third examples have the same issue - you're not doing > > > a single reshape but a collection of reshapes. But in those examples > > > your last dimension of x is zero. So when the interpreter does its > > > test run on the verb to see what kinds of results it's getting instead > > > of doing: > > > > > > (,0) $ 1 2 3 > > > > > > like it was in the first example, it is instead doing > > > > > > '' $ 1 2 3 > > > 1 > > > > > > In other words there trial result doesn't have a shape, so you get one > > > less dimension in your result than you had in x. > > > > > > As for your last example: > > > > > > <"1 x=: 1 $ 0 > > > +-+ > > > |0| > > > +-+ > > > $ <"1 x=: 1 $ 0 > > > > > > ...so your final example does do the reshaping, but you are reshaping > > > with an empty shape, so you still get an empty result. > > > > > > Does this make sense? > > > > > > Thanks, > > > > > > -- > > > Raul > > > > > > On Fri, Feb 5, 2016 at 3:06 AM, Matthew Baulch <[email protected]> > > > wrote: > > > > Original vocab says: > > > > "The shape of x$y is x,siy where siy is the shape of an item of y." > > > > > > > > NuVoc says (for x $ y): > > > > "If y is an atom or a list, the shape of the result is x", and "the > shape > > > > of [the result of x$y] is always x,}.$y". > > > > > > > > Let y =: 1 2 3 for all that follows. > > > > > > > > (1) > > > > x =: 1 0 1 $ 0 > > > > x$y has no atoms, shape 1 0 0 > > > > (2) > > > > x =: 1 0 0 $ 0 > > > > x$y has no atoms, shape 1 0 > > > > (3) > > > > x =: 1 0 $ 0 > > > > x$y has a single atom: 1, and shape 1 > > > > (4) > > > > x =: 1 $ 0 > > > > x$y has no atoms, shape 0 > > > > > > > > Examples (1)-(3) appear to violate the definitions. Only example (4) > > > > agrees. Can anyone shed some light on this? (3) strikes me as > > > particularly > > > > strange. I'm sure I must have missed something. > > > > > > > > My head is spinning! > > > > > ---------------------------------------------------------------------- > > > > 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 > > -- > regards, > ==================================================== > GPG key 1024D/4434BAB3 2008-08-24 > gpg --keyserver subkeys.pgp.net --recv-keys 4434BAB3 > gpg --keyserver subkeys.pgp.net --armor --export 4434BAB3 > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
