> What I want to do is to see how the fit has changed over time You can do this with 2 (1 lsfit ::0: |:)\ |: t . The ::0: part is required because t is randomly generated and so some infixes of it are not in the domain of lsfit .
The problem is one of rank, as you surmised. You can see it immediately if you compare the results of 2 <\"1 t : > 2 <\"1 t > ┌───┬───┬───┬───┬───┬───┬───┐ > │0 7│7 8│8 9│9 2│2 2│2 5│5 5│ > ├───┼───┼───┼───┼───┼───┼───┤ > │5 3│3 8│8 0│0 0│0 5│5 9│9 6│ > └───┴───┴───┴───┴───┴───┴───┘ to desired input of 0 7,:5 3 . Note that that array is composed of 4 numbers, but each box contains just 2 numbers. You've done a 2-infix on rows. What you want is the 2-infixes on columns. Hence |: . > Any thoughts on the process, too? Personally, when I want to partition along different (non-leading) axes, I turn to ;. first. In this case, I probably would have written 2 2 (1&lsfit ::0:);._3 t . It's no shorter than the \ formulation, but it feels cleaner to me, as it avoids the transposition (and the attendant un-transpositions). Or were you asking about the process of lsfit or multivariate something-or-other? If so, I can't offer any advice. I don't even know what the verb's supposed to do :) -Dan ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
