Result shape

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In the previous sections the question of the shape of the result was glossed over. For a monad the frame of the result is the same as the frame of the argument. For a dyad the frame of the result is the frame of the longer of the frames of the arguments (or either frame if they are the same).

With a verb like + that has an atom result for each atom argument this is straightforward. Things get more interesting with verbs that have more complicated behavior.

Consider the verb $. Look it up in NuVoc and you'll see it has rank of _ for monads which indicates an infinite (unbounded) rank and means that the verb applies to the entire argument. If you think about the monad $ with a result that is the shape of its entire right argument this makes sense. The dyad left rank is 1 and this means that it applies to lists from the left argument. The dyad right rank is unbounded _ and so applies to the entire right argument.

   2 4 $ i.3
0 1 2 0
1 2 0 1
   2 4 $"1 0 i.3
0 0 0 0
0 0 0 0

1 1 1 1
1 1 1 1

2 2 2 2
2 2 2 2

The first example is what you have seen before, but what is going on in the second? The $"1 0 means that $ will get cell arguments as a list (1-cells) on the left and as an atom (0-cell) on the right. The left frame is empty (nothing is left of the shape of the left argument after a 1-cell is taken) and the right frame is 3 (there are 3 0-cells in the right argument). So the result frame is 3.

2 4 $ 0   gives   0 0 0 0   left 1-cell $ right 0-cell 
                  0 0 0 0

... $ 1   gives   1 1 1 1   repeat 1-cell $ next 0-cell
                  1 1 1 1

... $ 2   gives   2 2 2 2   repeat 1-cell $ next 0-cell
                  2 2 2 2


The frame of the result is 3 and the things in that frame are 2 by 4 tables, so the shape of the final result is 3 2 4.

   $ 2 4 $"1 0 i.3
3 2 4


Rank (noun rank, verb rank, frames, cells, and the rank conjunction) applies to all verbs and greatly increases the ways in which you can use any verb.

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