On 4 February 2014 04:17, Elias Mårtenson <loke...@gmail.com> wrote: > Thank you. This made me aware of the fact that I still don't understand > this. You are right that the rank is zero, shown by the fact that ⍴,/1 2 3 > returns an empty result: > > 8⎕CR ⍴,/1 2 3 > ┌⊖┐ > │0│ > └─┘ > > > Can you explain to me why that is the case? When I look at the result from > ,/1 2 3 it looks like an array that contains a single element: another array > with the values 1 2 3 in it.
Yup, that's right. It's a scalar (a rank-zero array) whose single item is the vector (rank-one array) 1 2 3. > But it isn't. The result is rank zero, and I > can't take the first element from it: > > (,/1 2 3)[1] > RANK ERROR > (,/1 2 3)[1] This is a mis-use of bracket indexing. With bracket indexing [x;y;z], the number of indices inside the brackets has to match the rank of the array you're indexing into, so [1] will only work on a vector. That's why you get an error. > But wait, I can take the first element from it: > > ↑,/1 2 3 > 1 2 3 This works because ↑ will give you the first item of an array of any rank. In fact ↑,/ is a common idiom for catenating a bunch of vectors. Jay.
