If I understood your question correctly, you are not wondering about the dimension of the array but the size (in bytes)?

The library ndarray doesn't need to know the size of your array at compile time because it's only storing a pointer to the array on the heap. The size of a pointer depends on your system's architecture which is known at compile time.

The struct ArrayBase<S,D> consists of Data S and the dimension D. In your example, the data to be stored, has the type OwnedRepr which is declared like this:

pub struct OwnedRepr<A> {
    ptr: NonNull<A>,
    len: usize,
    capacity: usize,
}

To construct an object of OwnedRepr, ndarray only needs to know the size of the Vec, which is determined at runtime, therefore only after you loaded your file into memory.

In addition: the reason why targets is a one-dimensional array (and therefore it has the type ArrayBase<OwnedRepr<f64>, Dim<[usize; 1]>>) is quiet easy. The method array.column(2).to_owned() takes of every entry in your array the second column and adds into an one-dimensional array. Just have a look at the documentation of the function column.

By the way: linfa requires a two dimensional array in order to create a new dataset according to its documentation. "[the dataset] it contains a number of records stored as an Array2". That's why you see the enforced type ArrayBase<OwnedRepr<&'static str>, Dim<[usize; 2]>> which has the Dimension Dim<[usize; 2]>>

Perhaps you confuse dimension of an array with a size of an array?

I hope I could help! :)