Re: [Bug-apl] How does a vector become a matrix?

[Elias Mårtenson]

The only way to make it more consistent is to make behaviours that today are allowed into an error.

What would you expect ⊃'333' '55555' to do?

Regards,

Elias

On 12 May 2014 22:43, Blake McBride <address@hidden> wrote:

Thanks.  I have to say, with no reflection on present company, I am about as frustrated and disgusted with nested arrays, as defined by IBM, as I could be.   Having enclose do one thing for all arrays and another for scalars has caused me endless hours of frustration.  (Isn't a scalar just a zero dimension array?)  How much time has one to spend making enclose do what comes naturally to ones mind?  Now I find that disclose actually modifies data beyond the ability to reconstruct it.  In your example, if one string were a different length than the other, APL will lengthen it to match the longest upon disclose.  The original length of each string is lost forever.  Why stop there?  Why not change a 4 to a 7?

Having enclose and disclose uniformly add and remove layers of boxing only is simple, consistent, predictable, useful, and easy to understand.  If I add 3 and then subtract 3 I end up with the same number.  But if I enclose and then disclose, I end up with something different - sometimes.  Imagine that!

      '333' '55555'

┌→────────────┐

│┌→──┐ ┌→────┐│

││333│ │55555││

│└───┘ └─────┘│

└∊────────────┘

      ⊃'333' '55555'

┌→────┐

↓333  │

│55555│

└─────┘

      (⊃'333' '55555')[1;]

┌→────┐

│333  │

└─────┘

      ⍴(⊃'333' '55555')[1;]

┌→┐

│5│

└─┘

There are ways to rationalize almost anything.  IMO, the IBM nested array approach is confusing, unpredictable, and renders it a tool of very careful last resort.

I know there has been debate about this in the past, and I am not looking to resurrect it.  It is a real shame IBM chose the path it chose.

Blake

On Mon, May 12, 2014 at 5:08 AM, Jay Foad <address@hidden> wrote:

APL2's Disclose (Dyalog calls it Mix) will convert a vector of vectors
into a matrix:

      ⊃'timor' 'mortis'
┌→─────┐
↓timor │
│mortis│
└──────┘

Your second application of Disclose is applied to a 1-vector of
1-vectors (,⊂,7), so it returns a 1x1 matrix.

Jay.

On 12 May 2014 06:03, Blake McBride <address@hidden> wrote:
>       ⊃⊃⊂,⊂,7
> ┌→┐
> ↓7│
> └─┘
>       ⍴⊃⊃⊂,⊂,7
> ┌→──┐
> │1 1│
> └───┘
>