Ok, that clears up things a bit; thanks.

I'll stick with @. though because the verbs I use are much
more complicated than 1: , 2:

On Sat, Oct 8, 2011 at 8:43 PM, Marshall Lochbaum <[email protected]>wrote:

> $i.10 is a 1-element list, while #i.10 is a scalar. You can determine this
> by looking at the shape:
>
>   $ $i.10
> 1
>   $ #i.10
>
>   #@$ $i.10
> 1
>   #@$ #i.10
> 0
>
> A scalar is defined as having zero dimensions, that is, the shape is an
> empty list. One important thing to know is that $ always returns a list and
> # always returns a scalar.
>
> # is equivalent to {.!.1@$  , which is the first element of the shape, or
> 1
> if it is empty.
>
> @. tends to not work too well, being a complicated conjunction. In this
> case
> I would recommend using (1+2<#) , which accomplishes the task much more
> simply, and can also be adjusted to work with rank.
>
> Marshall
>
> On Sat, Oct 8, 2011 at 1:24 PM, EelVex <[email protected]> wrote:
>
> > Thanks,
> >
> > (2<#) suits me fine
> > but I still don't understand why (2<$) is not working here.
> > Why
> > (2<$) i.10
> > doesn't have the same rank as
> > (2<{.@$) i.10
> > (and how should I know?)
> >
> > On Sat, Oct 8, 2011 at 2:45 PM, Aai <[email protected]> wrote:
> >
> > > Look at the result of e.g.
> > >     (2<$) i.2 3 3
> > > 0 1 1
> > >
> > > and agenda @. uses the result of verb v in m @. v to select one of
> gerund
> > m
> > >
> > > You probably want something like one of the following:
> > >
> > >    (2<{.@$) i.10
> > > 1
> > >
> > >    (2<#@$) i.10
> > > 0
> > >
> > > or use #
> > >
> > >    (2<#) i.10
> > > 1
> > >
> > >
> > >    1:`2:@.(2<{.@$) i.10
> > > 2
> > >
> > >      1:`2:@.(2<#@$) i.10
> > > 1
> > >
> > >  1:`2:@.(2<#) i.10
> > > 2
> > >
> > >
> > >
> > >
> > > Hallo EelVex, je schreef op 08-10-11 12:35:
> > > > Hi all.
> > > >
> > > > Why does this:
> > > >
> > > >     1:`2: @. (2<$) i.10
> > > >
> > > > give a rank error?
> > > >
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