4 : '({. -: {:) y ]\ x' does ({. -: {:) after doing y ]\ x , which is
({. -: {:)@: <something like y ]\ x>
y ]\ x is (]\)~
So can write:
({. -: {:) @: (]\)~
2009/11/10 <[email protected]>:
> Hi Ian,
>
>> t1 =: 4 : '(_1{a) -: (0{ a=. y >\/ (x))'
>
> Uuuh... let's _simplify_ this.
>
> (x) doesn't gain you anything, x will do. (Too much C macros? :-)
>
> In y >\/ x what is the / supposed to do? Was it an attempt to
> make "... tl 2 3 4" compute three matrices with different widths?
> I'd really like to know. However, it does actually nothing in this
> context, and I strongly suggest to change the simplified y >/ x
> into a y ]\ x so readers of your code don't start searching for
> boxes. You want to have a list of you unaltered sliding windows,
> and the identity function expresses this clearly.
>
> In your problem statement, you wrote "want to compare the first
> element with the last element". Lucky you: J has "first" and
> "last" as primitives {. and {: so there's no need for fidgeting
> around with indexes. We'll also reorder their application to match
> your original problem statement more closely:
>
> tl =: 4 : '({.a) -: ({:a =. y ]\ x)'
>
> Repeated occurences of anything in the code can usually be factored
> out in some way. ({.a) -: ({:a) is crying out loud to be re-written
> with a fork: ({. -: {:) a and at that point you don't even need
> your temp variable anymore:
>
> tl =: 4 : '({. -: {:) y ]\ x'
>
> To quote Laurie Anderson: "much, Much, MUCH... better."
> BTW, this might give you some ideas, too:
>
> (3 ,: _3) {. 'abcdefghijklmnopqrstuvwxyz'
> abc
> xyz
>
> Martin Neitzel
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