Here's a sentence that I think would be more helpful in the vocabulary:
]A=:'one';'two';'three'
┌───┬───┬─────┐
│one│two│three│
└───┴───┴─────┘
#&>A
3 3 5
([:#>)">A
3 3 5
^
NB. Then shouldn't the vocabulary say u&v y -> ([: u v y)"v
A similar sentence for @ might save beginners a lot of anguish.
Linda
-----Original Message-----
From: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of km
Sent: Saturday, January 12, 2013 4:20 AM
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] joining to an empty list
You have quoted the vocabulary entry correctly, but must interpret the first
sentence in the light of the third. (Look at the entry again and count periods
to locate the first and third sentences.)
Your usual tools can help:
f =: 13 : '#&> y'
f
#&>
g =: 13 : '# > y'
g
[: # >
Because > has ranks 0 0 0 and monad #&> is equivalent to #@>, monad #&> is
equivalent to
h
([: # >)"0
f 'one';'two';'three'
3 3 5
h 'one';'two';'three'
3 3 5
g 'one';'two';'three'
3
Kip Murray
Sent from my iPad
On Jan 12, 2013, at 1:37 AM, "Linda Alvord" <lindaalv...@verizon.net> wrote:
> My issue is with the vocabulary. Here is compose:
>
> u&v y ↔ u v y . Thus +:&- 7 is _14 (double the negation). Moreover, the
> monads u&v and u@v are equivalent.
>
> f=: 13 :'#&>y'
> f 'one';'two';'three'
> 3 3 5
>
> g=: 13 :'#>y'
> g 'one';'two';'three'
> 3
>
> I play the part of a perpetual beginner. When a statement is in the
> vocabulary I should be able to trust that it will work. As I see it, it does
> not work.
>
> My question is why not?
>
> Linda
>
> -----Original Message-----
> From: programming-boun...@forums.jsoftware.com
> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Linda Alvord
> Sent: Saturday, January 12, 2013 2:19 AM
> To: programm...@jsoftware.com
> Subject: Re: [Jprogramming] joining to an empty list
>
> Fortunately it shows up in my sent messages in small print.
>
> Linda
>
> -----Original Message-----
> From: programming-boun...@forums.jsoftware.com
> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Raul Miller
> Sent: Friday, January 11, 2013 10:14 PM
> To: programm...@jsoftware.com
> Subject: Re: [Jprogramming] joining to an empty list
>
> On Fri, Jan 11, 2013 at 10:07 PM, Linda Alvord <lindaalv...@verizon.net>
> wrote:
>> I don't understand why f and g do not agree?
>
> Because they are doing something different.
>
>> #&>'one';'two';'three'
>> 3 3 5
>> f=: 13 :'#&>y'
>> f 'one';'two';'three'
>> 3 3 5
>
> f counts the number of items in each box.
>
>> NB. Compose: u&v y ↔ u v y
>>
>> g=: 13 :'#>y'
>> g=: 13 :'#>y'
>> g 'one';'two';'three'
>> 3
>
> g counts the number of items after unboxing.
>
> Let's try replacing # with <
>
> <@> 'one';'two';'three'
> +---+---+-----+
> |one|two|three|
> +---+---+-----+
> <> 'one';'two';'three'
> +-----+
> |one |
> |two |
> |three|
> +-----+
>
> Or, let's change the data and use $
>
> $@> i.each 2 2; 3 3; 4 4
> 2 2
> 3 3
> 4 4
> $> i. each 2 2; 3 3; 4 4
> 3 4 4
>
> Remember that > has rank 0 so F@> is going to run an independent instance of
> F for each box that you are unpacking.
>
> Does this help?
>
> --
> Raul
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