[Jprogramming] joining to an empty list

[Linda Alvord]

That link is about  &.  Not &

Linda

-----Original Message-----
From: programming-bounces at forums.jsoftware.com [mailto:programming-bounces at forums.jsoftware.com] On Behalf Of km
Sent: Saturday, January 12, 2013 10:26 AM
To: programming at jsoftware.com
Subject: Re: [Jprogramming] joining to an empty list

Proof read your NB !

Sent from my iPad


On Jan 12, 2013, at 8:51 AM, "Linda Alvord" <lindaalvord at verizon.net> wrote:

> 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-bounces at forums.jsoftware.com 
> [mailto:programming-bounces at forums.jsoftware.com] On Behalf Of km
> Sent: Saturday, January 12, 2013 4:20 AM
> To: programming at 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" <lindaalvord at 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 at 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-bounces at forums.jsoftware.com 
>> [mailto:programming-bounces at forums.jsoftware.com] On Behalf Of Linda 
>> Alvord
>> Sent: Saturday, January 12, 2013 2:19 AM
>> To: programming at 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-bounces at forums.jsoftware.com 
>> [mailto:programming-bounces at forums.jsoftware.com] On Behalf Of Raul 
>> Miller
>> Sent: Friday, January 11, 2013 10:14 PM
>> To: programming at jsoftware.com
>> Subject: Re: [Jprogramming] joining to an empty list
>> 
>> On Fri, Jan 11, 2013 at 10:07 PM, Linda Alvord <lindaalvord at 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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