In my tests, on longer sequences, Henry Rich's approach seemed about
the same speed as mine.
Thanks,
--
Raul
On Sat, Dec 7, 2019 at 6:33 AM R.E. Boss wrote:
> Forcrand's solution is more or less the version I came up with,
> unfortunately, it's far from linear (what is not compensated by its
As always Brian, I am impressed; and probably will be even more so after the
time it will take me to understand the documents you and Jimmy referenced.
Long ago I concluded that
x (exp1;exp2;...;expn) y
was just a way for smart j'ers to let beginners like me to compare the results
of related e
The expression X exp1;exp2;exp3 Y needs ()'s to be properly constructed as a 5
element "train": X (exp1;exp2;exp3) Y. As near as I can tell it is referenced
in the following link.
https://www.jsoftware.com/help/dictionary/dicte.htm
On that page the reference is to the following page:
https://ww
> is the following what you want ?
>
>1 2 3 (+ ; * ; -) 6 7 8
Not quite. The example in my original email was probably overly minimal to be
completely unhelpful.
This is closer to what I am trying to do:
(+1:)`*`%`(+2:)@.(0 , (1 + 2&|) , 3:)"0 [ 5 6 NB. fictional usage
0.857143
Thank you.
I may reply more in a year ot two when I have fully grokked Roger’s note in
your reply!
Wanting to be snippy in a complaint I was making to my local bank, I wanted to
compare how many prime numbers my iPhone (using J) could calculate in the 15
seconds their cash machine took to read
Hi,
The first vocabulary is full of information. So much so that it is easy to
overlook lots of "tasty tidbits".
The answer your question " where is that first explained to the J Beginner
? " , it is not so much explained but rather demonstrated.
I had to change my reading style from "passive assi
As shown throughout the Vocabulary.
But where is that first explained to the J Beginner?
> On Dec 7, 2019, at 1:08 PM, Jimmy Gauvin wrote:
>
> Hi,
>
> note also that the shape of the results can vary considerably :
>
> 1 2 3 (+ ; * ; - ; +/ ; ,&4 5) 6 7 8
> ┌──┬───┬┬───┬─
Hi,
note also that the shape of the results can vary considerably :
1 2 3 (+ ; * ; - ; +/ ; ,&4 5) 6 7 8
┌──┬───┬┬───┬─┐
│7 9 11│6 14 24│_5 _5 _5│7 8 9│6 7 8 4 5 0 0 0 0│
│ │ ││8 9 10│6 7 8 4 5 4 5 0 0│
│ │ ││9 10 11│6 7
Does this shed any light on the mystery??
J Http Server
|ill-formed name: null_
*** response not sent for ~addons/docs/help/dictionary/null
*** html409 Conflict
I had not seen any of the Safari/Reader errors included in my original message
until just now, when I replied to it
> On Dec 7, 20
Another rabbit hole…
This expression format is used throught the original Iverson/Hui Vocabulary,
and I had always been bothered that I had to figure out what was going on
before I could get much value from the Vocabulary examples.
To investigate what I had missed long ago, I was in JHS, and
Lovely! Not at all obvious (to me)until you do it with simple verbs.
> On Dec 7, 2019, at 9:30 AM, Jimmy Gauvin wrote:
>
> Hi,
>
> is the following what you want ?
>
> 1 2 3 (+ ; * ; -) 6 7 8
> ┌──┬───┬┐
> │7 9 11│6 14 24│_5 _5 _5│
> └──┴───┴┘
>
>
>
>> On
What amazes me most is the influence (of your Moral: Always append to the end
of arrays) on the performance.
ts' d 1 _1 2 _2 {~ 1 ?.@# 4'
0.0032329 528576
ts' d 1 _1 2 _2 {~ 10 ?.@# 4'
0.0313838 4198080
ts' s 1 _1 2 _2 {~ 1 ?.@# 4'
0.0178007 855744
ts' s 1 _1 2 _2 {~
You are right, it doesn't allow trains when v is a verb.
Henry Rich
On 12/6/2019 9:54 PM, ethiejiesa via Programming wrote:
Am I just doing something silly? Or does @. really not support building trains
when the right operand is a verb? Here is an overly minimal example of what I
want:
Hi,
is the following what you want ?
1 2 3 (+ ; * ; -) 6 7 8
┌──┬───┬┐
│7 9 11│6 14 24│_5 _5 _5│
└──┴───┴┘
On Fri, Dec 6, 2019 at 9:54 PM ethiejiesa via Programming <
programm...@jsoftware.com> wrote:
> Am I just doing something silly? Or does @. really not
First some bad news:
s=: ,`(}.@])@.(= -@{:)/
s 1 _1 0
1 _1 0
s=: ,`(}.@])@.(= -@{.)/
s 1 _1 0
0
s 0 1 _1
So our solutions don't work if the accumulator array goes empty before the end.
Fortunately this can be fixed (see below).
Now some good news:
s=: ,`(}.@])@.(0:`(= -@{.)@.
Forcrand's solution is more or less the version I came up with, unfortunately,
it's far from linear (what is not compensated by its elegance).
foo=: ,`(}.@])@.(0=(+{.))/
ts'foo 1 _1 2 _2{~1?.@#4'
0.0166915 854720
ts'foo 1 _1 2 _2{~10?.@#4'
1.9109807 6818496
Miller's solutio
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