And there are structural solutions as well.
boxdraw_j_ 1
f_silly =: (}.@:[ [ smoutput@:,)&.:>
(f_silly/ , {:)^:4'initial value';'constant'
initial valueconstant
nitial valueconstant
itial valueconstant
tial valueconstant
+-++
|ial value|constant|
+-++
Good day, Jers
J is fantastic world. It's make a lot for deep journey inside any subject
area. Just as a good long game.
My question: does anybody knows somebody or a place (University,
Instituate, Laboratory, Company...) who/where doing BCI research and also
used J? Maybe BCI + neurofeedback res
((+])^:10) 1
1024
(+]^:10) 1
2
Does this help?
- Bo.
Den 20:16 søndag den 2. august 2015 skrev Raul Miller
:
I think Pascal has pretty much already covered it, but these are equivalent:
f^:5 n
f f f f f n
And these are also equivalent:
m f^:3 n
m f m f m f n
Thanks,
Hello,
Many thanks for you answer.
Try:
require 'web/gethttp'
gethttp url
It is perfect.
And yes, my previous install of J was not finish... good advice :
install all
now labs works correctly.
now, diving time in J ! :D
Yves
-
I think Pascal has pretty much already covered it, but these are equivalent:
f^:5 n
f f f f f n
And these are also equivalent:
m f^:3 n
m f m f m f n
Thanks,
--
Raul
On Sun, Aug 2, 2015 at 11:26 AM, Jon Hough wrote:
> I'm attempting some functions that use verb power (^:) , but
Running dissect on a line, and then clicking around in the display,
might help visualization.
Henry Rich
On 8/2/2015 12:12 PM, 'Pascal Jasmin' via Programming wrote:
f =: (+^:10)&0
f 5
50
the f below, uses x as the "original y", and y is an additional number to be
added to (set to 0 here to
f =: (+^:10)&0
f 5
50
the f below, uses x as the "original y", and y is an additional number to be
added to (set to 0 here to be equivalent to no parameter).
in ^:, the x parameter to the resulting verb is a constant.
- Original Message -
From: Jon Hough
To: "programm...@jsoftwa
Thanks, yes my attempt doesn't work, and it was a bit silly. I was just trying
to give the simplest possible example of my problem. I think my example hasn't
shown the issue I'm talking about.
I want to do some iteration on y, where each step of the iteration has access
to the original y value.
first you are using + monadially here.
A conjunction (most of them) becomes a verb after it has 2 arguments bound to
it. So,
f =: +^:10
is a dyadic (or monadic verb, but monadic + is not that interesting)
5 (+^:10) 0
50
5 (+^:10) 2
52
A way to double 10 times, that looks a bit simila
I'm attempting some functions that use verb power (^:) , but I'm a little lost
when it comes to the concept of what is getting iterated.
For example, in pseudo code:
function f(y){
a = y; //cache initial value
counter = 10;
while(counter-->0){
y = y + a;
}
return y;}
In the above code, y was in
Although I don't really have anything to contribute, I would like to say that I
have enjoyed reading past JoJ issues. Personally, it would be great to see it
continue. Sorry if this reply is not helpful, but I thought I'd express some
appreciation.
Thanks, Jon
> From: mikelpa...@hotmail.es
> To
Journal of J was intended to extend the knowledge of J.
J community is not large and for this reason the cooperation of everyone is
necessary.
After four years, the group of editors is thinking of closing the project. The
main reason is the lack of collaboration in the form of papers, articles,
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