In a tacit definition you can access x and y with [ and ]. So in your
example, y (0=|) x gets your current output, and y (] #~ 0=|) x gets
your wanted output, as it will be parsed as (] #~ (0 = |)), so two forks.
On Thu Jul 7, 2022 at 1:24 PM CEST, wrote:
> Hi,
>
> as a first step into the "land
> Now why steps 10-13 result in the next row of a sierpinski triangle, I
> don't know, I guess it has to do with generating a Pascal triangle.
It's not that important for learning J, but it gives a hint why the
seemingly confusing `72#:~8#2` was used. `bm{~3#.\row` maps three consecutive
bits to b
-.@~: gets a mask of duplicates, I. then their indices.
With them you can get a list of (index, value) pairs:
((],.{~) I.@:-.@~:) a
5 2
8 3
11 3
12 2
With 0 ({ (],.{~) I.@:-.@~:)) a you can 0-index into this list.
If you are only interested into the first duplicate of a value,
you could filter d
With mixed bases, for e.g. 3:
3 2 1 #: i. 6
An equivalent function of permh is thus:
(>:@i.@- #: i.@!) 3
If you want to map arbitrary n to a permutation with unknown length,
you could use something like this (I love J for things like !^:_1):
(#:~ [:>:@i.@-@<.@>: !^:_1)"0 (2+i.4) NB. needs specia
Oh, or even simpler I guess:
(<;.1~ 1, }.~:}:) 0 1 0 1 1 1 0 0
On Thu Oct 21, 2021 at 7:14 PM CEST, xash wrote:
> (
> On Thu Oct 21, 2021 at 7:01 PM CEST, Raoul Schorer wrote:
> > Hi,
> >
> > Really trivial question I am stuck on:
> >
> > given a bit ve
( Hi,
>
> Really trivial question I am stuck on:
>
> given a bit vector
> 0 1 0 1 1 1 0 0
>
> how to easily sequentially box 1s & 0s? Such as:
> |0|1|0|1 1 1|0 0|
>
> Thanks!
> Raoul
> --
> For information about J forums see http:/
> I feel like there should be a way to avoid this undesirable effect in
> order
> to stack all files one onto the other. Or is the more idiomatic way to
> first stack the data and then to filter out the fill rows?
Box each file, then raze it:
; <@loadcsv"0 filenames
---
On Thu May 20, 2021 at 3:22 PM CEST, Hauke Rehr wrote:
> I strongly expect one could do better than to
> {.+i.@-.@-/
Maybe a little bit cleaner: (}.i.@>:)/ 2 6
(and good catch on that (<+_1*>) -> (<->), thanks!)
--
For informatio
f=:((#~ 1&|.) ,&.> <;.1~) 0 1, 0= 2+/\ }: (<+_1*>) }.
The first part assigns _1 to elements that are greater than the next one,
and 1 to elements that are less than the next one. Where _1 1 or 1 _1
occur, a new group starts. Thus 0= 2+/\.
This gets us the starting indices for <;.1. Because the bo
Most k version support dictionaries. Might be worth to take a look.
But basically you operate on them with the same array symbols
and they do the expected beheaviour. So if x d: y would create a dict
with x for keys and y for values, k supports:
dict1=:(a;b) d: 1 2
dict2=:(b;c) d: 3 4
dict1
You can get every substring in a list with ; <@<\\. s
If k is quite large, your approach with every k : #) <@<\ ]
tab=: (<:@# ; }: ; {:)&>
count=: ~. ,. <@#/.~
f=: count@:tab@:subs
4 f s
--
For information about J
Without I.:
q2=: (1r4 3r4 <.@* #) -/@:{ \:~
And a fun one:
q3=: \:~ -&({~ # <.@% 4:) /:~
--
For information about J forums see http://www.jsoftware.com/forums.htm
A minor improvement to your fast version is to work on both sides with
the transposed data (which as you noted is faster, as more data in a
row is usually better than a lot of short rows). -/"1&|: will create two
big tables for 2d data quite efficient. (If you can store the data with
shape 2 100 i
kM2 takes in `data`, not the `edist data`. Otherwise, it should work.
Though after I realized what this all does, taking in the distance matrix
makes more sense. Thus:
kM2_step=: 4 : 0
groups=. (i. <./)@:{"1
sub=. {"1 {&x
best=. [ {~ [:(i. <./) [:+/"1 sub
(y groups x) best/. (i. # x)
)
kM2=: ] kM2
closest mediods.
On Sat Feb 13, 2021 at 2:47 PM CET, xash wrote:
> This is already very good code! Some improvements other than edist:
>
> Using `each` for rank "0 like in the last line of kM_step should be
> avoided, as it just boxes/unboxes the result unecessarily.
>
> Ins
This is already very good code! Some improvements other than edist:
Using `each` for rank "0 like in the last line of kM_step should be
avoided, as it just boxes/unboxes the result unecessarily.
Instead of building up the groups per hand (I. & (= & s)), you can just
use `/.` to group the indices.
l =: _1 _2 0 1 2 _1 4 5 _6
p =: <&0
(<;.1~ (~:_,}:)@p) l
┌─┬─┬──┬───┬──┐
│_1 _2│0 1 2│_1│4 5│_6│
└─┴─┴──┴───┴──┘
On Thu Jan 7, 2021 at 10:12 AM CET, Justin Paston-Cooper wrote:
> I am sure this has been asked and formulated somewhere else. I don't
> know what the name of it is.
>
Even simpler:
M e. V
:-)
On Sat Dec 26, 2020 at 1:18 AM CET, Thomas Bulka wrote:
> Hi Hauke,
>
> thank you very much. This works perfectly. Looks, like I’ve been
> thinking to complicated, again...
>
> Regards,
>
> Thomas
>
>
> On 26 Dec 2020, at 0:58, Hauke Rehr wrote:
>
> > +/ M e."_ 0 V
> > d
|. ('()'i.n)} ')','(',: n
or tacit
([:|. ('()'&i.)`(')','(',:])}) n
On Fri Dec 18, 2020 at 10:35 PM CET, Skip Cave wrote:
> Using curlyrt: (amend in place)
>
> n=.'1+(2*3)+(4*(5+6))'
>
>
> |.(+/1 2*'()'=/n)}n,(17#')'),:(17#'(')
>
> ((6+5)*4)+(3*2)+1
>
>
>
> Skip Cave
> Cave Consulting LLC
>
>
> O
A bit more concise, but the same approach - mostly to show off &. :-)
If sep is implemented the same or has a proper obverse, you could of
course use it instead of 10&#.inv.
#m=.(#~ (1 */@p: (|.~ ,.@i.@#)&.(10&#.inv))"0) i. 1e6
On Sat Oct 31, 2020 at 3:10 AM CET, Skip Cave wrote:
> Here's the pro
This is much simpler; groups based on the first element, recursively call
itself until no path is left, and links the result back together. Well, was at
least a nice exercise with L:n yesterday. :-)
] leaves =. 'g';1
] paths =. (,0);1 0
group =: {.&>@[ )@[ $:&.> group)@.(0<#@;@[)
pat
Using L: to apply /. on each leaf (splitting paths) until finished, then
rebuilding the paths via {:: and using them to index into the elements.
] leaves =. 'g';1
] paths =. (,0);1 0
NB. tree based on paths with <'' as elements:
structure=: ({.@> )^:(*@#@;)L:1^:_
NB. <'' -> path ->
Yet another variation on the overlap version:
'AA' (] #~ #@[ */\ (<:@##1:)@[ , -.@E.) 'abcAAAdefAAA'
On Sun, 30 Aug 2020 21:03:37 -0400
Raul Miller wrote:
> Well, if you want the alternate semantics you suggested in that comment:
>
>'AA' (] rplc '';~[) 'abcAAAdefAAA'
> abcAdefA
>
> Or, if
… and depending on what you want, there might be a better approach: parse the
numbers, raze the strings, itemize the characters, and interpret each as a
digit.
<@( [: ". @ ,. @ ; <@":"0)\ *: >: i. 15
On Fri, 21 Aug 2020 12:46:24 -0500
Skip Cave wrote:
> Definitions:
> ea =: each
>
> rab =: ]#
]m=:<\ *: >: i.15
(sep @ ". @ (,&'x') @ rab @ ": @ ,)each m
Appends a 'x' after each number string, so it will get interpreted as an
extended number. Otherwise big numbers get converted to floats, e.g.
149162536496481100121 -> 1.49163e20.
On Fri, 21 Aug 2020 12:46:24 -0500
Skip Cave wrote:
>
(0 0;1 1)}&(2 2$0)"1 (arr1 ,. arr2)
On Sat, 08 Aug 2020 10:13:08 +0200
Thomas Bulka wrote:
> Hello everyone,
>
> I'm stuck with a problem, which (I think) should be really easy to solve
> in J, but I somehow am not able to do it. This is what I want to do:
>
> Let's assume I have two arra
For your original format, with a transition matrix to simplify finding
"catenaries" for each column:
tm=:3 4 $ 0 1 0 1 2 2 2 2 3 1 3 1
start=: 0,~ = + 2*(=>:) NB. same height = 1, direct children = 2, others = 0
step=: {::&tm@, NB. apply tm
nub=: ] ,:~ 4* 2= ] NB. '-'
Hello everyone,
I'm new to J and while working with 1d indices feels quite natural, using
multidimensional indices not so. For example, finding the indices
(3 2$1 2,1 5,4 1) of #'s in this bit mask^Wmap:
map=:'#'= [;._2 (0 : 0)
..
..#..#
..
..
.#
..
)
With 1d indices it
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