...@jsoftware.com
[mailto:programming-boun...@jsoftware.com] On Behalf Of Alexander Mikhailov
Sent: Friday, June 01, 2012 1:35 PM
To: programming@jsoftware.com
Subject: Re: [Jprogramming] rank trouble
-
Without the nub gives an arguably more correct answer:
pi (]-.]-.[) e
2 1 2 1 3
e
(20 $])~ pi
3 3 1 1 1 1 2 2
R.E. Boss
-Oorspronkelijk bericht-
Van: programming-boun...@jsoftware.com
[mailto:programming-boun...@jsoftware.com] Namens Linda Alvord
Verzonden: zaterdag 2 juni 2012 11:24
Aan: 'Programming forum'
Onderwerp: Re: [Jprogramming] rank trouble
You could also do this:
e ([:~.]-.]-.[) pi
3 1 2
pi ([:~.]-.]-.[) e
2 1 3
Without the nub gives an arguably more correct answer:
pi (]-.]-.[) e
2 1 2 1 3
e (]-.]-.[) pi
3 1 1 2
/:~1 e,:pi
1 1 2 2 3 7 8 8 8
1 1 2 3 4 5 5 6 9
On Thu, May 31, 2012 at 10:48 AM, Graham Parkhouse
-
Without the nub gives an arguably more correct answer:
pi (]-.]-.[) e
2 1 2 1 3
e (]-.]-.[) pi
3 1 1 2
-
Why not
e ([ -. -.) pi
2 1 2 1 3
?
--
For information about J forums see
Thank you Devon and Alexander! -. does intersection most elegantly. I shall
use it.
On Fri, 1 Jun 2012 12:50:43 -0400 Devon McCormick devon...@gmail.com
wrote:
You could also do this:
e ([:~.]-.]-.[) pi
3 1 2
pi ([:~.]-.]-.[) e
2 1 3
Without the nub gives an
Can anyone help me to understand where I am going wrong?
e=: 2 7 1 8 2 8 1 8 3
pi=: 3 1 4 1 5 9 2 6 5
isecn=: ([: /:~ [: ~. e. # [)1 NB. Intersection of the two sets
e isecn pi
1 2 3
pi isecn e
1 2 3
isecn/ e,:pi
1 2 3
isecn/1 e,.pi
This last sentence give me 9
On Thu, May 31, 2012 at 10:48 AM, Graham Parkhouse
graham.parkho...@ntlworld.com wrote:
e=: 2 7 1 8 2 8 1 8 3
pi=: 3 1 4 1 5 9 2 6 5
isecn=: ([: /:~ [: ~. e. # [)1 NB. Intersection of the two sets
...
isecn/1 e,.pi
This last sentence give me 9 blank lines. I expected and
Perhaps this will help:
e=: 2 7 1 8 2 8 1 8 3
pi=: 3 1 4 1 5 9 2 6 5
e,.pi
2 3
7 1
1 4
8 1
2 5
8 9
1 2
8 6
3 5
]/1 e,.pi
3 1 4 1 5 9 2 6 5
[/1 e,.pi
2 7 1 8 2 8 1 8 3
On 5/31/2012 10:48, Graham Parkhouse wrote:
Can anyone help me to understand where I am going wrong?
isecn/ y inserts verb isecn between the items of y
Consider
|: e ,. pi NB. items are e and pi
2 7 1 8 2 8 1 8 3
3 1 4 1 5 9 2 6 5
isecn/ |: e ,. pi
1 2 3
On 5/31/2012 9:48 AM, Graham Parkhouse wrote:
Can anyone help me to understand where I am going wrong?
e=: 2 7 1 8
Thank you everybody! I was just thinking that, using 1, I was just swapping
rows for columns - sloppy thinking; transpose is indeed necessary!
Using [ and ] as test functions on structure is very helpful, like using
to demonstrate partitioning conjunctions.
Regards
Graham
On Thu, May 31, 2012
I have been trying to think up a useful response here, but I just do not
know enough from this description to make good choices.
Typically, I build up the expressions that I find useful first and then
come up with names for them later. Sometimes I re-arrange things that work
so that they are
Raul,
On Fri, Nov 18, 2011 at 12:02:47PM -0500, Raul Miller wrote:
I have been trying to think up a useful response here, but I just do not
know enough from this description to make good choices.
Well, I appreciate the attempt. :) I don't know enough to know what
information would be necessary
I think that representative data, along with a description of the result
and purpose of the code would work. (Or, if not, that should be enough for
us to ask intelligent questions.)
--
Raul
On Fri, Nov 18, 2011 at 12:45 PM, Daniel Lyons fus...@storytotell.orgwrote:
Raul,
On Fri, Nov 18,
On Fri, Nov 18, 2011 at 12:49:55PM -0500, Raul Miller wrote:
I think that representative data, along with a description of the result
and purpose of the code would work. (Or, if not, that should be enough for
us to ask intelligent questions.)
I've been disinclined to share the problem because
Raul,
On Fri, Nov 18, 2011 at 03:53:07PM -0500, Raul Miller wrote:
Ok.. I hope we have not ruined anything here (to prevent people from
googling this and studying up ahead of time, it would be best if the
terminology were somewhat obfuscated). But, anyways...
I don't think it will be a
I did not see anything attached, but forgot about that before I was ready
to reply.
And here's an overview of how I treated angle differences:
degdif/~ 0 90 180 270 360
0 _90 _180 900
900 _90 _180 90
_180 900 _90 _180
_90 _180 900 _90
0 _90 _180 900
On Fri, Nov 18, 2011 at 04:34:24PM -0500, Raul Miller wrote:
I did not see anything attached, but forgot about that before I was ready
to reply.
It looks like the mailing list has stripped the attachments.
--
Daniel Lyons
--
In general, structuring J code is an exercise in picking the right words.
That said, in this case: adding 1 on a verb that assumes a rank 1
argument is only meaningful for verbs which do the right thing on equal
length arguments.
--
Raul
On Wed, Nov 16, 2011 at 11:36 PM, Daniel Lyons
I think that people very often misunderstand what they are doing when
setting/changing rank.
I very rarely see people set more than one number after the as in 1 and
on an occasion 0 1 and not realizing that you actually need three numbers
to set the ranks when the ranks are for a dyadic verb.
From the Vocabulary page on Rank un
The verb un applies u to each cell as specified by the rank n . The
full form of the rank used is 3$.|.n . For example, if n=:2 , the three
ranks are 2 2 2 , and if n=: 2 3, they are 3 2 3 .
On 11/17/2011 9:13 AM, Björn Helgason wrote:
I think that people
On Thu, Nov 17, 2011 at 09:49:32AM -0500, Raul Miller wrote:
In general, structuring J code is an exercise in picking the right words.
Well, in case you can help with that, here's my situation. For this
particular problem, I need to calculate a distance, so I have an
origin and a destination.
Since you know you want the function to applied with rank 1, you should
probably build it into the definition, e.g.
timeToMove=: ((timeToRotate : azimuth) . (timeToAscend :
elevation))1
left timeToMove right
0.15 0.45 1.15
On Tue, Nov 15, 2011 at 9:39 AM, Daniel Lyons
On Nov 16, 2011, at 8:49 AM, Devon McCormick wrote:
Since you know you want the function to applied with rank 1, you should
probably build it into the definition, e.g.
timeToMove=: ((timeToRotate : azimuth) . (timeToAscend :
elevation))1
left timeToMove right
0.15 0.45 1.15
I
I cannot duplicate your error. From your definitions I get
left timeToMove right
0.25 0.45
You might notice
left
0 1
2 3
4 5
elevation left
2 3
elevation1 left
1 3 5
Which do you want?
On 11/15/2011 1:08 AM, Daniel Lyons wrote:
I suspect this is a classic beginner misstep,
I cannot duplicate it either.
I suspect that a definition which is different from what was posted.
--
Raul
On Tue, Nov 15, 2011 at 7:27 AM, Kip Murray k...@math.uh.edu wrote:
I cannot duplicate your error. From your definitions I get
left timeToMove right
0.25 0.45
You might notice
On Nov 15, 2011, at 5:27 AM, Kip Murray wrote:
I cannot duplicate your error.
Trying it again this morning, I see I must have had a bad definition in my
session. Loading from the file I get the same behavior as you and Raul. Sorry
about that!
From your definitions I get
left
I suspect this is a classic beginner misstep, but I don't see what I'm doing
wrong.
I have a verb which doesn't work unless I use the rank-1 adverb. Here are the
definitions:
diff =: -`(-~) @.
azimuth =: 0 {::
elevation =: 1 {::
minRotationDistance =: - . : (360|) -~
timeToRotate =:
Becomes even clearer if you define:
plus=: (4 : '(:x),''+'',(:y)')0
plus/a
┌───┬───┬───┬───┐
│0+8+16 │1+9+17 │2+10+18│3+11+19│
├───┼───┼───┼───┤
│4+12+20│5+13+21│6+14+22│7+15+23│
└───┴───┴───┴───┘
plus/1 a
┌───┬───┐
│0+1+2+3
Hi,
I have just started with J and I have a question concerning rank (I think).
I don't understand why the following does what it does:
a=. 2 2 $ i.4
b=.+/a
c=.+/1 a
After executing these lines
a contains
0 1
2 3
b contains 2 4 and
c contains 1 5
I understand where b is coming
You almost had it. I've fixed the error below.
Henry Rich
On 6/8/2011 5:41 PM, Squint6 wrote:
Hi,
I have just started with J and I have a question concerning rank (I think).
I don't understand why the following does what it does:
a=. 2 2 $ i.4
b=.+/a
c=.+/1 a
After executing
On 8 Jun 2011, at 15:41, Squint6 wrote:
a=. 2 2 $ i.4
b=.+/a
c=.+/1 a
After executing these lines
a contains
0 1
2 3
b contains 2 4 and
c contains 1 5
I understand where b is coming from but c's value confuses me. Here's how my
(faulty) reasoning about c=.+/1 a goes.
Thank you, Kent, for posing a newbie question that crystallises the
essence of Rank (). And thanks to those who've offered an answer.
I was just about to supply the content for the NuVoc entry on Rank:
http://www.jsoftware.com/jwiki/Vocabulary/quote
Your micro-example spotlights exactly what a
Thanks for the answers, Henry and Dan. It seems kind of obvious now, but it
had me scratching my head before you enlightened me.
Ian, always glad when my ignorance can be of service.
Kent.
--
For information about J forums
I wonder how the following is explained with rank:
i. 2 2
┌───┐
│0 1│
│2 3│
└───┘
1 i. 2 2
┌───┬───┐
│0 1│2 3│
└───┴───┘
The posted explanations seem to be missing something...
On 6/8/2011 11:21 PM, Ian Clark wrote:
Thank you, Kent, for posing a newbie question that crystallises the
essence
Thanks for pointing out that Dic explains it better than Voc.
I accept unreservedly that there is no shortage of documentation on
Rank, and that it is adequately explained somewhere. But let's ask: is
the delivery of this material effective?
...No, I can't answer that. It takes an articulate J
Ian, always glad when my ignorance can be of service.
You may laugh, but articulate ignorance is a precious commodity. A
perishable one.
On Thu, Jun 9, 2011 at 3:54 AM, Squint6 squi...@aol.com wrote:
Thanks for the answers, Henry and Dan. It seems kind of obvious now, but it
had me
The default rank of the monadic verb (Box) is infinite so it will
operate on the whole of its right argument at once. Hence:
i. 2 2
+---+
|0 1|
|2 3|
+---+
by adding 1 to the verb we are feeding the argument to the verb a
1-cell at-a-time.
As has already been shown, the 1-cells of ( i.2 2 )
A source of confusion for a newcomer can be that the items of an array are
shown
in a vertical list except when the array is a vector the items are shown in a
horizontal list. Thus in the array
]a =: 3 2 4 $ i. 24
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
16 17 18 19
20 21 22
Thank you R.E. Boss and Brian. The reference material is so good. Yes, I'm
clearer on agreement now. And I'm not sure it's worth troubling you all with
this, but my underlying problem was linked with the empty vector. What I was
wanting was a verb c with these properties:
0 1 c 0 1 2
0 0
0 1
0
'',1 2 3
1 2 3
NB. OK
'',0 [1 2 3
|length error
| '',0[1 2 3
NB. Is this because there are NO atoms in LH argument?
NB. I would have preferred to have seen 1 2 3
1 2 3,0/ 4 5 6
1 4
1 5
1 6
2 4
2 5
2 6
3 4
3 5
3 6
NB. OK
'',0/ 4 5 6
NB. I would have preferred to
Parkhouse
Verzonden: zaterdag 28 augustus 2010 12:32
Aan: programming@jsoftware.com
Onderwerp: [Jprogramming] Rank zero append (,0)
'',1 2 3
1 2 3
NB. OK
'',0 [1 2 3
|length error
| '',0[1 2 3
NB. Is this because there are NO atoms in LH argument?
NB. I would have
I find your explanation very helpful, especially because of the links
you have given to the primer, which I had not studied carefully. Refer
to the following link, please, where we read, For a dyad the left
rank of the verb and the rank of the left argument determine the frame
of the left
I see that the example I meant to show literal arguments was not
included so I am adding it now, though it is certainly not noteworthy.
''(,0) frame '123'
┌─┬─┐
│0│3│
└─┴─┘
And the next examples show a monadic verb and a dyadic verb with
literal arguments.
+ frame 2 3 4
┌─┐
│3│
ĵaŭ, 18 Mar 2010, Zsbán Ambrus skribis:
2010/3/18 Frederick W Mellem mell...@frontiernet.net:
i..(0) 4 2 3
T---T-┐
│0 1 2 3 │0 1 │0 1 2 │
L---+---+--
i..(0) 4 2 3
--T-T-T-┐
│0 │1 │2 │3 │
+-+-+-+-+
│0 │1 │ │ │
+-+-+-+-+
i.. 4 2 3
T---T-┐
│0 1 2 3 │0 1 │0 1 2 │
L---+---+--
i..(0) 4 2 3
T---T-┐
│0 1 2 3 │0 1 │0 1 2 │
L---+---+--
i..(0) 4 2 3
--T-T-T-┐
│0 │1 │2 │3 │
+-+-+-+-+
│0 │1 │ │ │
+-+-+-+-+
│0 │1│2 │ │
L-+-+-+--
2010/3/18 Frederick W Mellem mell...@frontiernet.net:
i..(0) 4 2 3
T---T-┐
│0 1 2 3 │0 1 │0 1 2 │
L---+---+--
i..(0) 4 2 3
--T-T-T-┐
│0 │1 │2 │3 │
+-+-+-+-+
│0 │1 │ │ │
+-+-+-+-+
│0 │1│2 │ │
L-+-+-+--
Please explain the scalar
Hi Frederick
Because i..(0) 4 2 3 means (i..)(0) 4 2 3
- Bo
--- Den tors 18/3/10 skrev Frederick W Mellem mell...@frontiernet.net:
Fra: Frederick W Mellem mell...@frontiernet.net
Emne: [Jprogramming] Rank amateur question
Til: programming@jsoftware.com
Dato: torsdag 18. marts 2010 22.48
I was revisiting some code from a few years ago when I was first coming to
grips with J.
It basically reformats a table formatted with one column per variable for each
record (spreadsheet format) to having one record for each variable (database
format).
]tst=: (;:'tag yr weight dob') , 0 i.
:25 PM
To: programming@jsoftware.com
Subject: Re: [Jprogramming] rank by a key
In the following example, collection 0 is 10 30 20 20 and has ranks 3
0
1 1, while collection 1 is 200 100 300 and has ranks 1 2 0. The
problem
is the extra 0 appended to make them the same size, [...]
The usual way
On Fri, Nov 13, 2009 at 1:10 PM, Tirrell, Jordan (Consultant) key=:
1 1,1 0,1 1,1 0,1 0,0 1,:1 1
data=: 0 1,1 1,0 2,0 2,2 0,0 0,:1 1
key rankbykey1 data
2 1 1 2 0 0 0
key rankbykey2 data
2 1 1 2 0 0 0
I cannot figure out how to use ~: to express this function as Raul
Miller suggested,
It works great, and in a few seconds less.
Thanks!
Jordan
-Original Message-
From: programming-boun...@jsoftware.com
[mailto:programming-boun...@jsoftware.com] On Behalf Of Raul Miller
Sent: Friday, November 13, 2009 2:34 PM
To: Programming forum
Subject: Re: [Jprogramming] rank
Given a key (in the sense of /.), I'd like to obtain the rank (the
i.~\:~ kind, not the kind) of each item within the collection
corresponding to its key. My first instinct was to try (i.~\:~)/. but
this is not exactly what I want.
In the following example, collection 0 is 10 30 20 20 and has
On Tue, Nov 10, 2009 at 4:23 PM, Tirrell, Jordan (Consultant)
jtirr...@thomasnet.com wrote:
Given a key (in the sense of /.), I'd like to obtain the rank (the
i.~\:~ kind, not the kind) of each item within the collection
corresponding to its key. My first instinct was to try (i.~\:~)/. but
In the following example, collection 0 is 10 30 20 20 and has ranks 3 0
1 1, while collection 1 is 200 100 300 and has ranks 1 2 0. The problem
is the extra 0 appended to make them the same size, [...]
The usual way to avoid those extra fills is to box the interim values
and to flatten
i.~ 8 4 4 4 8 8 8
0 1 1 1 0 0 0
8 4 4 4 8 8 8 (;@(@(i.~\:~)/.) /: i...@[) 10 200 100 300 30 20 20
3 1 2 0 0 1 1
All in all, this is not very different from what you did. Slighltly
less boxing/unboxing. Can you measure any difference?
or
8 4 4 4 8 8 8 (;@(@(i.~\:~)/.) \:
Is this what you wanted?
$('abc')=.4 2 7 3 9
5
or this?
$('a b c')=.4 2 7
2009/2/18 David Vincent-Jones davidvincentjo...@gmail.com
OK .. I know this is a rank problem but I still have difficulties
understanding how to solve this simple question.
($'abc')=4 2 7 3 9
|length error
|
The result of $ in this case is a one-element vector,
which is a mismatch with a 5-element vector.
You can use # instead of $ .
- Original Message -
From: David Vincent-Jones davidvincentjo...@gmail.com
Date: Wednesday, February 18, 2009 11:55
Subject: [Jprogramming] Rank problem
-
From: David Vincent-Jones davidvincentjo...@gmail.com
Date: Wednesday, February 18, 2009 11:55
Subject: [Jprogramming] Rank problem
To: JSoftware programming@jsoftware.com
OK .. I know this is a rank problem but I still have difficulties
understanding how to solve this simple question
18, 2009 12:32
Subject: Re: [Jprogramming] Rank problem
To: Programming forum programming@jsoftware.com
Thank you .. yes # is what I should have used ... obvious in
hindsight.
David
On Wed, 2009-02-18 at 12:04 -0800, Roger Hui wrote:
The result of $ in this case is a one-element vector
On Wed, Oct 29, 2008 at 8:17 PM, June Kim [EMAIL PROTECTED] wrote:
I was surprised since I originally thought the result of (1 + *)/
should be instead:
1 1 1
1 2 3
1 3 5
(1+*)0/~i.3
1 1 1
1 2 3
1 3 5
From the dictionary:
In general, each cell of x is applied to the entire of y
Can somebody tell me how to get the following
behavior that does not use Bond but with Bond.
$(3 3$0 0 0 0 2 0 0 0 0)|. _1 i. 3 4 5 3
3 4 5 3
$(3 3$0 0 0 0 2 0 0 0 0)|.1 _1 i. 3 4 5 3
3 4 5 3
$(3 3$0 0 0 0 2 0 0 0 0)|.1 _1 _1 i. 3 4 5 3
3 4 5 3
When I use Bond as below
On 3/6/08, Brian Schott [EMAIL PROTECTED] wrote:
Can somebody tell me how to get the following
behavior that does not use Bond but with Bond.
$(3 3$0 0 0 0 2 0 0 0 0)|. _1 i. 3 4 5 3
3 4 5 3
The verb you want to bond with is |._1
In your example, you were bonding with the verb |.
Dear J Forum:
I am applying +/ to arrays that can be of either rank 1 or rank 2. The
following approach works fine for rank 2:
+/ i. 2 5
5 7 9 11 13
But for a rank 1 array, I want just the array itself. I can do that with
+/_1:
+/_1 i. 5
0 1 2 3 4
However +/_1 i.2 5 collapses along
In validate.ijs find ismatrix. (Or change = to :)
ismatrix=: 2 = [EMAIL PROTECTED]
]`(+/)@.ismatrix i. 2 5
5 7 9 11 13
]`(+/)@.ismatrix i. 5
0 1 2 3 4
]`(+/)@.ismatrix 5
5
On Mon, 18 Feb 2008, Leigh J. Halliwell wrote:
+ Dear J Forum:
+
+ I am applying +/ to arrays that
Other possibilities:
+/_1@|:
+/@(,:^:([EMAIL PROTECTED]))
+/^:(1 [EMAIL PROTECTED])
FYI,
--
Raul
--
For information about J forums see http://www.jsoftware.com/forums.htm
Either of these:
+./1 i. 2 5
5 7 9 11 13
+./1 i. 5
0 1 2 3 4
dos=: ] ` (+/) @. (2=[:#$)
dos i. 5
0 1 2 3 4
dos i. 2 5
5 7 9 11 13
The latter is probably more efficient.
On 2/18/08, Leigh J. Halliwell [EMAIL PROTECTED] wrote:
Dear J Forum:
I am applying +/ to arrays that can
Message -
From: Joey K Tuttle [EMAIL PROTECTED]
Date: Friday, February 15, 2008 22:12
Subject: Re: [Jprogramming] Rank of ,.
To: Programming forum programming@jsoftware.com
Better to ask what version of J. Interesting change between
j504 and j601 - more fodder to chew...
version ''
j504
Rank 1 verbs treat rows as items.
Transpose turns columns into rows.
Adverse lets you try one thing and if that fails try the other.
So I suspect that this might work:
nfpt2 ::nfpt1 |: csvdata
--
Raul
--
For information
Joey K Tuttle:
Better to ask what version of J.
Indeed. Somehow this expression was too cute to be true...
I fixed my version now. Thanks to Roger for the step-by-step
equivalency which was helpful in chasing out the bug quickly.
neitzel 8 jfront -q
$ 'abc' ,. . ,. '123'
3 2
9!:14''
On Fri, Feb 15, 2008 at 2:26 AM, Yoel Jacobsen [EMAIL PROTECTED] wrote:
a ,@,.(0 _) b
Note that this gives the same result for your sample data:
a ,@,0 _ b
--
Raul
--
For information about J forums see
,2 a,0/b
1 10 1 20 1 30
2 10 2 20 2 30
3 10 3 20 3 30
--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
--
For information about J forums see http://www.jsoftware.com/forums.htm
(reformatted the original)
Hi,
I have a question about composing functions. I have the following scenario.
1. I read in a csv file.
2. I split the array its constituent columns.
3. The columns are a mix of text and number data.
4. I wrote two verbs that gives me the nub, frequency, and
February 2008 10:56
To: Programming forum
Subject: RE: [Jprogramming] Rank Question
(reformatted the original)
Hi,
I have a question about composing functions. I have the
following scenario.
1. I read in a csv file.
2. I split the array its constituent columns.
3. The columns
Yoel Jacobsen programming@jsoftware.com wrote:
I have experimented with ,. and ,: without success.
What did I miss?
.
a ,. . ,. b
1 10 1 20 1 30
2 10 2 20 2 30
3 10 3 20 3 30
Martin
I am not sure this is what you want but, for the
common parts of the two verbs I think you can use the
following.
(@#/.~) NB. for frequency
(#/.~ @% # ) NB. for %
+
+ nfpt=:([: 1 ~.) ,. ([: 0 #/.~) ,. [: 0 #/.~ % # NB. For Text
columns
+ nfpt2=:([: 0 ~.) ,. ([: 0 #/.~) ,. [: 0
Better to ask what version of J. Interesting change between
j504 and j601 - more fodder to chew...
version ''
j504/2005-03-16/15:30
Running in: Linux
Linux version 2.4.20-28.8 ([EMAIL PROTECTED]) (gcc
version 3.2 20020903 (Red Hat Linux 8.0 3.2-7)) #1 Thu Dec 18
12:53:39 EST 2003
1 2 3
How can I append each element of the RHS list to each element of the LHS list?
Example:
a =. 1 2 3
b =. 10 20 30
I would like a to achieve: a something b that will result in:
1 10 1 20 1 30
2 10 2 20 2 30
3 10 3 20 3 30
I have experimented with ,. and ,: without success.
What did I miss?
Thank you.
I eventually did this:
a ,@,.(0 _) b
Yoel
On Fri, Feb 15, 2008 at 9:18 AM, Chris Burke [EMAIL PROTECTED] wrote:
Yoel Jacobsen wrote:
How can I append each element of the RHS list to each element of the LHS
list?
Example:
a =. 1 2 3
b =. 10 20 30
I would
Yoel Jacobsen wrote:
How can I append each element of the RHS list to each element of the LHS list?
Example:
a =. 1 2 3
b =. 10 20 30
I would like a to achieve: a something b that will result in:
1 10 1 20 1 30
2 10 2 20 2 30
3 10 3 20 3 30
If you append each element of the RHS to each
To rotate a matrix by its axes separately one does
(=i.2)|.i.3 4
4 5 6 7
8 9 10 11
0 1 2 3
1 2 3 0
5 6 7 4
9 10 11 8
This works as expected.
However
(=i.2)|.2 i.3 4 5
|length error
| (=i.2)|.2 i.3 4 5
This is strange, since it should be equal to
(=i.2)|.2 i.3 4
13 september 2007 8:22
Aan: 'Programming forum'
Onderwerp: [Jprogramming] Rank of rotate
To rotate a matrix by its axes separately one does
(=i.2)|.i.3 4
4 5 6 7
8 9 10 11
0 1 2 3
1 2 3 0
5 6 7 4
9 10 11 8
This works as expected.
However
(=i.2)|.2 i.3 4 5
On 9/13/07, R.E. Boss [EMAIL PROTECTED] wrote:
(=i.2)|.2 i.3 4 5
|length error
| (=i.2)|.2 i.3 4 5
...
What's wrong?
Looks like a bug. Replacing |. with a functionally equivalent
verb (such as |.]) works.
--
Raul
The second expression should not have signalled error.
This bug has now been fixed for the next release.
Thank you for finding and reporting it.
- Original Message -
From: R.E. Boss [EMAIL PROTECTED]
Date: Wednesday, September 12, 2007 23:22
Subject: [Jprogramming] Rank of rotate
http://www.jsoftware.com/books/help/dictionary/d300.htm states that the rank
of u.v is 2 _ _
However:
(+/.*) b. 0
_ _ _
--
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Samuel Tardieu wrote:
http://www.jsoftware.com/books/help/dictionary/d300.htm states that the
rank
of u.v is 2 _ _
However:
(+/.*) b. 0
_ _ _
(+/ .*) b. 0
2 _ _
+/.*
+/. *
John
--
For information about J
John == John Randall [EMAIL PROTECTED] writes:
John(+/ .*) b. 0
John 2 _ _
Ooops :-)
--
For information about J forums see http://www.jsoftware.com/forums.htm
Hello
Firstly, have a look at my quick and dirty code:
t=: 0 : 0
1 aaa 10 0.2
2 bbb 2 1
3 ccc 5 5
4 ddd 22 0.987
5 eee 1 0.987
6 fff 100 1
7 ggg 20 0.4
)
t2=:;:;._2
r=:] i.~ [: ~. \:~
keyf=:[: {:1 @(.each)@ {:1
tt=:keyf (\:@[ { :each@(0)@:@[EMAIL PROTECTED] ,. ]) ]
sp=: [ , ' ' , ]
The Vocabulary page for Evoke Gerund says, for the first defined case:
m `: 0Append Appends the results of the individual verbs;
the ranks
are the maxima over their ranks
The second example given
(+ b.0) ; (%. b.0) ; (+`%.`:0 b.0)
+-+-+-+
|0 0 0|2 _ 2|_ _
Subject: [Jprogramming] Rank of Evoke Gerund
The Vocabulary page for Evoke Gerund says, for the first defined case:
m `: 0 Append Appends the results of the individual verbs; the ranks
are the maxima over their ranks
The second example given
(+ b.0) ; (%. b.0) ; (+`%.`:0 b.0
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