The "calendar" verb from J's "stdlib" has an optional left argument that
allows you to specify the starting day of the week:
calendar=: 3 : 0
0 calendar y
:
a=. ((j<100)*(-100&|){.6!:0'')+j=. {.y
b=. (a-x)+-/<.4 100 400%~<:a
r=. 28+3,(~:/0=4 100 400|a),10$5$3 2
r=. (-7|b+0,+/\}:r)|."0 1 r(]&:>:*"1
Is a two-step process too ugly?
a
=:6 7$ 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 1011
12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 0 0 0 0 0
(a='_')}' ',:~a=: ":(a=0)}_,:~a
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
3
osts.
On 09/28/2017 08:00 AM, programming-requ...@forums.jsoftware.com wrote:
Date: Wed, 27 Sep 2017 17:57:54 +0200
From: Rudolf Sykora
To:programm...@jsoftware.com
Subject: [Jprogramming] print an array
Message-ID:
Content-Type: text/plain; charset="UTF-8"
Hello,
I have an ar
On 28 September 2017 at 15:43, Raul Miller wrote:
> But, yes, this works:
>
>r=: , LF,.~": _7 ]\ 0 >. i.&.(+&5) 32
So this is much in the way I wanted it at the very begining.
Thanks.
Ruda
--
For information about J foru
Yes, you're right in a way about @: , but I'd skipped defining a verb
to do the job.
What I had in mind, but didn't explicitly show, was something like
remz =: ($$rplc & (' 0';' ')) @ ": NB. @ also works ok!
so allowing
remz a
etc.
And yes, the $$ restores the shape of the rave
I did not write stringreplace, so am not prepared to discuss its
implementation tradeoffs.
But, yes, this works:
r=: , LF,.~": _7 ]\ 0 >. i.&.(+&5) 32
r
0 0 0 0 0 0 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31 0 0 0 0 0
r rpl
On 28 September 2017 at 14:50, Raul Miller wrote:
> Another way to work around this problem would be to use stringreplace"1
>
>r=: _7 ]\ 0 >. i.&.(+&5) 32
>(":r) rplc"1 ' 0';' '
Ok. Thanks. So is there a good reason for txt=. ,y in stringreplace?
Or, can the whole of r be represented as
If you look at the definition of stringreplace, the first line is
txt=. ,y
So it's working with ,r
Another way to work around this problem would be to use stringreplace"1
(Though, personally, I prefer rplc over stringreplace, because that
leaves more room for other stuff on my lines. Also, ne
On 27 September 2017 at 19:13, 'Mike Day' via Programming
wrote:
>($$rplc & (' 0';' ')) @: ": a
Yes. (I guess the @: is unnecessary here, right?)
I actually had tried something similar first:
(' 0';' ') stringreplace ":r
which, however, yielded
1 2 3 4 5 6 7 8 9 10 11 12 13 14 1516 17 18
On 27 September 2017 at 19:44, chris burke wrote:
> a=. (a<32) * a=. 0 >. _5 + i.6 7
>
>
> 'b<>3.0' (8!:2) a
Thanks, this is probably the expected form, though
I still have to find out, what exactly it does. :))
Ruda
---
On 28 September 2017 at 01:21, Roger Hui wrote:
> If you are making a calendar you may find the following of interest:
> http://www.jsoftware.com/papers/eem/qq101.htm . It's APL but the ideas
> translate pretty readily into J.
Thanks.
Actually, I am not making a calendar :), I just wanted to see
On 28 September 2017 at 01:09, Devon McCormick wrote:
> Assuming a is a numeric matrix:
>
>a=. ":a
>(a='0')}a,:' '
>1
> 2 3 4 5 6 7 8
> 9 1 11 12 13 14 15
> 16 17 18 19 2 21 22
> 23 24 25 26 27 28 29
> 3 31
>
>
see the missing zero in 10, 20, 30; isolated ze
And there is a calendar verb in the standard library that may be able to
produce what you want.
calendar 2017 4
┌─┐
│ Apr │
│ Su Mo Tu We Th Fr Sa│
│1│
│ 2 3 4 5 6 7 8│
│ 9 10 11 12 13 14 15│
│ 16 17 18 19 20 21 22│
│ 23 24 25 2
If you are making a calendar you may find the following of interest:
http://www.jsoftware.com/papers/eem/qq101.htm . It's APL but the ideas
translate pretty readily into J.
On Wed, Sep 27, 2017 at 8:57 AM, Rudolf Sykora
wrote:
> Hello,
>
> I have an array of integers
>
> 0 0 0 0 0 0 1
Assuming a is a numeric matrix:
a=. ":a
(a='0')}a,:' '
1
2 3 4 5 6 7 8
9 1 11 12 13 14 15
16 17 18 19 2 21 22
23 24 25 26 27 28 29
3 31
On Wed, Sep 27, 2017 at 1:44 PM, chris burke wrote:
> a=. (a<32) * a=. 0 >. _5 + i.6 7
>
>
> 'b<>3.0' (8!:2) a
a=. (a<32) * a=. 0 >. _5 + i.6 7
'b<>3.0' (8!:2) a
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31
On Wed, Sep 27, 2017 at 10:13 AM, 'Mike Day' via Programming <
programm...@jsoftware.com> wrote:
> Assuming a
Assuming a is an integer array, not character,
how about:
($$rplc & (' 0';' ')) @: ": a
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31
?
(Leading zeros should only appear for zero itself).
Though isn't there a print/format fun
a =: 0 : 0
0 0 0 0 0 0 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31 0 0 0 0 0
)
$b =: ".;._2 a
6 7
>,each ;/(3&":`(' '"_))@.(0=])"0"1 b
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25
Hello,
I have an array of integers
0 0 0 0 0 0 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31 0 0 0 0 0
and I want to have it printed (to a string, say) without
the zeros. I.e, I want to replace each isolated 0 with a space.
How can I do i
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