(I hope this displays alright, but Thunderbird or
something will probably chuck in some line-throws)
Consider
(i.3) + table ,0
+-+-+
|+|0|
+-+-+
|0|0|
|1|1|
|2|2|
+-+-+
(i.3) + table 1 2
+-+---+
|+|1 2|
+-+---+
|0|1 2|
|1|2 3|
|2|3 4|
+-+---+
and
0 + table i.3
+-+----------+
|+|0 |
+-+----------+
|0| 1 2 0 1 2|
+-+----------+
(,0) + table i.3
+-+-----+
|+|0 1 2|
+-+-----+
|0|0 1 2|
+-+-----+
These demonstrate that table requires vector arguments.
This can force them to be so:
0 (+ table)&, i.3
+-+-----+
|+|0 1 2|
+-+-----+
|0|0 1 2|
+-+-----+
0 (+ table)&,~ i.3
+-+-+
|+|0|
+-+-+
|0|0|
|1|1|
|2|2|
+-+-+
I wondered if I was correct in saying "vector", but I
think this shows they should be; if, for example, we
have a matrix argument, the results are not too
helpful:-
(i.3) (+ table)i.2 2
+-+---+
|+|0 1|
| |2 3|
+-+---+
|0|0 1|
|1|2 3|
|2| |
| |1 2|
| |3 4|
| | |
| |2 3|
| |4 5|
+-+---+
(i.3) (+ table)~i.2 2
+---+-----+
|+ |0 1 2|
| |0 0 0|
+---+-----+
|0 1|0 1 2|
|2 3|1 2 3|
| | |
| |2 3 4|
| |3 4 5|
+---+-----+
whereas
(i.3) (+ table)&, i.2 3
+-+-----------+
|+|0 1 2 3 4 5|
+-+-----------+
|0|0 1 2 3 4 5|
|1|1 2 3 4 5 6|
|2|2 3 4 5 6 7|
+-+-----------+
Any help?
Mike
On 09/10/2014 06:29, Linda Alvord wrote:
Try this instead:
(i.3) +/ 0
(i.3) +/1
(i.3) +/1 2
(i.3) + table 0
(i.3) + table 1
(i.3) + table 1 2
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Linda Alvord
Sent: Thursday, October 09, 2014 1:25 AM
To: [email protected]
Subject: [Jbeta] tab;e vs /
I expect them to agree. Is this how table should look?
(i.3) +/ 0
0 1 2
(i.3) +/1
1 2 3
(i.3) +/1 2
1 2
2 3
3 4
(i.3) + table 0
--T------┐
│+│0 │
+-+------+
│0│ 0 1 2│
│1│ │
│2│ │
L-+-------
(i.3) + table 1
--T------┐
│+│1 │
+-+------+
│0│ 1 2 3│
│1│ │
│2│ │
L-+-------
(i.3) + table 1 2
--T---┐
│+│1 2│
+-+---+
│0│1 2│
│1│2 3│
│2│3 4│
L-+----
Linda
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