Hi David,
thanks, fixed in SVN 1046.
/// Jürgen
On 04/20/2018 10:16 PM, David Tran
wrote:
Hi,
a ← 2 3 5 6
b ← 4 5 5 6
a -.= b
¯2
Hi,
a ← 2 3 5 6
b ← 4 5 5 6
a -.= b
¯2 ⍝ bug? the value should be zero
-/a=b
0
Version info:
~ $ apl --version
BUILDTAG:
-
Project:GNU APL
Version / SVN: 1.7 / Unversioned directory
Build Date: 2017-10-05 09:52:49 UTC
Build OS:
Thank you.
Forgot to mention scalar extension. When a scalar is supplied as an
argument, it's lengthened to match the inner dimension of the other
argument.
(3 4⍴'cattrattfatt') +.= 't'
2 2 2
(3 4⍴'cattrattfatt') +.= ''
2 2 2
'a' +.= 3 4⍴'catratfatbat'
1 1 1 1
'aaa' +.
very nice work - thanks you for spending the time .. everthing here holds
when replacing +.x with +.= (which is of more interest to me)
On Fri, 17 Mar 2017 22:01:26 +0100
Nick Lobachevsky wrote:
> The key to understanding inner product is that the inner dimensions of
> the arguments have t
The key to understanding inner product is that the inner dimensions of
the arguments have to be the same. The inner dimension here is 3.
a←2 3⍴⍳6
b←3 4⍴⍳12
a
0 1 2
3 4 5
b
0 1 2 3
4 5 6 7
8 9 10 11
a+.×b
20 23 26 29
56 68 80 92
To solve this, first transpose th
Hi
I assume you mean the vectors both have same length as their single dimension
(no first or last - the same)
making them into 1 4⍴ arrays then requires the ⍉ use
don't you treat vectors differntly in the gnuapl code from 'true arrays' with 2
(or more dimensions)?
On Fri, 17 Mar 2017 18:1
Hi,
that would be more like an outer product:
a←b←5
3⍴⍳8
⍴a∘.=b
5 3 5 3
The inner product reduces away the two middle axes of the outer
product.
In the ⍉b case you have
⍴a∘.=⍉b
5 3 3 5
Hi,
yes. Because the first dimension of a is 4 and the last
dimension of b is also 4.
/// Jürgen
On 03/17/2017 06:05 PM, enz...@gmx.com
wrote:
yes last dimension of a must be = to first dimension of b
but then
x←1 2
thanks
i have no clue how the result of a+.=⍉b as 5x5 result is obtained from a 5x3
array and 3x5 array shouldn't it be a 15x15 result ... but i guess i
really just want to use +/a=b :)
On Fri, 17 Mar 2017 12:26:42 -0400
Christian Robert wrote:
>a+.=⍉b
> 3 0 0 0 0
> 0 3 0 0 0
yes last dimension of a must be = to first dimension of b
but then
x←1 2 3 4
y←1 3 3 3
x+.=y works
On Fri, 17 Mar 2017 17:29:24 +0100
Juergen Sauermann wrote:
> Hi,
>
> the number of columns in a must match the number of rows in b (or the other
> way around,
> I don't quite remember).
Hi,
the number of columns in a must match the number of rows
in b (or the other way around,
I don't quite remember).
Try:
a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3
/// Jürgen
a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3
On 2017-03-17 12:17, enz...@gmx.com wrote:
Hi
what am i missing here?
a←b←5 3⍴⍳8
a=b
+/a=b
a+.=b length error
Hi
what am i missing here?
a←b←5 3⍴⍳8
a=b
+/a=b
a+.=b length error
Hi Xiao-Yong,
thanks, fixed in SVN 771.
/// Jürgen
On 07/05/2016 10:12 PM, Xiao-Yong Jin
wrote:
Hi, here goes another micro error.
2 3+.{⍺×⍵}2 3
LENGTH ERROR
μ-Z__A_LO_INNER_RO_B[5] (μ-IA μ-IB)←⊃μ-I[μ-N]
Hi, here goes another micro error.
2 3+.{⍺×⍵}2 3
LENGTH ERROR
μ-Z__A_LO_INNER_RO_B[5] (μ-IA μ-IB)←⊃μ-I[μ-N]
^ ^
)reset
2 3{⍺+⍵}.{⍺×⍵}2 3
LENGTH ERROR
μ-Z__A_LO_INNER_RO_B[5] (μ-IA μ-IB)←⊃μ-I[μ-N]
^ ^
)r
Al this gossip about outer product - I think inner product needs some exposure
X ← 1 3 5 7
Y ← 2 3 6 7
X +.= Y
2
Hi,
the standard says:
If the last-item in the shape-list of A1 is not the same as the
first-item in the shape-list
of B1, signal length-error.
Scalar A or B are turned into vectors, but vectors are not turned
into matrices.
Hi Bug APL,Does the inner product coerce a vector to a matrix only when (⍴right arg vector)≤ left arg number of columns (when it is a matrix)? rotatemat←{2 2 ⍴ (2○⍵)(-1○⍵)(1○⍵)(2○⍵)} rm←rotatmat 3VALUE ERROR rm←rotatmat 3 ^ rm←rotatemat 3 rm¯0.9899924966 ¯0.14112000
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