Thank you all so much for all your excellent suggestions. In fact, Bo's
idea was what I started with. But two things (not mentioned in my original
mail) prevented me from using that approach. First, my vector space has 32
dimensions, and most of my vectors will be sparse. Second, I have a
somewhat
you can also look at the base vectors using complex numbers
+. +/ 2.3 0j3.9 3.1
5.4 3.9
Don Kelly
On 2019-11-25 1:39 p.m., 'Bo Jacoby' via Programming wrote:
Enumerate your base vectors 1 0 and 0 1 etc.
(2.3*1 0)+(3.9*0 1)+(3.1*1 0)
5.4 3.9
Thanks.
Bo.Den mandag den 25. novem
Enumerate your base vectors 1 0 and 0 1 etc.
(2.3*1 0)+(3.9*0 1)+(3.1*1 0)
5.4 3.9
Thanks.
Bo.Den mandag den 25. november 2019 19.45.04 CET skrev Raul Miller
:
Or, slightly more efficient (if that matters):
V1=: |:(2 3 4;2.3),(1 2;3.9),: 2 3 4;3.1
V2=: (~.@{.,:+/&.:>/./)V1
Or, slightly more efficient (if that matters):
V1=: |:(2 3 4;2.3),(1 2;3.9),: 2 3 4;3.1
V2=: (~.@{.,:+/&.:>/./)V1
V2-:|:(2 3 4;5.4),: 1 2;3.9
(Caution: email clients have become hostile to copy&paste, and may not
represent code accurately.)
Thanks,
--
Raul
On Mon, Nov 25, 2019 at 1:1
You've already got a good reply from Raul, but I'll offer these variants,
having spent a little while looking at the problem.
The first one possibly overuses what I think of as each (&.>) and every
(&>) :
f1 =: (({: every) ((~.@[);~each ((+/)/.)) >@({.every) )
It does the key operation
Your approach is exactly what my f does!
That said, there's other ways of phrasing this. Also, for this
specific calculation, it would be easier to work with a representation
of the form
V1=: |:|.@> v1
V2=: |:|.@> v2
Then:
V2-:(~.@{.,:+&.>//./)V1
1
It may help to use the dissect utilit
Thanks, Raul. You are always so prompt with your help! I am yet to decipher
your solution, but your suggestion regarding 2 column table is bang on
point.
My own approach was something like this: find nub of the 2nd column, then
for each element in the nub collect all occurences of it, and sum over
Yes, definitely.
For example, consider:
<"1 ({:"1 (+&.>/@:({."1), (<0 1){])/. ])>v1
Or,
f=: 13 :'<"1 ({:"1 (+&.>/@:({."1), (<0 1){])/. ])>y'
f v1
That said, this example may fall apart for your general case? (I don't
know what that is...)
Specifically, I don't know why the top level str
Dear all,
I am trying to implement a geometric algebra system in J. I have done
much of the stuff, but is getting stuck at one point. Hence this email.
Basically, I have a list of boxes like
v1=: (<2.3; 2 3 4), (<3.9; 1 2), <3.1; 2 3 4
This represents a (multi)vector whose math representatio
