Have you decided whether your code does implement the PAM algorithm?
xash suggested it might be doing the Voronoi method.
Thanks,
Mike
On 15/02/2021 18:16, Emir U wrote:
Hey again xash, I've improved the algo some following your example (especially
the use of ./). It has reduced the algo (at
Well, it's fixed for the next maintenance release of 9.02 as well, but I
don't know when that will be.
Henry Rich
On 2/15/2021 1:31 PM, 'Michael Day' via Programming wrote:
"Fixed for the next beta" suggests you might have overlooked that this
problem seems to occur in J9.02 as well.
Apologie
"Fixed for the next beta" suggests you might have overlooked that this
problem seems to occur in J9.02 as well.
Apologies if I'm wrong.
Cheers,
Mike
On 15/02/2021 17:16, Henry Rich wrote:
It comes down to this:
<. 17575011601890. 0 NB. bug!
17575011601891 0
<. 17575011601890.
175750
Hey again xash, I've improved the algo some following your example (especially
the use of ./). It has reduced the algo (at least with random initialisation)
to 3 lines ...
kM_step=: 4 : 0
vec=. (i. <./) each ;/ x {"1 y
argm=. 4 : '{ & x (i. <./) (+/ (<(
Subject: Re: k-mediods PAM algo, help wi
It comes down to this:
<. 17575011601890. 0 NB. bug!
17575011601891 0
<. 17575011601890.
17575011601890
When I rewrote >. to allow inplace operation, I forgot that comparison
tolerance causes trouble when the numbers get big. Singletons follow a
different path.
Fixed for next beta.
kM2 takes in `data`, not the `edist data`. Otherwise, it should work.
Though after I realized what this all does, taking in the distance matrix
makes more sense. Thus:
kM2_step=: 4 : 0
groups=. (i. <./)@:{"1
sub=. {"1 {&x
best=. [ {~ [:(i. <./) [:+/"1 sub
(y groups x) best/. (i. # x)
)
kM2=: ] kM2
Sorry, slightly careless in drafting - I'll correct the errors inline.
Please discard my earlier post.
On 15/02/2021 13:16, 'Michael Day' via Programming wrote:
Yes, it does seem to misbehave.
I looked at what happens with @ rather than @:
t0=: <.@ -:@ (* >:)
t1 =: <.@:-:@:(* >:)
(t0
Yes, it does seem to misbehave.
I looked at what happens with @ rather than @:
t0=: <.@ -:@ (* >:)
t =: <.@:-:@:(* >:)
(t0,:t1) 5928745 5928745 NB. compare @ with @:
17575011601885 17575011601885
17575011601886 17575011601886
I thought [: might fix it - but no:
tbr0=: [: <. [: -:
Thanks for your comments xash. You're right about average not mattering. Did
your kM2_step and kM functions run correctly for you? It doesn't look like it
runs to me so far as I can see the group function seems to assume the data to
be in the wrong shape for it to work. Can you confirm?
If you have
f=: 4 : '*: x (-"1) y'
And you want an explicit ed, defined in terms of f, which is equivalent to
ed=:+/&.:*:@:-"1
Then I would use:
ed=: 4 : '%: +/ x f y'"1
Note that "1 needs to be outside the explicit definition for ed/~ to
recognize that ed consumes rank 1 items.
I hope
Emir,
Maybe something like this
ed2=:4 : '+/ *: x - y'
ed2m=: ed2"1 _"_ 1~
ed2m 1 2 ,:4 6
0 25
25 0
Note table is just the verb with some rank. You can say something like
apply ed2 for all vectors on the left and all vectors on the right with
something two ranks like the above.
Best, Cliff
Dear all,
I had an issue when calculating the triangular number with the formula
<.@:-:@:(*
>:) in list context. Starting from 5928745 the outcome in list context is
larger than the correct value (additionally calculated with the binomial
coefficient in the first box):
t=: <.@:-:@:(* >:)
((2
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