Sorry for previous post a few minutes ago - I pressed send before typing!
Yes, true - about wasting the last element. Also, I realise I'd
defined ∆S wrongly
in my attempt at analysing the process, misinterpreting (}. % {. as (}.
% }:) ) .
Apologies for both...
Mike
On 27/08/2020 08:02, 'Bo Jacoby' via Programming wrote:
Mike wrote: "I don't like losing the last element of s - we're just wastingthat
calculation."
The last element of vector s does not enter into the matrix m, but (%.m)s needs
the last element of s.
Thanks!
Bo
Den tirsdag den 25. august 2020 04.07.55 CEST skrev 'Bo Jacoby' via Programming
<[email protected]>:
Thanks to Mike and Raul!
Mike asked: "Can you provide a reference for your generalised statistics?" No sir, I gave
myself the task: "find 3 numbers having the same mean powers as those of a given
dataset", and I also solved it myself. I appreciate the help from the programming forum for
cleaning up my smelly code.
U=.summary@tomb NB. for brevity
power=.(^/>:@i.)~
mean=.+/%#
3 mean@power #~i.5
3 10 35.4
3 mean@power 3 U #~i.5
3 10 35.4
3 U #~i.5 NB. these 3 numbers solve the problem
1.61014 3.4686 3.92126
Raul's example:
3 mean@power a=.0 1 2 3 4 5 6 7 8 9 10 20 20 20 20 20
9.6875 149.063 2689.06
3 U a NB. this answers the question
1.92566 7.43347 19.7034
3 mean@power 3 U a NB. the same mean powers as a .
9.6875 149.063 2689.06
3 mean@power 4.77812 13.4155 10.8689 NB. not right
9.68751 106.98 1269.18
1 U a NB. mean value
9.6875
4 U a
1.90665 3.98091 12.1419 20.7206
3 mean@power 4 U a NB. the same mean powers as a .
9.6875 149.062 2689.06
Thanks.
Bo.
Den mandag den 24. august 2020 20.37.04 CEST skrev 'Michael Day' via Programming
<[email protected]>:
More particularly - re - Bo Jacoby's original statement on 21/8/20.
Probably not
relevant if you haven't followed this branch of the thread.
On the code-smelliness topic - no anosmia here! - I've been nurdling
away at
your (Bo's) functions for generalisation of summary stats, even though
not yet
understanding what mathematical problem you're solving.
So far I've got: (using fixed font...)
NB. stomb to replace s @ tomb
stomb =: (% #) * (+/@:(^/) >:@i.)~
NB. Ma to replace m
NB. managing to treat i. as a common element,
NB. albeit needing to modify it for some steps
Ma =: |:@(((_1 ^ [) * (0 - [) |.!.0"0 1 (,"0 1~ >:)~) }:)~ i.@#
cf
m f. NB. original
|:@(|: |.!.0"0 1~ -@i.@#)@((* (_1 ^ i.@#))"1@(>:@i.@# , }:)@|:@($~ (2 # #)))
[data =. #~ i.5
1 2 2 3 3 3 4 4 4 4
3 stomb data
9 30 106.2
3 s @ tomb data
9 30 106.2
3 Ma @ stomb data
1 0 0
9 _2 0
30 _9 3
3 m @ s @ tomb data
1 0 0
9 _2 0
30 _9 3
But - I don't like losing the last element of s - we're just wasting
that calculation.
As for the maths - you've got numbers ni associated with indices i e. i.N
Let Sk = sum over i of ni.i^k for k e. i. m + 1
Then ∆Sk = m Sk % S[k-1] for k e. 1 + i.m ( ∆ is Delta if the
formatting fails!)
matrix M is [ 1 0 0 0 ...]
[∆S1 -2 0 0 ...]
[∆S2 -∆S1 3 0 ...]
[∆S3 -∆S2 ∆S1 -4 ...]
............
[..... ..... ...]
You then solve for vector b in
∆S = M matprod b
so we have
∆S1 = b1
∆S2 = b1 ∆S1 - 2 b2
∆S3 = b1 ∆S2 - b2 ∆S1 + 3 b3
....
You then append 1 to b, presumably defining a coefficient, b0 = 1 ,
and then find - and negate - the roots of the polynomial
b0 x^m + b1 x^(m-1) + ... + bm = 0 (I think)
Why? I don't see a direct relation to Newton interpolation (say),
though there
are some similarities. Can you provide a reference for your generalised
statistics?
Thanks,
Mike
On 23/08/2020 15:58, 'Bo Jacoby' via Programming wrote:
Raul wrote: "Generalized how?"
The special case n=1 of n&(summary@tomb) is the mean value (+/%#).So the general
case n&(summary@tomb) is a generalized mean value.
Correctness:
1 summary@tomb a=. (i.11) ,5#20
9.6875
(+/%#) a
9.6875
1 summary@tomb 3 summary@tomb a9.6875
((+/%#) 3 summary@tomb a) = (+/%#) a
1
3 summary@tomb a1.92566 7.43347 19.7034
(+/%#)1.92566 7.43347 19.70349.68751
1 summary@tomb 1.92566 7.43347 19.7034
9.68751
Thanks!
Bo
Den søndag den 23. august 2020 10.01.24 CEST skrev Raul Miller
<[email protected]>:
For example, consider this crude mechanism:
partmean=:4 :0
m=. (+/%#) y
b=. <.(x%n)*i.#y
w=. b (+/%#)/. /:~ y
x * m * w%+/w
)
3 partmean 0 1 2 3 4 5 6 7 8 9 10 20 20 20 20 20
2.38217 7.62295 19.0574
Correct? Incorrect? How would I know?
Note that the calculation for b could be replaced, for example:
arbmean=:4 :0
m=. (+/%#) y
b=. ?. (#y) # x
w=. b (+/%#)/. y
x * m * w%+/w
)
3 arbmean 0 1 2 3 4 5 6 7 8 9 10 20 20 20 20 20
4.77812 13.4155 10.8689
(+/%#)3 arbmean 0 1 2 3 4 5 6 7 8 9 10 20 20 20 20 20
9.6875
Thanks,
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