Hi Ian. Here is your solution.
ABC=: 3 : 0
c=.C y
y=.(0,>:c){RT"1&(,H&#)y
y=.y,(%/1{"1 y)*1{y
y=.y,(0{y)-2{y
y=.y,(#|:y){.{.{:y
y=.4 2{{:"1 RT"1 y
y,c
)
ABC 101 103 100 210 230 200
101.15 115.101 2
- Bo
>________________________________
> Fra: Ian Clark <earthspo...@gmail.com>
>Til: Programming forum <programming@jsoftware.com>
>Sendt: 6:46 mandag den 25. juni 2012
>Emne: Re: [Jprogramming] Kalman filter in J?
>
>Interesting approach, Bo.
>
>Turned my attention to CUSUM (an adaptive filter) which is rather easy
>to compute and gives good results, but needs parameters setting "by
>eye". I've yet to try a Bayesian approach, which I suspect will be the
>most sensitive of all.
>
>But I did play with your FT, using *:@(10&o.) to plot a power
>spectrum. I'm using rather noisier data than your example: my step is
>only roughly (sigma) high. Nevertheless with the onset of the step I
>can clearly see a high-frequency spike appear at the far end of the
>transformed time-series, due to the transformed Heaviside fn. Same
>problem as with the CUSUM statistic however: designing a detector
>which can be adjusted for false negatives and positives, and doesn't
>take too many subsequent samples to detect the step.
>
>Will try out your C function in my context, and try to tweak it. Your
>solution detects when the step happens, solving 1 in my stated
>objectives: 1-4. I fear however that I am more interested in 2-4.
>
>
>
>On Sun, Jun 24, 2012 at 6:06 PM, Bo Jacoby <bojac...@yahoo.dk> wrote:
>> Hi Ian
>> The Fourier Transform FT transforms a complex n-vector X into a complex
>> n-vector Y.If X is real, then Y is symmetric. And if X is symmetric then Y
>> is real. So I think it is a good idea to symmetrize X before taking the FT.
>> (This makes a slow program even slower. Optimize later!)
>>
>>
>> NB.sm=: symmetrize
>> sm=:[:,}:"1&(,:|.)
>> sm i.6 NB. see what I mean
>>
>> 0 1 2 3 4 5 4 3 2 1
>> NB.RT=: Real Transform. (9 o. removes every trace of complexity)
>> RT=:9 o.#{.FT&sm
>> NB.H=:Heaviside steps
>> H=:i.&<:</i.
>>
>> NB.C=:Index of the last item before the change
>> C=:[:(i.<./)[:+/"1[:*:[:}.[:(-"1{.)[:}."1[:(%{."1)[:}."1[:RT"1],[:H#
>> NB.test
>> C 101 204 202 200 203
>> 0
>> C 101 104 202 200 203
>> 1
>> C 101 104 102 200 203
>> 2
>> C 101 104 102 100 203
>> 3
>>
>> This is not a Kálmán filter, but it solves the problem.
>> - Bo
>>
>>
>>
>>>________________________________
>>> Fra: Ian Clark <earthspo...@gmail.com>
>>>Til: Programming forum <programming@jsoftware.com>
>>>Sendt: 3:20 fredag den 22. juni 2012
>>>Emne: Re: [Jprogramming] Kalman filter in J?
>>>
>>>Thanks, Bo. I'll look at it when I get back, probably over the weekend.
>>>
>>>Ian
>>>
>>>On Fri, Jun 22, 2012 at 12:44 AM, Bo Jacoby <bojac...@yahoo.dk> wrote:
>>>> Hi Ian.
>>>> The following is a tool, but not yet a solution.
>>>>
>>>> FT is a slow Fourier Transform. When the problem has been solved it is
>>>> time to optimize. This FT has the nice property that FT^:_1 is identical
>>>> to FT.
>>>>
>>>> NB.FT =: Fourier Transformation
>>>> FT =: +/&(+*((%:%~_1&^&+:&(%~(*/])&i.))&#))
>>>> NB.rd =: round complex number.
>>>> rd =: (**|)&.+.
>>>> NB.H =: Heaviside step function
>>>> H =: (]#0:),-#1:
>>>> 10 H 3
>>>> 0 0 0 1 1 1 1 1 1 1
>>>> NB. FT = FT^:_1
>>>> rd FT FT 10 H 3
>>>> 0 0 0 1 1 1 1 1 1 1
>>>>
>>>>
>>>> NB. For fixed C the values of A and B to minimize +/*:X-A+B*(#X)H C are
>>>> found by solving (2{.FT X) = 2{.FT A+B*(#X)H C
>>>>
>>>> - Bo
>>>>>________________________________
>>>>> Fra: Ian Clark <earthspo...@gmail.com>
>>>>>Til: Programming forum <programming@jsoftware.com>
>>>>>Sendt: 18:15 torsdag den 21. juni 2012
>>>>>Emne: Re: [Jprogramming] Kalman filter in J?
>>>>>
>>>>>Yes, Bo, that looks like a fair statement of the problem, in terms of
>>>>>least-squares minimization.
>>>>>
>>>>>
>>>>>On Thu, Jun 21, 2012 at 4:46 PM, Bo Jacoby <bojac...@yahoo.dk> wrote:
>>>>>> Consider the step functions
>>>>>>
>>>>>> H=:(]#0:),-#1:
>>>>>> 10 H 0
>>>>>> 1 1 1 1 1 1 1 1 1 1
>>>>>> 10 H 3
>>>>>> 0 0 0 1 1 1 1 1 1 1
>>>>>> 10 H 10
>>>>>> 0 0 0 0 0 0 0 0 0 0
>>>>>> The problem is to find constants A,B,C to minimize the expression
>>>>>>
>>>>>> +/*:X-A+B*(#X)H C
>>>>>> Right?
>>>>>> -Bo
>>>>>>>________________________________
>>>>>>> Fra: Ian Clark <earthspo...@gmail.com>
>>>>>>>Til: Programming forum <programming@jsoftware.com>
>>>>>>>Sendt: 14:34 torsdag den 21. juni 2012
>>>>>>>Emne: Re: [Jprogramming] Kalman filter in J?
>>>>>>>
>>>>>>>Thanks, Bo and Ric. Yes, I'd tried searching jwiki on "Kalman" and
>>>>>>>Pieter's page was the only instance.
>>>>>>>
>>>>>>>I neglected to mention that in the most general case I can't really be
>>>>>>>confident that X1 and X2 have the same distribution function, let
>>>>>>>alone the same variance. But looking at it again, I see that under the
>>>>>>>restrictions I've placed the problem simplifies immensely to fitting a
>>>>>>>step-function H to: X=: K+G+H. If I can just do that, I'll be happy
>>>>>>>for now.
>>>>>>>
>>>>>>>Repeated application of FFT should allow me to subtract the noise
>>>>>>>spectrum F(G), or at least see a significant change in the overall
>>>>>>>spectrum emerge after point T, and that might handle the more general
>>>>>>>cases as well.
>>>>>>>
>>>>>>>Anyway it's simple-minded enough for me, and worth a try. FFT is,
>>>>>>>after all, "fast" :-)
>>>>>>>
>>>>>>>But won't an even faster transform do the trick, such as (+/X)? On the
>>>>>>>above model, X performs a drunkard's walk around a value M1 until some
>>>>>>>point T, after which it walks around M2. Solution: simply estimate M1
>>>>>>>and M2 on an ongoing basis.
>>>>>>>
>>>>>>>I get the feeling I ought to be searching on terms like "edge
>>>>>>>detection", "step detection" and CUSUM.
>>>>>>>
>>>>>>>Anyway, there's enough here to try.
>>>>>>>
>>>>>>>On Thu, Jun 21, 2012 at 10:28 AM, Ric Sherlock <tikk...@gmail.com> wrote:
>>>>>>>> Hi Ian,
>>>>>>>>
>>>>>>>> A quick search of the J wiki finds this:
>>>>>>>> http://www.jsoftware.com/jwiki/Stories/PietdeJong
>>>>>>>> Sounds like he might have what you're after?
>>>>>>>>
>>>>>>>> Cheers,
>>>>>>>> Ric
>>>>>>>>
>>>>>>>> On Thu, Jun 21, 2012 at 4:33 PM, Ian Clark <earthspo...@gmail.com>
>>>>>>>> wrote:
>>>>>>>>> Can anyone help? Has anyone written a Kalman filter in J?
>>>>>>>>>
>>>>>>>>> I'm not a specialist in either statistics or control theory, so I'm
>>>>>>>>> only guessing a Kalman filter is what I need. Though I do have a
>>>>>>>>> passing acquaintance with the terms: stochastic control and linear
>>>>>>>>> quadratic Gaussian (LQG) control. I am aware that a "Kalman filter"
>>>>>>>>> (like ANOVA) is more a topic than a black-box.
>>>>>>>>>
>>>>>>>>> So let me explain what I want it for.
>>>>>>>>>
>>>>>>>>> I have a time series X which I am assuming can be modelled like this:
>>>>>>>>>
>>>>>>>>> X=: K + G + (X1,X2)
>>>>>>>>>
>>>>>>>>> where
>>>>>>>>>
>>>>>>>>> K is constant
>>>>>>>>> G is Gaussian noise
>>>>>>>>> X1 is a random variable with mean: M1 and variance: V1
>>>>>>>>> X2 is a random variable with mean: M2 and variance: V2
>>>>>>>>>
>>>>>>>>> Typically X is a sequence of sensor readings, but may also be
>>>>>>>>> measurements from a series of user trials conducted on a working
>>>>>>>>> prototype, which suffers a design-change at a given point T.
>>>>>>>>>
>>>>>>>>> Simplifying assumptions (which unfortunately I may need to relax in
>>>>>>>>> due course):
>>>>>>>>>
>>>>>>>>> (a) X is not multivariate
>>>>>>>>> (b) X1 and X2 are Gaussian
>>>>>>>>> (c) V1=V2 (only the mean value changes, not the variance).
>>>>>>>>>
>>>>>>>>> The problem:
>>>>>>>>>
>>>>>>>>> 1. Estimate T=: 1+#X1 -- the point at which X1 gives way to X2.
>>>>>>>>>
>>>>>>>>> 2. Given T, estimate (M2-M1) -- the "underlying improvement", if any,
>>>>>>>>> of the change to the prototype.
>>>>>>>>>
>>>>>>>>> 3. (subcase of 2.) Given T, test the null hypothesis: M1=M2, viz that
>>>>>>>>> there has been no underlying improvement.
>>>>>>>>>
>>>>>>>>> 4. Estimate U=: #X2 -- the minimum number of samples needed after T in
>>>>>>>>> order to achieve 1-3 above with 95% confidence.
>>>>>>>>>
>>>>>>>>> In other words, detect the signal-in-noise: M1-->M2, and do so in
>>>>>>>>> real-time.
>>>>>>>>>
>>>>>>>>> Because of 4, the need to estimate T and (M2-M1) on an ongoing basis,
>>>>>>>>> I can't do a randomised block design. I gather that a Kalman filter,
>>>>>>>>> or some sort of adaptive filter, will handle this problem.
>>>>>>>>>
>>>>>>>>> But maybe something simpler will turn out good enough?
>>>>>>>>>
>>>>>>>>> Supposing I can get hold of a "black box" Kalman filter, I propose to
>>>>>>>>> test it out on generated data and compare its performance to some
>>>>>>>>> simple-minded approach, like estimating M1 / M2 from a simple moving
>>>>>>>>> average of the last U samples, or applying the F-test to 2 sets of U
>>>>>>>>> samples taken either side of T.
>>>>>>>>>
>>>>>>>>> But since the technique aims to be published, or at least critically
>>>>>>>>> scrutinised (and maybe incorporated in a software product), I'd rather
>>>>>>>>> depend on a state-of-art packaged solution than reinvent the wheel: a
>>>>>>>>> large and very well-turned wheel it appears to me.
>>>>>>>>>
>>>>>>>>> Ian Clark
>>>>>>>>> ----------------------------------------------------------------------
>>>>>>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>>>>>>----------------------------------------------------------------------
>>>>>>>For information about J forums see http://www.jsoftware.com/forums.htm
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>> ----------------------------------------------------------------------
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>>>>>----------------------------------------------------------------------
>>>>>For information about J forums see http://www.jsoftware.com/forums.htm
>>>>>
>>>>>
>>>>>
>>>> ----------------------------------------------------------------------
>>>> For information about J forums see http://www.jsoftware.com/forums.htm
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>>>
>>>
>>>
>> ----------------------------------------------------------------------
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>
>
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