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,
--
Raul
On Sun, Aug 23, 2020 at 3:52 AM Raul Miller <[email protected]> wrote:
>
> Ah, ignore that numeric example, I had overlooked the use of 0j1":
>
> Still... I am curious about how to judge the correctness of the partitioning.
>
> Thanks,
>
> --
> Raul
>
> On Sun, Aug 23, 2020 at 3:31 AM Raul Miller <[email protected]> wrote:
> >
> > Generalized how?
> >
> > Specifically, how should I judge correctness?
> >
> > (+/%#)0 1 2 3 4 5 6 7 8 9 10 20 20 20 20 20
> > 9.6875
> > (+/%#)1.9 7.4 19.7
> > 9.66667
> >
> > Thanks,
> >
> > --
> > Raul
> >
> > On Sat, Aug 22, 2020 at 2:38 PM 'Bo Jacoby' via Programming
> > <[email protected]> wrote:
> > >
> > >
> > > Hello Raul.
> > > My work is not differential equations. It is simpler.
> > > It is about generalizing the concept of one mean value into two or more
> > > values, characteristics of a dataset.
> > > Here are some simple examples.
> > > ] a =. (i.11) , 5#20 NB. some test data. 16 numbers
> > > 0 1 2 3 4 5 6 7 8 9 10 20 20 20 20 20
> > >
> > > 0j1":b =. 3 summary@tomb a NB. 16 numbers summarized into 3 numbers.
> > > 1.9 7.4 19.7
> > >
> > > 0j1":3 summary@tomb b NB. 3 numbers summarized into the same 3 numbers.
> > > 1.9 7.4 19.7
> > >
> > >
> > >
> > > 0j1":2 summary@tomb a NB. 16 numbers summarized into 2 numbers
> > > 2.3 17.1
> > >
> > > 0j1":2 summary@tomb b NB. 3 numbers summarized into the same 2 numbers
> > > 2.3 17.1
> > >
> > > 0j1":1 summary@tomb a NB. 16 numbers summarized into 1 number
> > > 9.7
> > >
> > > 0j1":1 summary@tomb b NB. 3 numbers summarized into the same number
> > > 9.7
> > >
> > >
> > >
> > > Thank you!
> > > Bo
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > Den lørdag den 22. august 2020 18.57.38 CEST skrev Raul Miller
> > ><[email protected]>:
> > >
> > > Hmm...
> > >
> > > Trying to bring myself up to speed on what you seem to be saying here
> > > first lead me to https://mathworld.wolfram.com/PowerMean.html and
> > > https://en.wikipedia.org/wiki/Newton%27s_identities and then to
> > > https://www.jsoftware.com/papers/MSLDE1.htm
> > >
> > > I don't have anything meaningful to add right now -- I am not even
> > > sure if I can form a relevant question. Maybe if I spend a half hour a
> > > day on this for long enough, I'll be able to come back to it
> > > intelligently.
> > >
> > > Until then, good luck.
> > >
> > > --
> > > Raul
> > >
> > > On Sat, Aug 22, 2020 at 11:20 AM 'Bo Jacoby' via Programming
> > > <[email protected]> wrote:
> > > >
> > > > Thanks to Raul and Brian!
> > > > tomb=:(^/ i.@>:)~ , NB. It works miracuously! Thank you!
> > > >
> > > > Raul wrote: "It sort of smells like you were working with matrices
> > > > designed for inversion and polynomial fitting"
> > > > Well, I want to find n numbers such that their mean powers are equal to
> > > > the mean powers of my input data set. I have n nonlinear equations in
> > > > n unknowns. The trick is to use the symmetry of the power sums. (A
> > > > power sum does not depend on the order of the variables). So the n
> > > > unknowns are roots in a polynomial of degree n. The coefficients of
> > > > this polynomial are solutions to n linear equations in n unknowns. See
> > > > Newton's identities
> > > >
> > > > |
> > > > |
> > > > | |
> > > > Newton's identities
> > > >
> > > > Let x1, ..., xn be variables, denote for k ≥ 1 by pk(x1, ..., xn) the
> > > > k-th power sum:
> > > > |
> > > >
> > > > |
> > > >
> > > > |
> > > >
> > > >
> > > > So yes, I compute the n*n-matrix m from the n power sums s, and solve
> > > > the linear equation m +/ .* e = s (by e=.(%.m)s), and the the
> > > > nonlinear equation e p. x = 0 (by p. e). It is not polynomial fitting.
> > > > Thank you for your help and for your interest!
> > > > Bo
> > > >
> > > >
> > > >
> > > > Den lørdag den 22. august 2020 15.44.33 CEST skrev Raul Miller
> > > ><[email protected]>:
> > > >
> > > > I don't really follow what you are doing here -- for example, I look
> > > > at stuff like 11 summary@hist i.11 and do not understand how that
> > > > result corresponds to "size of the dataset" -- but even without
> > > > knowing your intentions, I can point out that you can simplify tomb:
> > > >
> > > > tomb=:(^/ i.@>:)~ ,
> > > >
> > > > It sort of smells like you were working with matrices designed for
> > > > inversion and polynomial fitting, but that doesn't tell me much (if
> > > > anything) about your destination.
> > > >
> > > > Good luck,
> > > >
> > > > --
> > > > Raul
> > > >
> > > > On Sat, Aug 22, 2020 at 5:35 AM 'Bo Jacoby' via Programming
> > > > <[email protected]> wrote:
> > > > >
> > > > > Thanks to Michael and to ethiejiesa!
> > > > > The program is now:
> > > > >
> > > > > tomb=.[ (^/ i.@>:)~ ,@] NB. Powers of a tombola
> > > > >
> > > > > hist=.]* ((^/&i. >:)~ #) NB. Powers of a histogram
> > > > > s=.(*#)@(}.%{.)@(+/) NB. summation
> > > > > f1=.$~2 # # NB. n*n matrix
> > > > > f2=.>:@i.@#,}: NB. insert S0, remove Sn
> > > > > f3=.*_1^i.@# NB. change signs
> > > > > f4=.|:|.!.0"0 1~-@i.@# NB. shift in zeroes
> > > > > m=.|:@f4@(f3"1@f2@|:@f1) NB. Matrix
> > > > > e=.1,(%.m) NB. solve lin.eqns.
> > > > > pol=._1 x: -@|.@>@{:@p.@|. NB. solve alg.eqn.
> > > > > hp=.x: :: ] NB. high precision
> > > > > summary=.pol@e@s@hp f. NB. complete program
> > > > > (I need the comma in tomb, so now it is smelly again.)
> > > > > Michael wrote: "Any clue as to the input to summary, or what you're
> > > > > summarising? Sorry if it's obvious!"
> > > > > No sir, it is not obvious if it is not obvious to you.
> > > > > Left input to summary@tomb and summary@hist is the size of the output
> > > > > dataset.
> > > > > Right input to summary@tomb is the input dataset described as a
> > > > > tombola. The same number may occur on many tickets. The order of the
> > > > > tickets is immaterial.
> > > > > Right input to summary@hist is the input dataset described as a
> > > > > histogram. the number of zeroes, the number of ones, the number og
> > > > > twos &c.
> > > > > Examples:
> > > > >
> > > > > 11 summary@tomb i.11
> > > > >
> > > > >
> > > > > 0 1 2 3 4 5 6 7 8 9 10
> > > > >
> > > > > 11 summary@tomb 11 NB. this wouldn't work without the comma in tomb
> > > > > 11 11 11 11 11 11 11 11 11 11 11
> > > > >
> > > > > 1 summary@tomb i.11 NB. mean value
> > > > > 5
> > > > >
> > > > > 2 summary@tomb i.11 NB. mean plus/minus std.dev.
> > > > > 1.83772 8.16228
> > > > >
> > > > >
> > > > > 3 summary@tomb i.11 NB. generalized into 3 numbers
> > > > >
> > > > > 1.12702 5 8.87298
> > > > >
> > > > >
> > > > >
> > > > > Explanations:
> > > > >
> > > > >
> > > > >
> > > > > 5 tomb i.11
> > > > >
> > > > > 1 0 0 0 0 0
> > > > >
> > > > > 1 1 1 1 1 1
> > > > >
> > > > > 1 2 4 8 16 32
> > > > >
> > > > > 1 3 9 27 81 243
> > > > >
> > > > > 1 4 16 64 256 1024
> > > > >
> > > > > 1 5 25 125 625 3125
> > > > >
> > > > > 1 6 36 216 1296 7776
> > > > >
> > > > > 1 7 49 343 2401 16807
> > > > >
> > > > > 1 8 64 512 4096 32768
> > > > >
> > > > > 1 9 81 729 6561 59049
> > > > >
> > > > > 1 10 100 1000 10000 100000
> > > > >
> > > > > +/ hp 5 tomb i.11
> > > > > 11 55 385 3025 25333 220825
> > > > >
> > > > > (}.%{.) +/ hp 5 tomb i.11
> > > > > 5 35 275 2303 20075
> > > > >
> > > > > (*#) (}.%{.) +/ hp 5 tomb i.11
> > > > > 25 175 1375 11515 100375
> > > > >
> > > > > s hp 5 tomb i.11
> > > > > 25 175 1375 11515 100375
> > > > >
> > > > > f1 s hp 5 tomb i.11
> > > > > 25 175 1375 11515 100375
> > > > >
> > > > > 25 175 1375 11515 100375
> > > > >
> > > > > 25 175 1375 11515 100375
> > > > >
> > > > > 25 175 1375 11515 100375
> > > > >
> > > > > 25 175 1375 11515 100375
> > > > >
> > > > > |: f1 s hp 5 tomb i.11
> > > > > 25 25 25 25 25
> > > > >
> > > > > 175 175 175 175 175
> > > > >
> > > > > 1375 1375 1375 1375 1375
> > > > >
> > > > > 11515 11515 11515 11515 11515
> > > > >
> > > > > 100375 100375 100375 100375 100375
> > > > >
> > > > > f2 |: f1 s hp 5 tomb i.11
> > > > > 1 2 3 4 5
> > > > >
> > > > > 25 25 25 25 25
> > > > >
> > > > > 175 175 175 175 175
> > > > >
> > > > > 1375 1375 1375 1375 1375
> > > > >
> > > > > 11515 11515 11515 11515 11515
> > > > >
> > > > > f3"1 f2 |: f1 s hp 5 tomb i.11
> > > > > 1 _2 3 _4 5
> > > > >
> > > > > 25 _25 25 _25 25
> > > > >
> > > > > 175 _175 175 _175 175
> > > > >
> > > > > 1375 _1375 1375 _1375 1375
> > > > >
> > > > > 11515 _11515 11515 _11515 11515
> > > > >
> > > > > f4 f3"1 f2 |: f1 s hp 5 tomb i.11
> > > > > 1 25 175 1375 11515
> > > > >
> > > > > 0 _2 _25 _175 _1375
> > > > >
> > > > > 0 0 3 25 175
> > > > >
> > > > > 0 0 0 _4 _25
> > > > >
> > > > > 0 0 0 0 5
> > > > >
> > > > > |: f4 f3"1 f2 |: f1 s hp 5 tomb i.11
> > > > > 1 0 0 0 0
> > > > >
> > > > > 25 _2 0 0 0
> > > > >
> > > > > 175 _25 3 0 0
> > > > >
> > > > > 1375 _175 25 _4 0
> > > > >
> > > > > 11515 _1375 175 _25 5
> > > > >
> > > > > m s hp 5 tomb i.11
> > > > > 1 0 0 0 0
> > > > >
> > > > > 25 _2 0 0 0
> > > > >
> > > > > 175 _25 3 0 0
> > > > >
> > > > > 1375 _175 25 _4 0
> > > > >
> > > > > 11515 _1375 175 _25 5
> > > > >
> > > > > (%.m) s hp 5 tomb i.11
> > > > > 25 225 875 1340 450
> > > > >
> > > > > |.1,(%.m) s 5 tomb i.11
> > > > > 450 1340 875 225 25 1
> > > > >
> > > > > p.|.1,(%.m) s hp 5 tomb i.11
> > > > > ┌─┬─────────────────────────────────────┐
> > > > >
> > > > > │1│_9.54306 _7.0882 _5 _2.9118 _0.456938│
> > > > >
> > > > > └─┴─────────────────────────────────────┘
> > > > >
> > > > > _1 x: - |. > {: p.|.1,(%.m) s hp 5 tomb i.11
> > > > > 0.456938 2.9118 5 7.0882 9.54306
> > > > >
> > > > > 5 summary@tomb i.11
> > > > > 0.456938 2.9118 5 7.0882 9.54306
> > > > >
> > > > >
> > > > >
> > > > > Thanks!
> > > > > Bo. Den fredag den 21. august 2020 17.01.49 CEST skrev 'Michael
> > > > > Day' via Programming <[email protected]>:
> > > > >
> > > > > Any clue as to the input to summary, or what you're summarising?
> > > > >
> > > > > Sorry if it's obvious!
> > > > >
> > > > > Note these (display ?) errors:
> > > > >
> > > > > pol =. _1 [space] x: ...
> > > > >
> > > > > hp =. x [space] :: ] ...
> > > > >
> > > > > Cheers,
> > > > >
> > > > > Mike
> > > > >
> > > > >
> > > > > On 21/08/2020 15:41, 'Bo Jacoby' via Programming wrote:
> > > > > >
> > > > > > tomb=.(^/i.@>:)~ NB. tombola lottery powers
> > > > > > hist=.({."1~>:)~(*[:^/~i.@#@,)NB. histogram powers
> > > > > > s=.(*#)@ (}.%{.) @ (+/) NB. summation
> > > > > > f1=.$~2# # NB. n*n matrix
> > > > > > f2=.>:@i.@#, }: NB. insert S0, remove Sn
> > > > > > f3=.*_1^i.@# NB. change sign
> > > > > > f4=.|:|.!.0"01~-@i.@# NB. shift zeroes in
> > > > > > m=.|:@f4@(f3"1@f2@|:@f1) NB. matrix
> > > > > > e=.1,(%.m) NB. solve linear equations
> > > > > > pol=._1x: -@|.@>@{:@p.@|. NB.solve alg.eqn.
> > > > > > hp=.x::: ] NB. high precision
> > > > > > summary=.pol@e@s@hp f. NB. complete program
> > > > > > sorry for the still missing carriage returns.
> > > > > > Den fredag den 21. august 2020 16.37.19 CEST skrev Bo Jacoby
> > > > > ><[email protected]>:
> > > > > >
> > > > > > tomb=.(^/i.@>:)~ NB. tombola lottery
> > > > > >powershist=.({."1~>:)~(*[:^/~i.@#@,)NB. histogram powerss=.(*#)@
> > > > > >(}.%{.) @ (+/) NB. summationf1=.$~2# # NB. n*n matrixf2=.>:@i.@#, }:
> > > > > >NB. insert S0, remove Snf3=.*_1^i.@# NB. change
> > > > > >signf4=.|:|.!.0"01~-@i.@# NB. shift zeroes
> > > > > >inm=.|:@f4@(f3"1@f2@|:@f1) NB. matrixe=.1,(%.m) NB. solve linear
> > > > > >equationspol=._1x: -@|.@>@{:@p.@|. NB.solve alg.eqn.hp=.x::: ] NB.
> > > > > >high precisionsummary=.pol@e@s@hp f. NB. complete program
> > > > > > This should be better. Sorry.
> > > > > > Bo. Den fredag den 21. august 2020 16.29.35 CEST skrev 'Bo
> > > > > > Jacoby' via Programming <[email protected]>:
> > > > > >
> > > > > > Thank you!
> > > > > > I was trying to accomplish this
> > > > > > tomb=.(^/i.@>:)~ NB. tombola lottery
> > > > > > powershist=.({."1~>:)~(*[:^/~i.@#@,)NB. histogram powerss=.(*#)@
> > > > > > (}.%{.) @ (+/) NB. summationf1=.$~2# # NB. n*n matrixf2=.>:@i.@#,
> > > > > > }: NB. indsæt S0, fjern Snf3=.*_1^i.@# NB. change
> > > > > > signf4=.|:|.!.0"01~-@i.@# NB. shift zeroes
> > > > > > inm=.|:@f4@(f3"1@f2@|:@f1) NB. matrixe=.1,(%.m) NB. solve linear
> > > > > > equationspol=._1x: -@|.@>@{:@p.@|. NB.solve alg.eqn.hp=.x::: ] NB.
> > > > > > high precisionsummary=.pol@e@s@hpf. NB. complete program
> > > > > > 1&(summary@tomb) computes the mean value of a dataset.
> > > > > >
> > > > > > 1 summary@tomb i.11
> > > > > > 5
> > > > > >
> > > > > >
> > > > > > The mean value of 99 zeroes and 1 one is
> > > > > >
> > > > > > 1 summary@hist 99 1
> > > > > >
> > > > > > 0.01
> > > > > >
> > > > > >
> > > > > >
> > > > > > The mean value plus/minus the standard deviation is computed like
> > > > > > this
> > > > > >
> > > > > > 2 summary@tomb i.11
> > > > > >
> > > > > > 1.83772 8.16228
> > > > > >
> > > > > > 2 summary@hist 99 1
> > > > > > _0.0894987 0.109499
> > > > > >
> > > > > >
> > > > > >
> > > > > > Of course these two numbers have the correct mean value
> > > > > >
> > > > > > 1 summary@tomb 2 summary@tomb i.11
> > > > > >
> > > > > > 5
> > > > > >
> > > > > > 1 summary@tomb 2 summary@hist 99 1
> > > > > > 0.01
> > > > > >
> > > > > >
> > > > > >
> > > > > > The computation is generalized to 3 numbers
> > > > > >
> > > > > > 3 summary@tomb i.11
> > > > > >
> > > > > > 1.12702 5 8.87298
> > > > > >
> > > > > > 3 summary@hist 99 1
> > > > > > _0.108204j0.16452 _0.108204j_0.16452 0.246409
> > > > > >
> > > > > > having the correct mean values and standarddeviations.
> > > > > >
> > > > > > 2 summary@tomb 3 summary@tomb i.11
> > > > > >
> > > > > > 1.83772 8.16228
> > > > > >
> > > > > > 2 summary@tomb 3 summary@hist 99 1
> > > > > > _0.0894987 0.109499
> > > > > >
> > > > > >
> > > > > >
> > > > > > Thus complicated probability distributions functions are summarized
> > > > > > by a few numbers. That is what I accomplished!
> > > > > > Thanks!
> > > > > > Bo.
> > > > > > Den fredag den 21. august 2020 15.33.50 CEST skrev ethiejiesa
> > > > > >via Programming <[email protected]>:
> > > > > >
> > > > > > Bo Jacoby <[email protected]> wrote:
> > > > > >> How to de-smell this:
> > > > > >> 3([* i.@#@,@[ ^/ i.@>:@])~ 0 0 1 1 1
> > > > > > 3 (]* ((^/&i. >:)~ #)) 0 0 1 1 1
> > > > > >
> > > > > > Maybe? That comes from just a little mechanical algebra:
> > > > > >
> > > > > >> 3([* i.@#@,@[ ^/ i.@>:@])~ 0 0 1 1 1
> > > > > > First notice the repeated (i.) on either argument of (^/). There is
> > > > > > a general
> > > > > > pattern, (f@u v f@w <--> u v&f w), which in this case specializes to
> > > > > >
> > > > > > 3([* #@,@[ ^/&i. >:@])~ 0 0 1 1 1
> > > > > >
> > > > > > Then, at least in this example the (@,) is superfluous, so we elide
> > > > > > it:
> > > > > >
> > > > > > 3([* #@[ ^/&i. >:@])~ 0 0 1 1 1
> > > > > >
> > > > > > Finally, here is an idiom: ( (u@[ v w@]) <--> ((v w)~ u)~ ), and
> > > > > > since
> > > > > > (x v~~ y <--> x v y), we have
> > > > > >
> > > > > > 3 (]* ((^/&i. >:)~ #)) 0 0 1 1 1
> > > > > >
> > > > > > The last idiom is really a matter of preference. I like the bare
> > > > > > verbs, but
> > > > > > code golf-wise it doesn't win.
> > > > > >
> > > > > > More importantly, however, I really have no idea what you are
> > > > > > trying to
> > > > > > accomplish. The above just performs some J syntax manipulations;
> > > > > > however, this
> > > > > > doesn't at all demonstrate the main point of avoiding code smell as
> > > > > > I see it,
> > > > > > which is more about finding places to improve the *algorithm*
> > > > > > rather than only
> > > > > > cleaning up the syntax.
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> > > > >
> > > > > --
> > > > > This email has been checked for viruses by Avast antivirus software.
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