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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