I cannot reproduce your result.
I get this:
comb=: 4 : 0
k=. i.>:d=.y-x
z=. (d$<i.0 0),<i.1 0
for. i.x do. z=. k ,.&.> ,&.>/\. >:&.> z end.
; z
)
peermm=:3 :0
Y=. ;(/: #@>)</.~y
arg=. </.~i.~Y
n1=. 0 i.~1=#@> arg
m=. (i.@! A. i.)n1
for_d.n1}.arg do.
u=.{.s=.>d
n=.i.(#s)+{:$m
e=.n e."1 s comb&#n
m=.,/(u*e)+"1/"_1 (1-e)#inv"1/m
end.
m{Y
)
peermm 'ababc'
bbaac
bbaca
bbcaa
babac
babca
bcbaa
baabc
bacba
bcaba
baacb
bacab
bcaab
abbac
abbca
cbbaa
ababc
abcba
cbaba
abacb
abcab
cbaab
aabbc
acbba
cabba
aabcb
acbab
cabab
aacbb
acabb
caabb
Thanks,
--
Raul
On Sun, Apr 26, 2015 at 2:58 AM, R.E. Boss <r.e.b...@outlook.com> wrote:
>
> peermm 'ababc'
> |length error: peermm
> | m=.,/(u*e)+"1/"_1(1-e) #inv"1/m
>
>
> R.E. Boss
>
>
>> -----Original Message-----
>> From: programming-boun...@forums.jsoftware.com [mailto:programming-
>> boun...@forums.jsoftware.com] On Behalf Of Raul Miller
>> Sent: zondag 26 april 2015 3:22
>> To: Programming forum
>> Subject: Re: [Jprogramming] Permutations with repeats (combinations?)
>>
>> That's an error on my part.
>>
>> It' not unique to boxed data, you can see the same flaw with:
>>
>> peermm 'ababc'
>>
>> Basically, I did not spend enough time exploring the boundaries
>> between the two abstractions I was using.
>>
>> I think this fixes that problem:
>>
>> comb=: 4 : 0
>> k=. i.>:d=.y-x
>> z=. (d$<i.0 0),<i.1 0
>> for. i.x do. z=. k ,.&.> ,&.>/\. >:&.> z end.
>> ; z
>> )
>>
>> peermm=:3 :0
>> Y=. ;(/: #@>)</.~y
>> arg=. </.~i.~Y
>> n1=. 0 i.~1=#@> arg
>> m=. (i.@! A. i.)n1
>> for_d.n1}.arg do.
>> u=.{.s=.>d
>> n=.i.(#s)+{:$m
>> e=.n e."1 s comb&#n
>> m=.,/(u*e)+"1/"_1 (1-e)#inv"1/m
>> end.
>> m{Y
>> )
>>
>> There might be a better (more concise and faster) way of setting this up?
>>
>> Thanks,
>>
>> --
>> Raul
>>
>>
>> On Sat, Apr 25, 2015 at 8:29 PM, 'Pascal Jasmin' via Programming
>> <programm...@jsoftware.com> wrote:
>> > this has wrong results
>> >
>> > peermm 1 1 ; 2 2 ; 1 1 ; 2 2 ;3
>> >
>> >
>> >
>> > ----- Original Message -----
>> > From: Raul Miller <rauldmil...@gmail.com>
>> > To: Programming forum <programm...@jsoftware.com>
>> > Cc:
>> > Sent: Saturday, April 25, 2015 8:24 PM
>> > Subject: Re: [Jprogramming] Permutations with repeats (combinations?)
>> >
>> > What problem does it have with boxed data? (It worked fine on boxed
>> > data when I tested it that way.)
>> >
>> > Thanks,
>> >
>> > --
>> > Raul
>> >
>> > On Sat, Apr 25, 2015 at 8:05 PM, 'Pascal Jasmin' via Programming
>> > <programm...@jsoftware.com> wrote:
>> >> Its even faster, and handles text, but not boxed data
>> >>
>> >> one use case for boxes is keeping some items together as a group
>> >>
>> >> ;"1 rperm2 1 2 ; 1 2 ; 3 ;3 ; 3
>> >> 3 3 1 2 1 2 3
>> >> 3 3 1 2 3 1 2
>> >> 3 3 3 1 2 1 2
>> >> 3 1 2 3 1 2 3
>> >> 3 1 2 3 3 1 2
>> >> 3 1 2 1 2 3 3
>> >> 1 2 3 3 1 2 3
>> >> 1 2 3 3 3 1 2
>> >> 1 2 3 1 2 3 3
>> >> 1 2 1 2 3 3 3
>> >>
>> >>
>> >>
>> >>
>> >>
>> >>
>> >>
>> >> ----- Original Message -----
>> >> From: Raul Miller <rauldmil...@gmail.com>
>> >> To: Programming forum <programm...@jsoftware.com>
>> >> Cc:
>> >> Sent: Saturday, April 25, 2015 2:02 PM
>> >> Subject: Re: [Jprogramming] Permutations with repeats (combinations?)
>> >>
>> >> My first attempt at implementing like that ran into a problem with
>> >> mix-matched types in r for some cases, so I reworked the thing to
>> >> support arguments of any type:
>> >>
>> >> comb=: 4 : 0
>> >> k=. i.>:d=.y-x
>> >> z=. (d$<i.0 0),<i.1 0
>> >> for. i.x do. z=. k ,.&.> ,&.>/\. >:&.> z end.
>> >> ; z
>> >> )
>> >>
>> >> peermm=:3 :0
>> >> arg=. (/: #@>)</.~i.~y
>> >> n1=. 0 i.~1=#@> arg
>> >> m=. (i.@! A. i.)n1
>> >> for_d.n1}.arg do.
>> >> u=.{.s=.>d
>> >> n=.i.(#s)+{:$m
>> >> e=.n e."1 s comb&#n
>> >> m=.,/(u*e)+"1/"_1 (1-e)#inv"1/m
>> >> end.
>> >> m{y
>> >> )
>> >>
>> >> I have no idea if this is actually faster. I'll let someone else worry
>> >> about that.
>> >>
>> >> --
>> >> Raul
>> >>
>> >> On Sat, Apr 25, 2015 at 12:58 PM, Roger Hui
>> <rogerhui.can...@gmail.com> wrote:
>> >>> The peermm function can be made faster by doing the actual
>> permutations of
>> >>> those cells of </.~y having only a single element.
>> >>>
>> >>>
>> >>> On Fri, Apr 24, 2015 at 10:46 PM, Raul Miller <rauldmil...@gmail.com>
>> wrote:
>> >>>
>> >>>> On Fri, Apr 24, 2015 at 7:03 PM, Roger Hui
>> <rogerhui.can...@gmail.com>
>> >>>> wrote:
>> >>>> > The details of implementing this algorithm are left as an exercise
> for
>> >>>> the
>> >>>> > reader. :-)
>> >>>>
>> >>>> Do I qualify as a reader?
>> >>>>
>> >>>> comb=: 4 : 0
>> >>>> k=. i.>:d=.y-x
>> >>>> z=. (d$<i.0 0),<i.1 0
>> >>>> for. i.x do. z=. k ,.&.> ,&.>/\. >:&.> z end.
>> >>>> ; z
>> >>>> )
>> >>>>
>> >>>> peermm=:3 :0
>> >>>> r=.i.1 0
>> >>>> for_d.</.~y do.
>> >>>> u=.{.s=.>d
>> >>>> n=.i.(#s)+{:$r
>> >>>> e=.n e."1 s comb&#n
>> >>>> r=.,/(u*e)+"1/"_1 (1-e)#inv"1/r
>> >>>> end.
>> >>>> )
>> >>>>
>> >>>> peermm 1 1 2 2
>> >>>> 2 2 1 1
>> >>>> 2 1 2 1
>> >>>> 2 1 1 2
>> >>>> 1 2 2 1
>> >>>> 1 2 1 2
>> >>>> 1 1 2 2
>> >>>>
>> >>>> --
>> >>>> Raul
>> >>>>
> ----------------------------------------------------------------------
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>> >>
>> >>>>
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