The 11 line fragment at the top is probably something like
tiny_toy_network=:3 :0
X=: #: 1 3 5 7
Y=: ,. 0 1 1 0
syn0=: (2*?3 4$0)-1
syn1=: (2*?4 1$0)-1
for_j. i.6 do.
l1=: %1+^-X +/ .* syn0
l2=: %1+^-l1 +/ .* syn1
l2_delta=: (Y - l2)*(l2*(1-l2))
l1_delta=: (l2_delta +
All,
I found this article on building a Neural Network in 11 lines of Python.
Does anyone know enough Python to translate this into J?
http://iamtrask.github.io/2015/07/12/basic-python-network/
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Thanks Gilles, my typo indeed !
> On 18 Nov 2017, at 4:57 am, Gilles Kirouac wrote:
>
> 1(0,+/\}.20 10 12)}42$0 NB. mask for length 10 12 20
>
> 1(0,+/\}:20 10 12)}42$0 NB. Drop last, not first; for 20 10 12
>
>
> Le 2017-11-17 à 00:24, Daniel Lyons a écrit :
>> Thanks Rob!
>>
>> I did feel
blockmake=: (-@[ {."1&.> ] <@({.,"0 1 }.)"0 1~ 1+])i.
(5 blockmake 4) -: 4 makeblock 5
1
Thanks,
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Raul
On Fri, Nov 17, 2017 at 1:14 PM, 'mike_liz@tiscali.co.uk' via
Programming wrote:
> Erling Helenas, Raul Miller, and others have come up with various
> methods to generate subse
a variation on Rob's method:
dat=: 3 42$'John Smith WA 418-Y11-4111' (restore the blanks)
msk=: (i.42) e. +/\0,20 10 12
('';msk) <;.1 dat
On Thu, Nov 16, 2017 at 9:24 PM, Daniel Lyons
wrote:
> Thanks Rob!
>
> I did feel that the general idea of the rolling sum generating the indic
Erling Helenas, Raul Miller, and others have come up with various
methods to generate subsets of “restricted generating functions” (RGFs)
suitable for the production of partitions of sets. Several of these
have used Ruskey’s algorithm.
I’ve found a fairly simple approach which has the benefi
1(0,+/\}.20 10 12)}42$0 NB. mask for length 10 12 20
1(0,+/\}:20 10 12)}42$0 NB. Drop last, not first; for 20 10 12
Le 2017-11-17 à 00:24, Daniel Lyons a écrit :
> Thanks Rob!
>
> I did feel that the general idea of the rolling sum generating the indices of
> the frets was not terrible, it ju
&.:
Henry Rich
On Nov 17, 2017 11:32 AM, "Raul Miller" wrote:
> ^. is rank 0, which means you would be taking the geometric mean of
> each number individually.
>
> You could do E&.(^."1) or E&.(^."2) or probably E&.(^."_) depending on
> what you are trying to accomplish.
>
> Since E&.:^. is equ
^. is rank 0, which means you would be taking the geometric mean of
each number individually.
You could do E&.(^."1) or E&.(^."2) or probably E&.(^."_) depending on
what you are trying to accomplish.
Since E&.:^. is equivalent to E&.(^."E) which, in turn, is equivalent
to E&.(^."_) you probably d
The arithmetic mean is
E=.+/ % #
For example E 2 3
2.5
The geometric mean is
E&.(2&^.) 2 3
2.44949
or
E&.:(^.) 2 3
2.44949
but (&.) does not work with (^.)
E&.(^.) 2 3
2 3
How come?
Thank! Bo.
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