Previews of coming attractions:
L4=: (]*=) + - % (^.%)
L4~ i.10
|NaN error: L4
| L4~i.10
|[-10]
Linda
-----Original Message-----
From: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Henry Rich
Sent: Sunday, April 20, 2014 3:28 PM
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] Logarithmic Mean
This has trouble when the numbers are close.
3 ((]*=) + - % ^.@:%) 3.000000000000001
5.66667
3 ((]*=!.0) + - % ^.@:%) 3.000000000000001
2.66667
3 ((]*=) + - % ^.@:%) 3.00000000000001
5.96774
3 ((]*=!.0) + - % ^.@:%) 3.00000000000001
2.96774
3 ((]*=) + - % ^.@:%) 3.0000000000001
6
3 ((]*=!.0) + - % ^.@:%) 3.0000000000001
3
3 ((]*=) + - % ^.@:%) 3.000000000001
2.99967
3 ((]*=!.0) + - % ^.@:%) 3.000000000001
2.99967
3 ((]*=) + - % ^.@:%) 3.00000000001
3
3 ((]*=!.0) + - % ^.@:%) 3.00000000001
3
Henry Rich
On 4/20/2014 1:52 PM, Roger Hui wrote:
> Actually, according to the Wikipedia page you quoted, the log mean is
> defined to be the difference divided by the log of their quotient. In
your
> L function, if I replace the difference of the logs by the (equivalent)
log
> of the quotient,
>
> L =: (]*=) + - % -&^.
> L3=: (]*=) + - % ^.@:%
>
> The problem of equal arguments is finessed, and is more efficient for
> having to compute only one log instead of two.
>
> 10 ts '(L |.)i.1e5'
> 0.0360588 5.24506e6
> 10 ts '(L3 |.)i.1e5'
> 0.00544869 5.24506e6
>
> 0 L3 i.10
> 0 0 0 0 0 0 0 0 0 0
> (i.10) L3 0
> 0 0 0 0 0 0 0 0 0 0
> L3~ i.10
> 0 1 2 3 4 5 6 7 8 9
>
>
>
>
>
>
> On Sat, Apr 19, 2014 at 8:53 PM, Raul Miller <rauldmil...@gmail.com>
wrote:
>
>> https://en.wikipedia.org/wiki/Logarithmic_mean defines logarithmic
>> mean as the the limit of difference of two numbers divided by the
>> difference of their (natural) logs. The limit is to deal with the case
>> where the two numbers are equal - in this case, we want an identity
>> function.
>>
>> I was tempted to implement logarithmic mean as
>> Lm=: (- % -&^.)^:~:
>>
>> But this has a problem with rank:
>> 2 Lm 2
>> 2
>> 2 3 Lm 3 2
>> 2.4663 2.4663
>> 2.4663 2.4663
>>
>> I could solve this with ("0) but that strikes me as inefficient.
>> L2=: (- % -&^.)^:~:"0
>>
>> So, instead:
>> L=: (]*=) + - % -&^.
>>
>> ((L-:L2) |.) i.1e5
>> 1
>>
>> Is that worth it?
>>
>> timespacex '(L |.) i.1e5'
>> 0.0365999 5.24506e6
>> timespacex '(L2 |.) i.1e5'
>> 0.0828581 3.14982e6
>>
>> Probably...
>>
>> A caution, though:
>> 0 L 0
>> |NaN error: L
>>
>> Because:
>> ^. 0
>> __
>>
>> So maybe the explicit rank implementation isn't such a bad idea after
all?
>>
>> Thanks,
>>
>> --
>> Raul
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>>
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