RMS=: (+/ % #)&.:*:
5!:4 <'RMS'
-- / --- +
-----+- %
-- &.: -+ L- #
L- *:
rms=:[:%:[:(+/%#)*:
5!:4 <'rms'
-- [:
+- %:
--+ -- [:
│ │ -- / --- +
L----+----+- %
│ L- #
L- *:
rms has a nice tree, and fewer concepts are required.
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Don Kelly
Sent: Monday, November 18, 2013 2:06 AM
To: [email protected]
Subject: Re: [Jprogramming] Novice problem
You sent an answer to my question before I asked it! Thanks- I suspected
but didn't know that this was it.
I think that I will stay with the earlier definition:
rms=:[:%:[:(+/%#)*:
which, while it does have brackets is somewhat more readable for my present
state of understanding. Yours is shorter and I expect it would be faster.
Now all of you have given me enough to digest for now.
Don
On 17/11/2013 7:23 AM, Raul Miller wrote:
> Here's another definition of rms
>
> Rms=: +/@:*: %:@% #
> Rms 1 2 1 2
> 1.58114
>
> Explanation:
>
> We do not need to square the numbers in the argument to #, we only
> need to square them in the argument to +/
>
> We only need the square root on the result of %
>
> Makes sense?
>
> Also, here's a partial explanation for the (+/%#)&.:*: definition of
> RMS
>
> &.: means "under" much like &. except that the derived verb has
> infinite rank - the verb on the left gets the entire array which
> resulted from the verb on the right, regardless of the rank of the
> verb on the right. In other words, it is equivalent to (+/ % #) &.
> (*:"_) In other words: square the numbers, add them up, divide by
> their sum, then do the inverse of squaring on the result.
>
> Thanks,
>
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
