The problem is that
rms =: %:@(+/%#)@*:
doesn't work in this case but replacing @ with @: does work
(+/%#)@:*: 1 2 1 2
1 4 1 4
(+/%#)@:*: 1 2 1 2
2.5
it appears that (+/%#)@ 1 2 1 2 leads to 1 4 1 4 -the summation taken over
individual elements
(i.e (+/1%1),(+/4%1)... rather than acting on the whole result of the squaring
(1+4+1+4)%4 (or mean square).
It's the same thing that happens with &. and &.: the &.: and @: have infinite
rank
See At in the dictionary -this is made clear there (why not elsewhere? where I
had problems ).
Thanks to Raul, I see what is going on.
As a known example:
s=. 1 o. 1p1*0.01*i.100
rms =: %:@:(+/%#)@:*:
rms s
0.707107
%%:2 rms value of a sine wave of magnitude 1
0.707107
Don
On 18/11/2013 3:53 AM, Piet de Jong wrote:
I always read/interpret "@" or "@:" as "of the"
Took me a while to get this "interpretation" but now I use it all the time
when I write J programs:
NB. square root "of the" mean "of the" squares.
On Mon, Nov 18, 2013 at 6:06 PM, Don Kelly <[email protected]> wrote:
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,
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