Re: filtering logical array

[Robert T. Short]

On 06/27/2012 08:12 AM, seektime wrote:
> y = (conv(a,conv([1 1 1],[1 1 1]))>5)(3:end-2)
>
>
> Hi Bob and Mike,
>
> Thanks for your input. I want to learn more about convolution - obviously
> it's a neat and elegant (and probably fast) way of coding and execution.
>
> However, after playing around for some time I still fail to see the
> underlying rules to configure the three constants (5, 3, 2) in the above
> line of code. The logical array a I provided was only an example. In reality
> I'm applying the code on much larger arrays, with considerably longer
> strings of 1's. So "y" needs to be calculated for strings of 20-100 1's. The
> solution I provided works for me. But it'll be great if you give me a hint
> on what rule the constants follow for convolution.
> Thanks again,
> Michael
Michael,

What you observed was a couple octave geeks trying to create a one-line 
solution to a problem.  Sometimes I wonder about a language in which 
efficiency relies so much on tricks (think APL) but when it is available 
you look for such solutions.

First, convolution isn't necessarily fast, but it is easy (at least if 
you have the conv routine available).  For this simple case though

y=conv(x,[1 1 1])

creates a running sum of the elements of the elements of x, i.e. y(n) = 
x(n) + x(n-1) + x(n-2).  The second convolution

z = conv(y,[1 1 1])

creates a running sum of y, and convolution is commutative/associative 
(etc.) so we can convolve the running sum elements first.

So, to filter a string of L ones, try

f1 = ones(1,L);
f   = conv(f1,f1);
T  = 2*L-1;
y  = (conv(x,f)>T)(L:end-L+1)

or, as a one-liner

y = (conv(x,conv(ones(1,L),ones(1,L)))>=2*L)(L:end-L+1)


I didn't actually try this - it very likely has syntax or other small 
errors, but I think the idea should work.

Bob