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Re: Re: Multiple application of LinearFilter

  • To: mathgroup at
  • Subject: [mg64858] Re: [mg64850] Re: Multiple application of LinearFilter
  • From: "Lea Rebanks" <lrebanks at>
  • Date: Mon, 6 Mar 2006 05:00:59 -0500 (EST)
  • Sender: owner-wri-mathgroup at

Many thanks Bill.

Very interesting. I will try that, however I feel I may run into problems.

Is it possible to email me your email address so I may respond directly
to you.

I have tried figuring out your cryptic email below with no success after
many tries. 

Speak soon.

Best Regards - Lea Rebanks...

-----Original Message-----
From: Bill Rowe [mailto:readnewsciv at] 
To: mathgroup at
Subject: [mg64858] [mg64850] Re: Multiple application of LinearFilter

On 3/4/06 at 2:35 AM, lrebanks at (Lea Rebanks) wrote:

>I have been using LinearFilter on my data with great success &
>achieve better results the more times I pass my data through the
>LinearFilter. ( I have had to adjust data length to maintain
>correct indexing with PadLeft - but this is not my question here.)

>Below is how I am applying the multiple LinearFilter.



>Data1=LinearFilter[originaldata, {1/2,1/2}];
>Data2=LinearFilter[Data1, {1/2,1/2}]; 
>Data3=LinearFilter[Data2, {1/2,1/2}];

>My question is - Is there a shorter way of writing multiple passes
>of the above.

Yes. But rather than using LinearFilter, I think the built-in function
ListConvolve is more efficient and flexible. For your specific
application, ListConvolve differs from LinearFilter only in the order of
the arguments, i.e.,



Repeated passes with a given kernel are equivalent to a single pass with
a longer kernel. That is 3 passes with the kernel (1/2,1/2) give the
same result as a single pass with the kernel {1/8,3/8,3/8,1/8}

Here is three passes using ListConvolve and the kernel {1/2,1/2}

ListConvolve[{1/2, 1/2}, 
   ListConvolve[{1/2, 1/2}, ListConvolve[{1/2, 1/2}, 
     {a, b, c, d, e, f}]]]//Simplify
{(1/8)*(a + 3*b + 3*c + d), (1/8)*(b + 3*c + 3*d + e), 
  (1/8)*(c + 3*d + 3*e + f)}

and here is the result using a single pass

ListConvolve[{1/8, 3/8, 3/8, 1/8}, {a, b, c, d, e, f}]//Simplify

{(1/8)*(a + 3*b + 3*c + d), (1/8)*(b + 3*c + 3*d + e), 
  (1/8)*(c + 3*d + 3*e + f)}

As you can see the results are the same.

The real advantage of ListConvolve over LinearFilter is the additional
flexibility ListConvolve gives, For example, to get an output the same
length as the input you can do

ListConvolve[{1/2, 1/2},  {a, b, c}, {1, 1}]

{a/2 + c/2, a/2 + b/2, b/2 + c/2}

or if you wanted the equivalent of


you could do

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