Re: Re: Multiple application of LinearFilter
- To: mathgroup at smc.vnet.net
- Subject: [mg64858] Re: [mg64850] Re: Multiple application of LinearFilter
- From: "Lea Rebanks" <lrebanks at netvigator.com>
- Date: Mon, 6 Mar 2006 05:00:59 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
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 earthlink.net] To: mathgroup at smc.vnet.net Subject: [mg64858] [mg64850] Re: Multiple application of LinearFilter On 3/4/06 at 2:35 AM, lrebanks at netvigator.com (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. >EG >originaldata >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., In[13]:= LinearFilter[{a,b,c,d},{1/2,1/2}]==ListConvolve[{1/2,1/2},{a,b,c,d}] Out[13]= True 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} In[14]:= ListConvolve[{1/2, 1/2}, ListConvolve[{1/2, 1/2}, ListConvolve[{1/2, 1/2}, {a, b, c, d, e, f}]]]//Simplify Out[14]= {(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 In[15]:= ListConvolve[{1/8, 3/8, 3/8, 1/8}, {a, b, c, d, e, f}]//Simplify Out[15]= {(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 In[18]:= ListConvolve[{1/2, 1/2}, {a, b, c}, {1, 1}] Out[18]= {a/2 + c/2, a/2 + b/2, b/2 + c/2} or if you wanted the equivalent of ListConvolve[{1/2,1/2},PadLeft[{a,b,c},4]] you could do ListConvolve[{1/2,1/2},{a,b,c},{1,1},0] -- To reply via email subtract one hundred and four