?runmed
-- Bert
On Sun, Mar 27, 2011 at 2:44 PM, David Winsemius <dwinsem...@comcast.net> wrote:
>
> On Mar 27, 2011, at 10:56 AM, chuan_zl wrote:
>
>> Hello,everybody. My name is Chuan Zun Liang. I come from Malaysia. I am
>> just
>> a beginner for R. Kindly to ask favor about median filter. The problem I
>> facing as below:
>>
>>
>>> x<-matrix(sample(1:30,25),5,5)
>>> x
>>
>> [,1] [,2] [,3] [,4] [,5]
>> [1,] 7 8 30 29 13
>> [2,] 4 6 12 5 9
>> [3,] 25 3 22 14 24
>> [4,] 2 15 26 23 19
>> [5,] 28 18 10 11 20
>>
>> This is example original matrices of an image. I want apply with median
>> filter with window size 3X# to remove salt and pepper noise in my matric.
>> Here are the script I attend to writing.The script and output shown as
>> below:
>>
>>> MedFilter<-function(mat,sz)
>>
>> + {out<-matrix(0,nrow(mat),ncol(mat))
>> + for(p in 1:(nrow(mat)-(sz-1)))
>> + {for(q in 1:(ncol(mat)-(sz-1)))
>> + {outrow<-median(as.vector(mat[p:(p+(sz-1)),q:(q+(sz-1))]))
>> + out[(p+p+(sz-1))/2,(q+q+(sz-1))/2]<-outrow}}
>> + out}
>>
>
> Noting that median is probably the "rate-limiting" factor, I looked for
> another way to get the "middle" of 9 items. Using order seem faster:
>
>> system.time( replicate(100000, x2s <- sort(x2)))
> user system elapsed
> 9.829 0.212 10.029
>> system.time( replicate(100000, x2m <- median(x2)))
> user system elapsed
> 7.169 0.126 7.272
>> system.time( replicate(100000, x2s <-x2[order(x2)[5] ]))
> user system elapsed
> 1.907 0.051 1.960
>
> So see if this is any faster. On my system it's about three times faster:
>
> x <- matrix(sample(364*364), 364,364)
> out <- matrix(0, 364,364)
> for(xi in 1:(nrow(x)-2)) {
> for(yi in 1:(ncol(x)-2) ) {
> xm <- x[xi+0:2, yi+0:2]
> d[xi+1, yi+1] <-xm[order(xm)[5] ]}}
>
> #---------tests ------
>> system.time(for(xi in 1:(nrow(x)-2)) {
> + for(yi in 1:(ncol(x)-2) ) {
> + xm <- x[xi+0:2, yi+0:2]
> + d[xi+1, yi+1] <-xm[order(xm)[5] ]}} )
> user system elapsed
> 3.806 0.083 3.887
>
>> system.time(MedFilter(x,3) )
> user system elapsed
> 11.242 0.202 11.427
>
>
>>> MedFilter(x,3)
>>
>> [,1] [,2] [,3] [,4] [,5]
>> [1,] 0 0 0 0 0
>> [2,] 0 8 12 14 0
>> [3,] 0 12 14 19 0
>> [4,] 0 18 15 20 0
>> [5,] 0 0 0 0 0
>>
>> Example to getting value 8 and 12 as below:
>>
>> 7 8 30 8 30 29
>> 4 6 12 (median=8) 6 12 5 (median=12)
>> 25 3 22 3 22 14
>>
>> Even the script can give output. However, it is too slow. My image size is
>> 364*364. It is time consumption. Is it get other ways to improving it?
>
> David Winsemius, MD
> West Hartford, CT
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Bert Gunter
Genentech Nonclinical Biostatistics
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