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}> 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? Best Wishes Chuan -- View this message in context: http://r.789695.n4.nabble.com/Asking-Favor-For-the-Script-of-Median-Filter-tp3409462p3409462.html Sent from the R help mailing list archive at Nabble.com.
( sorry if this is a duplicate, I am not sure if hotmail is dropping some of my posts. Thanks )>You obviously want to delegate inner loops to R packages that execute as native, hopefully optimized, code. Generally a google search that starts with "R CRAN" will help. In this case it looks like a few packages available,>http://www.google.com/search?sclient=psy&hl=en&q=R+cran+median+filter> > > > > > > > > > > > > > ---------------------------------------- >> Date: Sun, 27 Mar 2011 07:56:11 -0700 >> From: chuan_zl at hotmail.com >> To: r-help at r-project.org >> Subject: [R] Asking Favor For the Script of Median Filter >> >> 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} >> >>> 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? >> >> Best Wishes >> Chuan
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
On Mar 27, 2011, at 1:07 PM, Mike Marchywka wrote:> You obviously want to delegate inner loops to R packages that > execute as native, hopefully optimized, code. > Generally a google search that starts with "R CRAN" will help. > In this case it looks like a few packages available, >> > http://www.google.com/search?sclient=psy&hl=en&q=R+cran+median+filterDid you find any that include a 2D median filter? All the ones I looked at were for univariate data. -- David.>> >> ---------------------------------------- >>> Date: Sun, 27 Mar 2011 07:56:11 -0700 >>> From: chuan_zl at hotmail.com >>> To: r-help at r-project.org >>> Subject: [R] Asking Favor For the Script of Median Filter >>> >>> 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} >>> >>>> 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? >>> >>> Best Wishes >>> Chuan-- David Winsemius, MD West Hartford, CT
Oops. My error! You wanted a 2D filter. I suspect a search will find some implementation, but you may wish to consider ?loess instead (there are numerous others, no doubt). -- Bert On Sun, Mar 27, 2011 at 3:30 PM, David Winsemius <dwinsemius at comcast.net> wrote:> > On Mar 27, 2011, at 1:07 PM, Mike Marchywka wrote: > >> You obviously want to delegate inner loops to R packages that >> execute as native, hopefully optimized, code. >> Generally a google search that starts with "R CRAN" will help. >> In this case it looks like a few packages available, >>> >> http://www.google.com/search?sclient=psy&hl=en&q=R+cran+median+filter > > Did you find any that include a 2D median filter? All the ones I looked at > were for univariate data. > > -- > David. > > >>> >>> ---------------------------------------- >>>> >>>> Date: Sun, 27 Mar 2011 07:56:11 -0700 >>>> From: chuan_zl at hotmail.com >>>> To: r-help at r-project.org >>>> Subject: [R] Asking Favor For the Script of Median Filter >>>> >>>> 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} >>>> >>>>> 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? >>>> >>>> Best Wishes >>>> Chuan > > > -- > David Winsemius, MD > West Hartford, CT > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > 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
----------------------------------------> CC: chuan_zl at hotmail.com; r-help at r-project.org > From: dwinsemius at comcast.net > To: marchywka at hotmail.com > Subject: Re: [R] Asking Favor For the Script of Median Filter > Date: Sun, 27 Mar 2011 18:30:48 -0400 > > > On Mar 27, 2011, at 1:07 PM, Mike Marchywka wrote: > > > You obviously want to delegate inner loops to R packages that > > execute as native, hopefully optimized, code. > > Generally a google search that starts with "R CRAN" will help. > > In this case it looks like a few packages available, > >> > > http://www.google.com/search?sclient=psy&hl=en&q=R+cran+median+filter > > Did you find any that include a 2D median filter? All the ones I > looked at were for univariate data.I put almost zero thought or effort into that but an interested party could modify the words a bit and and, for example "image" and one of the first interesting hits is this, http://cran.r-project.org/web/packages/biOps/biOps.pdf x <- readJpeg(system.file("samples", "violet.jpg", package="biOps")) y <- imgBlockMedianFilter(x, 5)> > -- > David.
Here is one I wrote for the raster package. It searches a raster layer for
NA's and takes the median of the number of non NA adjacent cells determined
by neighbor count. You could turn your matrix into a raster to make it work
or change the code.
Hope you find it useful, Robert
neighbor.filter <- function(raster.layer,neighbor.count = 3) {
require(raster)
base.rast <- raster.layer
count <- 1
NA.ind <- which(is.na(base.rast[]))
median.vals <- matrix(NA,length(NA.ind),3)
for (j in 1:length(NA.ind)) {
row.ind.NA <- rowFromCell(base.rast, NA.ind[j])
col.ind.NA <- colFromCell(base.rast, NA.ind[j])
row.ind <- c(row.ind.NA-1,row.ind.NA,row.ind.NA+1)
col.ind <- c(col.ind.NA-1,col.ind.NA,col.ind.NA+1)
row.ind.check <- expand.grid(row.ind,col.ind)[,1]
col.ind.check <- expand.grid(row.ind,col.ind)[,2]
ind.del.1 <- c(which(row.ind.check >
dim(base.rast)[1]),which(row.ind.check
< 1))
if (length(ind.del.1) > 0) {
row.ind.check <- row.ind.check[-ind.del.1]
col.ind.check <- col.ind.check[-ind.del.1] }
ind.del.2 <- c(which(col.ind.check < 1),which(col.ind.check >
dim(base.rast)[2]))
if (length(ind.del.2) > 0) {
row.ind.check <- row.ind.check[-ind.del.2]
col.ind.check <- col.ind.check[-ind.del.2] }
if (length(which(base.rast[cellFromRowCol(base.rast, row.ind.check,
col.ind.check)] > 0)) >= neighbor.count) {
median.vals[count,c(1:3)] <- c(NA.ind[j],
median(base.rast[cellFromRowCol(base.rast,
row.ind.check, col.ind.check)], na.rm = T),
length(which(base.rast[cellFromRowCol(base.rast,
row.ind.check, col.ind.check)] > 0)))
count <- count + 1
}
}
median.vals <- median.vals[which(median.vals[,1] > 0),]
base.rast[median.vals[,1]] <- median.vals[,2]
return(base.rast) }
Robert Leaf, PhD
NOAA Narragansett Laboratory
--
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